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        <title>Question 1 Exercise 5.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-2-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Exercise 5.2

Solutions of Question 1 of Exercise 5.2 of Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)
$n$$1.2+2.2^2+3.2^3+4.2^4+\ldots$\begin{align}
&amp; S_n=1.2+2.2^2+3 \cdot 2^3+4 \cdot 2^4+\ldots +n \cdot 2^n....(i) \\
&amp; 2 S_n=1.2^2+2.2^3+3.2^4+4.2^5+\ldots +n \cdot 2^n....(ii)\end{align}\begin{align} (1-2) S_n&amp;=1 \cdot 2+(2-1) 2^2+(3-2) 2^2+(4-…</description>
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        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Exercise 1.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-3-p1?rev=1737476037&amp;do=diff</link>
        <description>Question 1, Exercise 1.3

Solutions of Question 1 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)
\begin{align}&amp;z-4w=3i\\ 
&amp;2z+3w=11-5i\end{align}\begin{align}z-4w&amp;=3i		…(i)\\
2z+3w&amp;=11-5i	…(ii)\end{align}$2$\begin{align}2z-8w&amp;=6i		…(iii)\end{align}\[\begin{array}{cccc}
2z&amp;-8w&amp;=6i  \\  
\mathop+\limits_{-}2z&amp;\mathop+\limits_{-}3w&amp;=\mathop-\limit…</description>
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        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 4 Exercise 4.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 4 Exercise 4.5

Solutions of Question 4 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 4(i)
$0 . \overline{8}$$$0 . \overline{8}=0.888888 \ldots$$\begin{align}0 . \overline{8}&amp;=0.8+0.08+0.008 \div 0.0008+ \ldots\\
\text { or } 0 . \overline{8}&amp;=0.8+(0.1)(0.8) +(0.1)^2(0.8)+\ldots \ldots \ldots \ldots .(\mathrm{i})\end{align}$$a_1=0.8, \…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 4 &amp; 5 Exercise 5.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-2-p3?rev=1737476038&amp;do=diff</link>
        <description>Question 4 &amp; 5 Exercise 5.2

Solutions of Question 4 &amp; 5 of Exercise 5.2 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 4
$5+\dfrac{7}{3}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots$\begin{align}
&amp; S_{\infty}=5+\dfrac{7}{3}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(i) \\
&amp; \dfrac{1}{3} S_{\infty}=\dfrac{5}{3}+\dfrac{7}{3^2}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(ii)
\e…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question, Exercise 10.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p4?rev=1737476039&amp;do=diff</link>
        <description>Question, Exercise 10.1

Solutions of Question 4 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\sin \alpha =-\dfrac{4}{5}$$\cos \beta =-\dfrac{12}{13}$$\alpha $$\beta $$\sin \left( \alpha -\beta  \right)$$\sin \alpha=-\dfrac{4}{5}$$\alpha$$\sin \beta=-\dfrac{12}{13}$$\beta$$$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$$\…</description>
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        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Exercise 3.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-2-p1?rev=1737476037&amp;do=diff</link>
        <description>Question 1, Exercise 3.2

Solutions of Question 1 of Exercise 3.2 of Unit 03: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question.1(i)
$\vec{a}=3\hat{i}-5\hat{j}$$\vec{b}=-2\hat{i}+3\hat{j}$$\vec{a}+2\vec{b}$\begin{align}\vec{a}+2\vec{b}&amp;=3\hat{i}-5\hat{j}+2(-2\hat{i}+3\hat{j})\\
&amp;=3\hat{i}-5\hat{j}-4\hat{i}+6\hat{j}\\
&amp;=-\hat{i}+\hat{j}\end{align}$\vec{a}=3\hat{i}-5\hat{…</description>
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        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5(i) &amp; 5(ii) Exercise 3.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-5-p3?rev=1737476038&amp;do=diff</link>
        <description>Question 5(i) &amp; 5(ii) Exercise 3.5

Solutions of Question 5(i) &amp; 5(ii) of Exercise 3.5 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 5(i)
$\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\quad$$\quad\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\quad$$\vec{a} \times \vec{b}\quad$$\vec{a} \times \vec{b}$$\vec{a}$$\vec{b}$$\vec{a} \times \vec{b}$$\vec{a}$$\vec{b}$$\vec{a} \times \v…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 &amp; 3 Exercise 4.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-4-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 2 &amp; 3 Exercise 4.4

Solutions of Question 2 &amp; 3 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 2
$27$$243$$$a_3=27 \quad\text{and}\quad a_5=243$$\begin{align}a_3&amp;=a_1 r^2=27\\
a_5&amp;=a_1 r^4=243.\end{align}\begin{align}\dfrac{a_1 r^4}{a_1 r^2}&amp;=\dfrac{243}{27}=9 \\
\Rightarrow r^2&amp;=9 \text { or } r= \pm 3 .\end{align}$$a_1(9)=27 \quad \te…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6 &amp; 7 Exercise 4.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-4-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 6 &amp; 7 Exercise 4.4

Solutions of Question 6 &amp; 7 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 6
$a_{10}=l, a_{13}=m$$a_{16}=n;\quad$$\ln =m^2$$a_n=a_1 r^{n-1}$\begin{align}a_{10}&amp;=a_1 r^9=l \\
a_{13}&amp;=a_1 r^{12}=m\\
\text{and} \quad a_{16}&amp;=a_1 e^{\mathbf{A 5}}=n\end{align}\begin{align}a_{10} \cdot a_{16}&amp;=\ln =(a_1 r^9)(a_1 r^{15})\\
…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 &amp; 6 Exercise 4.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p5?rev=1737476038&amp;do=diff</link>
        <description>Question 5 &amp; 6 Exercise 4.5

Solutions of Question 5 &amp; 6 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 5
$r$$S_{10}=244 S_5$$$S_n=\dfrac{a_1(r^n-1)}{r-1}$$$$S_{10}=\dfrac{a_1(r^{10}-1)}{r-1} \quad \text{and}\quad S_5=\dfrac{a_1(r^5-1)}{r-1}$$$S_{10}$$S_S$\begin{align}\dfrac{a_1(r^{10}-1)}{r-1}&amp;=244 \dfrac{a_1(r^5-1)}{r-1} \\
\Rightarrow r^{10}-…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 Exercise 6.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-3-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 2 Exercise 6.3

Solutions of Question 2 of Exercise 6.3 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$n$$r$${ }^n P_r=840$${ }^n C_r=35$\begin{align}
&amp;^n P_r=\dfrac{n !}{(n-r) !}=840 ....(i)\\
&amp;^n C_r=\dfrac{n !}{(n-r) ! r !}=35....(ii)\end{align}\begin{align}\dfrac{n !}{(n-r) !} \cdot \dfrac{(n-r) ! r !}{n !}&amp;=\dfrac{840}{35}\\
r!&amp;=24\\
\te…</description>
    </item>
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        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Exercise 1.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p1?rev=1737476037&amp;do=diff</link>
        <description>Question 1, Exercise 1.2

Solutions of Question 1 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1

If ${{z}_{1}}=2+i$${{z}_{2}}=1-i$${{z}_{1}}=2+i$${{z}_{2}}=1-i$$$z_1+z_2=z_2+z_1.$$\begin{align}z_1+z_2&amp;=(2+i)+(1-i)\\ 
&amp;=3 \ldots (i) \end{align}\begin{align} 
z_2+z_1&amp;=(1-i)+(2+i)\\
&amp;=3 \ldots (ii)\end{align}$$z_1 z_2=z_2 z_1.$$\begin{align}z_1 z_2 …</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3, Exercise 2.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p3?rev=1737476037&amp;do=diff</link>
        <description>Question 3, Exercise 2.1

Solutions of Question 3 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 3(i)
$A=\begin{bmatrix}x &amp; y &amp; z\end{bmatrix}$$B=\begin{bmatrix}a &amp; h &amp; g\\h &amp; b &amp; f\\g &amp; f &amp; c\end{bmatrix}$$C=\begin{bmatrix}x\\y\\z\end{bmatrix}$$\left( AB \right)C=A\left( BC \right)$$A=\begin{bmatrix}x &amp; y &amp; z\end{bmatrix}$$B=\begin{bmatri…</description>
    </item>
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        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3 &amp; 4, Exercise 3.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-2-p3?rev=1737476037&amp;do=diff</link>
        <description>Question 3 &amp; 4, Exercise 3.2

Solutions of Question 3 &amp; 4 of Exercise 3.2 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 3

If $\vec{r}=\hat{i}-9\hat{j}$$\vec{a}=\hat{i}+2\hat{j}$$\vec{b}=5\hat{i}-\hat{j}$$p$$q$$\vec{r}=p\vec{a}+q\vec{b}$$$\vec{r}=p\vec{a}+q\vec{b}.$$$\vec{r},\vec{a}$$\vec{b}$$$\hat{i}-9\hat{j}=p(\hat{i}+2\hat{j})+q(5\hat{i}-\hat{j})$$$$\implies \hat{i}-9\…</description>
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        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5(iii) &amp; 5(iv) Exercise 3.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-5-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 5(iii) &amp; 5(iv) Exercise 3.5

Solutions of Question 5(iii) &amp; 5(iv) of Exercise 3.5 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\quad$$\quad\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\quad$$(\vec{a}. \vec{b})^2,\quad|a|^2,\quad|b|^2$\begin{align}\vec{a} \cdot \vec{b}&amp;=(a_1 \hat{i}+a_2 \hat{j} + a_3 \hat{k}) \cdot(b_1 \hat{i}+b_2 \…</description>
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        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 and 2 Exercise 4.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-1-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 and 2 Exercise 4.1

Solutions of Question 1 and 2 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$2,4,6,8, \ldots ,50$$50 $$1,0,1,0,1, \ldots$$0$$1$$...,-4,0,4,8, \ldots, 60$$1,-\dfrac{1}{3}, \dfrac{1}{9},-\dfrac{1}{27}, \ldots,-\dfrac{1}{2187}$$a_n=\dfrac{n(n+1)}{2}$$$a_n=\dfrac{n(n+1)}{2}$$$n=1,$$$a_1=\dfrac{1(1+1)}{2}=1$$$n=2$$$a_2=\dfrac{2(2…</description>
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        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3 and 4 Exercise 4.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-1-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 3 and 4 Exercise 4.1

Solutions of Question 3 and 4 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\dfrac{1}{2}, \dfrac{2}{3} \dfrac{3}{4}, \dfrac{4}{5}, \ldots$$$\dfrac{1}{1+1}, \dfrac{2}{2+1}, \dfrac{3}{3+1}, \dfrac{4}{4+1},...$$$\dfrac{n}{n+1}$$2,-4,6,-8,10, \ldots$\begin{align}
&amp;(-1)^2 \cdot 2 \cdot 1, (-1)^3 \cdot 2 \cdot 2, (-1)^4 \cdot 2 \…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-2-p12?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 16 Exercise 4.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-2-p12?rev=1737476038&amp;do=diff</link>
        <description>Question 16 Exercise 4.2

Solutions of Question 16 of Exercise 4.2 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 16
$5$$8$$5$$8$$A_1, A_2, A_3, A_4, A_5$$5$$8$$5, A_1, A_2, A_3, A_4, A_5, 8$$$a_1=5 \text{ and } a_7=8.$$\begin{align}&amp;a_7=a+6d\\
\implies &amp;8=5+6d\\
\implies &amp;6d=8-5\\
\implies &amp;d=\dfrac{3}{6}=\dfrac{1}{2}.
\end{align}\begin{align}
A_1&amp;=a+d=5+\dfra…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 &amp; 6 Exercise 4.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-3-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 5 &amp; 6 Exercise 4.3

Solutions of Question 5 &amp; 6 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 5
$20$$120$$$a-2 d, a-d, a+d, a+2 d,$$$Condition-1$$20$\begin{align}a-3 d+a-d+a+d+a+3 d&amp;=20 \\
\Rightarrow 4 a&amp;=20\\
\Rightarrow a&amp;=5 .\end{align}$Condition-2$$120$\begin{align}(a-3 d)^2+(a-d)^2+(a+d)^2+(a+2 d)^2&amp;=120 \\
\Rightarrow a^2-6 a d+…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3 Exercise 4.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p3?rev=1737476038&amp;do=diff</link>
        <description>Question 3 Exercise 4.5

Solutions of Question 3 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 3
$a_2=2$$a_3=1$$a_1$$r$$$a_n=a_1 r^{n-1}$$$$a_2=a_1 r=2....(i)$$$$a_3=a_1 r^2=1...(ii)$$\begin{align}\dfrac{a_1 r^2}{a_1 r}&amp;=\dfrac{1}{2}\\
\Rightarrow r&amp;=\dfrac{1}{2} \text {, }\end{align}\begin{align}\dfrac{a_1}{2}&amp;=2\\
\Rightarrow a_1&amp;=4 \text {. …</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 11 &amp; 12 Exercise 4.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p8?rev=1737476038&amp;do=diff</link>
        <description>Question 11 &amp; 12 Exercise 4.5

Solutions of Question 11 &amp; 12 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$p^{t h}, q^{t h}$$r^{t h}$$a, b, c$$a^{q-r} b^{r-p} c^{p-q}=1$$a_n=a_1 r^{n-1}$$a_p=a_1 r^{p-1}=a \quad a_q=a_1 r^{q-1}=b$$a_r=a_1 r^{r-1}$\begin{align}a^{q-r}&amp;=(a_1 r^{p-1})^{q-r} . \\
b^{r-p}&amp;=(a_1 r^{q-1})^{r-p}, \text { and } \\
c^{p-q}&amp;=(a_1 r^…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 &amp; 6 Review Exercise</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 5 &amp; 6 Review Exercise

Solutions of Question 5 &amp; 6 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$5+12 x+19 x^2+26 x^3+\ldots$$n$\begin{align}S_n&amp;=5+12 x+19 x^2+26 x^3+\cdots+(7 n-2) x^{n-1}...(i)\\ 
x S_n&amp;=5 x+12 x^2+19 x^3+\cdots+(7 n-9) x^{n-1}+(7 n-1) x^n....(ii)\end{align}\begin{align}(1-x) S_n&amp;=5+(12-5) x+(19-12) x^2+\cdots\\
&amp;+[7 n-2-(…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-1-p1?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 and 2 Exercise 6.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-1-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 and 2 Exercise 6.1

Solutions of Question 1 and 2 of Exercise 6.1 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\dfrac{10 !}{3 ! .3 ! \cdot 4 !}$\begin{align}\dfrac{10 !}{3 ! \cdot 3 ! \cdot 4 !}&amp;=\dfrac{10.9 .8 \cdot 7 \cdot 6 \cdot 5.4 !}{3 ! \cdot 3 ! \cdot 4 !}\\
&amp;=\dfrac{10.9 .8 .7 .5}{3.2 .1}\\
&amp;=4200 \end{align}$\dfrac{3 !+4 !}{5 !-…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-1-p2?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3 &amp; 4 Exercise 6.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-1-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 3 &amp; 4 Exercise 6.1

Solutions of Question 3 &amp; 4 of Exercise 6.1 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\dfrac{1}{6 !}+\dfrac{2}{7 !}+\dfrac{3}{8 !}=\dfrac{75}{8 !}$\begin{align}\dfrac{1}{6 !}+\dfrac{2}{7 !}+\dfrac{3}{8 !}&amp;=\dfrac{1}{6 !}+\dfrac{2}{7.6 !}+\dfrac{3}{8.7 .6 !} \\
&amp; =\dfrac{56+16+3}{8 !}\\
&amp;=\dfrac{75}{8 !}\end{align}$\df…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p1?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 and 2 Exercise 6.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 and 2 Exercise 6.2

Solutions of Question 1 and 2 of Exercise 6.2 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$^6 P_6$\begin{align}^6 P_6&amp;=\dfrac{6 !}{(6-6) !}\\
&amp;=6 !=720\end{align}$^{20} P_2$\begin{align}^{20} P_2&amp;=\dfrac{20 !}{(20-2) !}\\
&amp;=\dfrac{20.19 .18 !}{18 !}\\
&amp;=20 \times 19=380\end{align}$^{16} P_3$\begin{align}^{16} P_3&amp;=\dfr…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Review Exercise 7</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/re-ex7-p1?rev=1737476039&amp;do=diff</link>
        <description>Question 1 Review Exercise 7

Solutions of Question 1 of Review Exercise 7 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Chose the correct option.
&lt;panel&gt;$O, P, Q, R, S, T, U$$2520$$9040$$5140$$4880$$\{1,2,3,4,5,6,7\}$$14$$42$$28$$21$$\{1,2,3,4,6,7,8\}$$3$$7$$120$$180$$144$$96$$\dfrac{(n+2) !(n-2) !}{(n+1) !(n-1) !}$$(n-3)$$(\dot{n}-1)$$\dfrac{n+1}{n…</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Unit 03: Vectors (Solutions)</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03?rev=1737476037&amp;do=diff</link>
        <description>Unit 03: Vectors (Solutions)

This is a third unit of the book Mathematics 11 published by Khyber Pakhtunkhwa Textbook Board, Peshawar, Pakistan. On this page we have provided the solutions of the questions.

After reading this unit the students will be able to$i$$j$$i$$j$$k$$O$$-A$$A$$i.i=j.j=k.k=1$$i.j=j.k=k.i=0$$i\times i =j\times j =k\times k=0$$i\times j = k$$j\times k =k\times j = i$$A \times B$$A$$B$$i.j\times k =j.k\times i=k.i\times j=1$$i.k\times j = J.i\times k=k.j\times i=-1$</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Exercise 1.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p1?rev=1737476037&amp;do=diff</link>
        <description>Question 1, Exercise 1.1

Solutions of Question 1 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)
${{i}^{9}}+{{i}^{19}}$\begin{align}{{i}^{9}}+{{i}^{19}}&amp;=i\cdot{{i}^{8}}+i\cdot{{i}^{18}}\\
&amp;=i\cdot{{\left( {{i}^{2}} \right)}^{4}}+i\cdot{{\left( {{i}^{2}} \right)}^{9}}\\
&amp;=i\cdot{{\left( -1 \right)}^{4}}+i\cdot{{\left( -1 \right)}^{9}}\\
&amp;=i\cdo…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p2?rev=1737476037&amp;do=diff">
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 &amp; 3, Exercise 1.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p2?rev=1737476037&amp;do=diff</link>
        <description>Question 2 &amp; 3, Exercise 1.1

Solutions of Question 2 &amp; 3 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 2
${{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$\begin{align}L.H.S.&amp;={{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}\\
&amp;=i\cdot i^{106}+i^{112}+i^{122}+i\cdot i^{152}\\
&amp;=i.{{\left( {{i}^{2}} \right)}^{53}}+{{\left( {{i}^{2}} \right)}^{56}}+…</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 4, Exercise 1.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p3?rev=1737476037&amp;do=diff</link>
        <description>Question 4, Exercise 1.1

Solutions of Question 4 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 4(i)
$\left( a,0 \right)\left( 2,-b \right)$\begin{align}&amp;\left( a,0 \right)-\left( 2,-b \right)\\
&amp;=\left( a+0i \right)-\left( 2-bi \right)\\
&amp;=\left( a-2 \right)+\left( 0+b \right)i\\
&amp;=\left( a-2 \right)+bi\end{align}$\left( -3,\dfrac{1}{2} \right)\le…</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5, Exercise 1.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p4?rev=1737476037&amp;do=diff</link>
        <description>Question 5, Exercise 1.1

Solutions of Question 5 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 5(i)
$8i+11,-7+5i$\begin{align}&amp;(8i+11)\times (-7+5i)\\
&amp;=\left( 11+8i \right)\times \left( -7+5i \right)\\
&amp;=\left( -77+40{{i}^{2}} \right)+\left( 55-56 \right)i\\
&amp;=\left( -77+40\left( -1 \right) \right)+\left( 55-56 \right)i\\
&amp;=\left( -77-40 \right)+…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p5?rev=1737476037&amp;do=diff">
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6, Exercise 1.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p5?rev=1737476037&amp;do=diff</link>
        <description>Question 6, Exercise 1.1

Solutions of Question 6 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 6(i)
$\dfrac{4+i}{3+5i}$$a+ib$\begin{align}\dfrac{4+i}{3+5i}&amp;=\dfrac{4+i}{3+5i}\times \dfrac{3-5i}{3-5i}\\
&amp;=\dfrac{\left( 12+5 \right)+\left( 3-20 \right)i}{9-25{{i}^{2}}}\\
&amp;=\dfrac{17-17i}{9+25}\\
&amp;=\dfrac{17}{34}-\dfrac{17}{34}i\\
&amp;=\dfrac{1}{2}-\dfr…</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7, Exercise 1.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p6?rev=1737476037&amp;do=diff</link>
        <description>Question 7, Exercise 1.1

Solutions of Question 7 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 7(i)
${{z}_{1}}=1+2i$${{z}_{2}}=2+3i$$|{{z}_{1}}+{{z}_{2}}|$$z_1=1+2i$$z_2=2+3i$\begin{align}
{{z}_{1}}+{{z}_{2}}&amp;=1+2i+2+3i\\
&amp;=1+2+2i+3i\\
&amp;=3+5i
\end{align}\begin{align}
|z_1+z_2|&amp;=\sqrt{3^2+5^2}\\
&amp;=\sqrt{9+25}\\ 
&amp;=\sqrt{34}\end{align}${{z}_{1}}=1+2…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p7?rev=1737476037&amp;do=diff">
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 8, Exercise 1.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p7?rev=1737476037&amp;do=diff</link>
        <description>Question 8, Exercise 1.1

Solutions of Question 8 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 8(i)
$\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}$$a+ib.$\begin{align}&amp;\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}\\
&amp;=\dfrac{\left( 3+2i \right)\left( 1-2i \right)+\left( 2+i \right)\left( 4-i \right)}{\left( 2+i \right)\left( 3+2i \right)}\\
&amp;=\dfrac{\left( 3+4+2i-6i …</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p9?rev=1737476037&amp;do=diff">
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 11, Exercise 1.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p9?rev=1737476037&amp;do=diff</link>
        <description>Question 11, Exercise 1.1

Solutions of Question 11 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 11(i)
${{z}_{1}}=2-i$${{z}_{2}}=-2+i$${\rm Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)$$z_1=2-i$$z_2=-2+i$$\overline{z_1}=2+i$\begin{align}
z_1 z_2&amp;=(2-i)(-2+i)\\ 
&amp;=-4+1+2i+2i\\
&amp;=-3+4i
\end{align}\begin{align}
\dfrac{z_1 z_2}{\…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p3?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3 &amp; 4, Exercise 1.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p3?rev=1737476037&amp;do=diff</link>
        <description>Question 3 &amp; 4, Exercise 1.2

Solutions of Question 3 &amp; 4 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 3
${{z}_{1}}=\sqrt{3}+\sqrt{2}i$${{z}_{2}}=\sqrt{2}-\sqrt{3}i$${{z}_{3}}=2+3i$${{z}_{1}}=\sqrt{3}+\sqrt{2}i$${{z}_{2}}=\sqrt{2}-\sqrt{3}i$${{z}_{3}}=2+3i$\begin{align}{{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)&amp;={{z}_{1}}{{z}_{2}}+{{z}_{1}}{{z}_{…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p4?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5, Exercise 1.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p4?rev=1737476037&amp;do=diff</link>
        <description>Question 5, Exercise 1.2

Solutions of Question 5 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 5(i)
${{z}_{1}}=2+4i$${{z}_{2}}=1-3i$$\overline{{{z}_{1}}+{{z}_{2}}}=\overline{{{z}_{1}}}+\overline{{{z}_{2}}}$${{z}_{1}}=2+4i$${{z}_{2}}=1-3i$$\overline{{{z}_{1}}}=2-4i$$\overline{{{z}_{2}}}=1+3i$\begin{align}z_1+z_2&amp;=2+4i+1-3i\\
&amp;=3+i \end{align}\begin…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p5?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6, Exercise 1.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p5?rev=1737476037&amp;do=diff</link>
        <description>Question 6, Exercise 1.2

Solutions of Question 6 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 6(i)
${{z}_{1}}$${{z}_{2}}$$|{{z}_{1}}{{z}_{2}}|=|{{z}_{1}}||{{z}_{2}}|$${{z}_{1}}=a+bi$${{z}_{2}}=c+di$$|z_1=\sqrt{a^2+b^2}|$$|z_2=\sqrt{c^2+d^2}|$\begin{align}
L.H.S.&amp;=|{{z}_{1}}{{z}_{2}}|\\
&amp;=|(a+bi)(c+di)|\\ 
&amp;=|ac-bd+(ad+bc)i|\\
&amp;=\sqrt{{{(ac-bd)}^{…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p6?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7, Exercise 1.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p6?rev=1737476037&amp;do=diff</link>
        <description>Question 7, Exercise 1.2

Solutions of Question 7 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 7(i)
$\dfrac{2+3i}{5-2i}$\begin{align}&amp;\dfrac{2+3i}{5-2i} \\
=&amp;\dfrac{2+3i}{5-2i}\times \dfrac{5+2i}{5+2i} \quad \text{by rationalizing} \\
=&amp;\dfrac{10-6+15i+4i}{25+4}\\
=&amp;\dfrac{4+19i}{29}\\
=&amp;\dfrac{4}{29}+\dfrac{19}{29}i \end{align}$=\dfrac{4}{29}$$=\…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p7?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 8, Exercise 1.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p7?rev=1737476037&amp;do=diff</link>
        <description>Question 8, Exercise 1.2

Solutions of Question 8 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 8(i)
$z+\overline{z}=2\operatorname{Re}\left( z \right)$$z=a+ib$$\overline{z}=a-ib$\begin{align}z+\overline{z}&amp;=\left( a+ib \right)+\left( a-ib \right)\\
&amp;=a+ib+a-ib\\
&amp;=2a\\
z+\overline{z}&amp;=2\operatorname{Re}\left( z \right)\end{align}$z-\overline{z}=2i…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p8?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 9, Exercise 1.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p8?rev=1737476037&amp;do=diff</link>
        <description>Question 9, Exercise 1.2

Solutions of Question 9 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 9(i)
$z=3+2i,$$-|z|\leq \operatorname{Re}\left( z \right)\leq |z|$$z=3+2i$$|z|=\sqrt{9+4}=\sqrt{13}$${\rm Re}z=3=\sqrt{9}$\begin{align} &amp;-\sqrt{13} \leq \sqrt{9} \leq \sqrt{13}\\
\implies &amp;-|z|\leq \operatorname{Re}\left( z \right)\leq |z|\end{align}$z=3…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-3-p2?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2, Exercise 1.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-3-p2?rev=1737476037&amp;do=diff</link>
        <description>Question 2, Exercise 1.3

Solutions of Question 2 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 2(i)
$P(z)$$$P\left( z \right)={{z}^{3}}+6z+20$$$$p\left( z \right)={{z}^{3}}+6z+20$$$(z-a)$$P(z)$$P(a)=0$$z=-2$\begin{align}
P(-2)&amp;=(-2)^3+6(-2)+20\\
&amp;=-8-12+20\\
&amp;=0\end{align}$z+2$${{z}^{3}}+6z+20$$$\begin{array}{c|cccc}
-2 &amp; 1 &amp; 0 &amp; 6 &amp; 20 \\  
 &amp; \d…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-3-p4?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5, Exercise 1.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-3-p4?rev=1737476037&amp;do=diff</link>
        <description>Question 5, Exercise 1.3

Solutions of Question 5 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 5(i)
${{z}^{2}}+z+3=0$$${{z}^{2}}+z+3=0.$$$a=1$$b=1$$c=3$\begin{align}z&amp;=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ 
&amp;=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)}\\
&amp;=\dfrac{-1\pm \sqrt{1-12}}{2}\\
&amp;=\…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-3-p5?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6, Exercise 1.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-3-p5?rev=1737476037&amp;do=diff</link>
        <description>Question 6, Exercise 1.3

Solutions of Question 6 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 6(i)
${{z}^{4}}+{{z}^{2}}+1=0$$$z^4+z^2+1=0$$$$z^4+2z^2+1-z^2=0$$$$( z^2+1 )^2-z^2=0$$$$( z^2+1+z)( z^2+1-z )=0$$$$( z^2+z+1 )( z^2-z+1 )=0$$$$(z^2+z+1 )=0$$$$z=\dfrac{-1\pm \sqrt{1-4}}{2}$$$$z=\dfrac{-1\pm \sqrt{3}i}{2}$$$$(z^2-z+1 )=0$$$$z=\dfrac{1\pm …</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/review-ex-1-p1?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Review Exercise 1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/review-ex-1-p1?rev=1737476037&amp;do=diff</link>
        <description>Question 1, Review Exercise 1

Solutions of Question 1 of Review Exercise 1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1
${{\left( \dfrac{2i}{1+i} \right)}^{2}}$$i$$2i$$1-i$$i+1$$\dfrac{5+2i}{4-3i}$$-\dfrac{7}{25}+\dfrac{26}{25}i$$\dfrac{5}{4}-\dfrac{2}{3}i$$\dfrac{14}{25}+\dfrac{23}{25}i$$\dfrac{26}{7}+\dfrac{23}{7}i$${{i}^{57}}+\frac{1}{{{i}^{25}}}$$0$$2i$$-2…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit01/review-ex-1-p2?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 &amp; 3, Review Exercise 1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit01/review-ex-1-p2?rev=1737476037&amp;do=diff</link>
        <description>Question 2 &amp; 3, Review Exercise 1

Solutions of Question 2 &amp; 3 of Review Exercise 1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.${{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}=0$$\forall n\in N$\begin{align}{{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}&amp;=0\\
L.H.S.&amp;={{i}^{n}}+{{i}^{n}}\cdot i+{{i}^{n}}\cdot {{i}^{2}}+{{i}^{n}}\cdot {{i}^{3}}\\
&amp;={{i}^{n}}\left( 1+i+{{i}^{2}}…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p1?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Exercise 2.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p1?rev=1737476037&amp;do=diff</link>
        <description>Question 1, Exercise 2.1

Solutions of Question 1 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)
$$\left[ \begin{matrix}
1 &amp; 2 &amp; 4  \\
\end{matrix} \right]
\left[ \begin{matrix}
1 &amp; 0 &amp; 2  \\
2 &amp; 0 &amp; 1  \\
0 &amp; 1 &amp; 2  \\
\end{matrix} \right]
\left[ \begin{matrix}
2  \\
4  \\
6  \\
\end{matrix} \right]$$\begin{align}&amp;\left[ \begin{matri…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p5?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 &amp; 6, Exercise 2.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p5?rev=1737476037&amp;do=diff</link>
        <description>Question 5 &amp; 6, Exercise 2.1

Solutions of Question 5 &amp; 6 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$A= \begin{bmatrix} 0 &amp; 2b &amp; -2  \\ 3 &amp; 1 &amp; 3  \\ 3a &amp; 3 &amp; -1 \end{bmatrix}$$a$$b$$A=\begin{bmatrix} 0 &amp; 2b &amp; -2  \\ 3 &amp; 1 &amp; 3  \\ 3a &amp; 3 &amp; -1 \end{bmatrix}$$$A^t=\left[ \begin{matrix}
   0 &amp; 3 &amp; 3a  \\
   2b &amp; 1 &amp; 3  \\
   -2 &amp; 3 &amp; -1  \\
\end{ma…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p7?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 8, Exercise 2.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p7?rev=1737476037&amp;do=diff</link>
        <description>Question 8, Exercise 2.1

Solutions of Question 8 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 8(i)
$A=\begin{bmatrix}1 &amp; 2 &amp; 0  \\3 &amp; -1 &amp; 4 \end{bmatrix}$$( A^t )^t=A$$$A=\left[ \begin{matrix}
   1 &amp; 2 &amp; 0  \\
   3 &amp; -1 &amp; 4  \\
\end{matrix}  \right]$$$$A^t=\left[  \begin{matrix}
   1 &amp; 3  \\
   2 &amp; -1  \\
   0 &amp; 4  \\
\end{matrix} \rig…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p8?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 9, Exercise 2.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p8?rev=1737476037&amp;do=diff</link>
        <description>Question 9, Exercise 2.1

Solutions of Question 9 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 9(i)
$A=\begin{bmatrix}2 &amp; -1 &amp; 3  \\1 &amp; \quad 0 &amp; 1 \end{bmatrix},$$B=\begin{bmatrix}1 &amp; 2  \\2 &amp; 2  \\ 3 &amp; 0 \end{bmatrix}$$( AB )^t=B^tA^t$$$A=\left[  \begin{matrix}
   2 &amp; -1 &amp; 3  \\
   1 &amp; \quad 0 &amp; 1  \\
\end{matrix}  \right],$$$$B=\left[…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p11?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 12, Exercise 2.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p11?rev=1737476037&amp;do=diff</link>
        <description>Question 12, Exercise 2.1

Solutions of Question 12 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 12(i)
$A=\begin{bmatrix}3 &amp; 2 &amp; 1  \\4 &amp; 5 &amp; 6  \\-2 &amp; 3 &amp; 4\end{bmatrix}$$A+A^t$$$A=\left[ \begin{matrix}
   3 &amp; 2 &amp; 1  \\
   4 &amp; 5 &amp; 6  \\
   -2 &amp; 3 &amp; 4  \\
\end{matrix} \right]$$$$A^t=\left[ \begin{matrix}
   3 &amp; 4 &amp; -2  \\
   2 &amp; 5 &amp; 3  \…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p12?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 13, Exercise 2.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-1-p12?rev=1737476037&amp;do=diff</link>
        <description>Question 13, Exercise 2.1

Solutions of Question 13 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 13(i)
$A$$3$$A+A^t$$$A=\left[ \begin{matrix}
   a_{11} &amp; a_{12} &amp; a_{13}  \\
   a_{21} &amp; a_{22} &amp; a_{23}  \\
   a_{31} &amp; a_{32} &amp; a_{33}  \\
\end{matrix} \right]$$$$A^t=\left[ \begin{matrix}
   a_{11} &amp; a_{21} &amp; a_{31}  \\
   a_{12} &amp; a_{22} …</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p2?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2, Exercise 2.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p2?rev=1737476037&amp;do=diff</link>
        <description>Question 2, Exercise 2.2

Solutions of Question 2 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 2(i)
$$\left| \begin{matrix}
   1 &amp; 2 &amp; 0  \\
   3 &amp; 1 &amp; 0  \\
   -1 &amp; 2 &amp; 0  \\
\end{matrix} \right|=0$$$\left| \begin{matrix}1 &amp; 2 &amp; 3  \\-8 &amp; 4 &amp; -12  \\2 &amp; -1 &amp; 3 \end{matrix} \right|=0$$$\left| \begin{matrix}
   1 &amp; 2 &amp; 3  \\
   -8 &amp; 4 &amp; -…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p4?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 4, Exercise 2.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p4?rev=1737476037&amp;do=diff</link>
        <description>Question 4, Exercise 2.2

Solutions of Question 4 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 4(i)
$\left| \begin{matrix}0 &amp; 1 &amp; 3  \\-1 &amp; 2 &amp; 1  \\2 &amp; 1 &amp; 1 \end{matrix} \right|.$\begin{align}&amp;\left| \begin{matrix}
   0 &amp; 1 &amp; 3  \\
   -1 &amp; 2 &amp; 1  \\
   2 &amp; 1 &amp; 1  \\
\end{matrix} \right| \\
=&amp;0\left( 2-1 \right)-1\left( -1-2 \right)+3\l…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p5?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5, Exercise 2.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p5?rev=1737476037&amp;do=diff</link>
        <description>Question 5, Exercise 2.2

Solutions of Question 5 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 5(i)
$\begin{vmatrix}a &amp; b &amp; c\\l &amp; m &amp; n\\x &amp; y &amp; z \end{vmatrix}=\begin{vmatrix}a &amp; l &amp; x\\b &amp; m &amp; y\\c &amp; n &amp; z \end{vmatrix}$\begin{align}L.H.S.&amp;=\begin{vmatrix}
a &amp; b &amp; c  \\
l &amp; m &amp; n  \\
x &amp; y &amp; z
\end{vmatrix}\\
&amp;=\begin{vmatrix}
a &amp; b &amp;…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p6?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6, Exercise 2.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p6?rev=1737476037&amp;do=diff</link>
        <description>Question 6, Exercise 2.2

Solutions of Question 6 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Questiopn 6(i)
$\left| \begin{matrix}a-b &amp; b-c &amp; c-a  \\b-c &amp; c-a &amp; a-b  \\c-a &amp; a-b &amp; b-c  \end{matrix} \right|=0$\begin{align} L.H.S&amp;=\left| \begin{matrix}
a-b &amp; b-c &amp; c-a  \\
b-c &amp; c-a &amp; a-b  \\
c-a &amp; a-b &amp; b-c  \\
\end{matrix} \right| \\ 
&amp;=\left| \…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p7?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7, Exercise 2.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p7?rev=1737476037&amp;do=diff</link>
        <description>Question 7, Exercise 2.2

Solutions of Question 7 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 7(i)
$\left| \begin{matrix}3860 &amp; 3861  \\3862 &amp; 3863 \end{matrix} \right|$$$\left| \begin{matrix}
   3860 &amp; 3861  \\
   3862 &amp; 3863  \\
\end{matrix} \right|=14911180-14911182$$$$=-2$$$\left| \begin{matrix}81 &amp; 82 &amp; 83  \\84 &amp; 85 &amp; 86  \\87 &amp; 8…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p9?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 11, Exercise 2.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p9?rev=1737476037&amp;do=diff</link>
        <description>Question 11, Exercise 2.2

Solutions of Question 11 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 11(i)
$\left[ \begin{matrix}7 &amp; 1 &amp; 3  \\6 &amp; 2 &amp; -2  \\5 &amp; 1 &amp; 1\end{matrix} \right]$$$A=\left[ \begin{matrix}
   7 &amp; 1 &amp; 3  \\
   6 &amp; 2 &amp; -2  \\
   5 &amp; 1 &amp; 1  \\
\end{matrix} \right]$$$$|A|=7(2+2)-1(6+10)+3(6-10)$$$$=28-16-12$$$$|A|=0$$$A$$\…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p11?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 13, Exercise 2.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p11?rev=1737476037&amp;do=diff</link>
        <description>Question 13, Exercise 2.2

Solutions of Question 13 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 13(i)
$x,$$\left| \begin{matrix}x &amp; 2 &amp; 3  \\0 &amp; -1 &amp; 1  \\0 &amp; 4 &amp; 5 \end{matrix} \right|=9$$$\left| \begin{matrix}
   x &amp; 2 &amp; 3  \\
   0 &amp; -1 &amp; 1  \\
   0 &amp; 4 &amp; 5  \\
\end{matrix} \right|=9$$$$x(-5-4)-2(0)+3(0)=9$$$$-9x=9$$$$x=-1$$$x,$$\left…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p14?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 18, Exercise 2.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-2-p14?rev=1737476037&amp;do=diff</link>
        <description>Question 18, Exercise 2.2

Solutions of Question 18 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 18(i)
$A$$B$$( A^{-1})^{-1}=A$$A$$2\times 2$$$A=\left[ \begin{matrix}
   a_{11} &amp; a_{12}  \\
   a_{21} &amp; a_{22}  \\
\end{matrix} \right]$$$$|A|=a_{11}a_{22}-a_{12}a_{21}$$$$AdjA=\left[ \begin{matrix}
   a_{22} &amp; -a_{12}  \\
   -a_{21} &amp; a_{11…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-3-p1?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Exercise 2.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-3-p1?rev=1737476037&amp;do=diff</link>
        <description>Question 1, Exercise 2.3

Solutions of Question 1 of Exercise 2.3 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)
$\begin{bmatrix}1 &amp; 3 &amp; -1  \\2 &amp; 1 &amp; 4  \\3 &amp; 4 &amp; -5\end{bmatrix}$\begin{align}&amp;\begin{bmatrix}
1 &amp; 3 &amp; -1  \\
2 &amp; 1 &amp; 4  \\
3 &amp; 4 &amp; -5 \end{bmatrix}\\
\underset{\sim}{R}&amp;\begin{bmatrix}
1 &amp; 3 &amp; -1  \\
0 &amp; -5 &amp; 6  \\
0 &amp; -5 &amp; -2 \end{bmat…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-3-p2?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2, Exercise 2.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-3-p2?rev=1737476037&amp;do=diff</link>
        <description>Question 2, Exercise 2.3

Solutions of Question 2 of Exercise 2.3 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 2(i)
$$\begin{bmatrix}4 &amp; -2 &amp; 5 \\ 2 &amp; 1 &amp; 0  \\ -1 &amp; 2 &amp; 3  \end{bmatrix}$$$$A=\begin{bmatrix}
4 &amp; -2 &amp; 5  \\
2 &amp; 1 &amp; 0  \\
-1 &amp; 2 &amp; 3 \end{bmatrix}.$$\begin{align}|A|&amp;=\begin{vmatrix}4 &amp; -2 &amp; 5  \\ 2 &amp; 1 &amp; 0  \\ -1 &amp; 2 &amp; 3 \end{vmatrix}\\
&amp;=…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-3-p3?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3, Exercise 2.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit02/ex2-3-p3?rev=1737476037&amp;do=diff</link>
        <description>Question 3, Exercise 2.3

Solutions of Question 3 of Exercise 2.3 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 3(i)
$$\left[ \begin{matrix}
   1 &amp; 0 &amp; -2  \\
   2 &amp; 2 &amp; 1  \\
   -1 &amp; 2 &amp; 3  \\
\end{matrix} \right]$$\begin{align}&amp;\begin{bmatrix}
1 &amp; 0 &amp; -2  \\
2 &amp; 2 &amp; 1  \\
-1 &amp; 2 &amp; 3 \end{bmatrix}\\
\underset{\sim}{R}&amp; \begin{bmatrix}
1 &amp; 0 &amp; -2  \\
0 &amp;…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-2-p2?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2, Exercise 3.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-2-p2?rev=1737476037&amp;do=diff</link>
        <description>Question 2, Exercise 3.2

Solutions of Question 2 of Exercise 3.2 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 2(i)

Find unit vector having the same direction as the vector $3\hat{i}.$$$\overset{\scriptscriptstyle\rightharpoonup}{a}=3\hat{i}$$$$|\overset{\scriptscriptstyle\rightharpoonup}{a}|=\sqrt{{{(3)}^{2}}}=3$$$$\hat{a}=\dfrac{{\overset{\scriptscriptstyle\rightharpo…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-2-p5?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7, Exercise 3.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-2-p5?rev=1737476037&amp;do=diff</link>
        <description>Question 7, Exercise 3.2

Solutions of Question 7 of Exercise 3.2 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 7(i)

Find the components and the magnitude of $\overrightarrow{PQ}$$P(-1,2)$$Q(2,-1)$\begin{align}\overrightarrow{PQ}&amp;=\overrightarrow{OQ}-\overrightarrow{OP}\\ 
&amp;=(2\hat{i}-\hat{j})-(-\hat{i}+2\hat{j})\\ 
&amp;=3\hat{i}-3\hat{j}\end{align}\begin{align}|\overrighta…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-2-p6?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7, Exercise 3.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-2-p6?rev=1737476037&amp;do=diff</link>
        <description>Question 7, Exercise 3.2

Solutions of Question 7 of Exercise 3.2 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 7(i)

Find the components and the magnitude of $\overrightarrow{PQ}$$P(-1,2)$$Q(2,-1)$\begin{align}\overrightarrow{PQ}&amp;=\overrightarrow{OQ}-\overrightarrow{OP}\\ 
&amp;=(2\hat{i}-\hat{j})-(-\hat{i}+2\hat{j})\\ 
&amp;=3\hat{i}-3\hat{j}\end{align}\begin{align}|\overrighta…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-2-p8?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 11, Exercise 3.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-2-p8?rev=1737476037&amp;do=diff</link>
        <description>Question 11, Exercise 3.2

Solutions of Question 11 of Exercise 3.2 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 11(i)

Find the position vectors of the point of division of the line segments joining point $C$$5\hat{j}$$D$$4\hat{i}+\hat{j}$$2:5$$C$$\overrightarrow{OC}=5\hat{j}$$D$$\overrightarrow{OD}=4\hat{i}+\hat{j}$$H$$\overline{CD}$$2:5$$H$\begin{align}\overrightarrow…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-3-p1?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Exercise 3.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-3-p1?rev=1737476037&amp;do=diff</link>
        <description>Question 1, Exercise 3.3

Solutions of Question 1 of Exercise 3.3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)

If $\vec{a}=3 \hat{i}+4 \hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}-5 \hat{k}$$\vec{a}\cdot \vec{b}$\begin{align}\vec{a} \cdot \vec{b}&amp;=(3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(\hat{i}-\hat{j}+3 \hat{k})\\
\Rightarrow &amp;=(…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-3-p2?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 and 3 Exercise 3.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-3-p2?rev=1737476037&amp;do=diff</link>
        <description>Question 2 and 3 Exercise 3.3

Solutions of Question 2 and 3 of Exercise 3.3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 2
$$\vec{a}=2 \hat{i} + 2 \hat{j}-5 \hat{k}, \quad \vec{b}=2 \hat{i}+\hat{j}-7 \hat{k}$$\begin{align}\vec{a}+\vec{b}&amp;=(2 \hat{i}+2 \hat{j}-5 \hat{k})+(2 \hat{i}+\hat{j}-7 \hat{k}) \\
\Rightarrow &amp;=4 \hat{i}+3 \hat{j}-12 \hat{k}\\
\Rightarrow|\vec{a}+\…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-3-p4?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6 Exercise 3.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-3-p4?rev=1737476037&amp;do=diff</link>
        <description>Question 6 Exercise 3.3

Solutions of Question 6 of Exercise 3.3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 6(i)

Let $\vec{a}=\hat{i}+3 \hat{j}-4 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$$m$$\vec{a}+m \vec{b}$$\vec{a}$\begin{align}
\vec{a}+m \vec{b}&amp; =\hat{i}+3 \hat{j}-4 \hat{k}+m(2 \hat{i}-3 \hat{j}+5 \hat{k}) \\
&amp; =(1+2 m) \hat{i}+(3-3 m) \hat{j}+(5 m-4) …</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-3-p5?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7 &amp; 8 Exercise 3.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-3-p5?rev=1737476037&amp;do=diff</link>
        <description>Question 7 &amp; 8 Exercise 3.3

Solutions of Question 7 &amp; 8 of Exercise 3.3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 7(i)
$\vec{a}$$\vec{b}$$\vec{a}=-\dfrac{3}{2} \hat{j}+\dfrac{4}{5} \hat{k} \cdot \vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}$$\vec{a}$$\vec{b}$$\vec{b}$$\vec{a}$$\vec{a}=-\dfrac{3}{2} \hat{j}+\dfrac{4}{5} \hat{k}\quad$$\vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}$\begin{a…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-3-p7?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 11, Exercise 3.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-3-p7?rev=1737476037&amp;do=diff</link>
        <description>Question 11, Exercise 3.3

Solutions of Question 11 of Exercise 3.3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 11 (i)

Show that the vectors $3 \hat{i}-2 \hat{j}+$$\hat{k} . \quad \hat{i}-3 \hat{j}-5 \hat{k}$$2 \hat{i}+\hat{j}-4 \hat{k}$$\vec{a}=3 \hat{i}-2 \hat{j}+\hat{k}$$\vec{b}=\hat{i}-3 \hat{j}+5 \hat{k}$$\vec{c}=2 \hat{i}+\hat{j}-4 \hat{k}$\begin{align}|\vec{a}|&amp;…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-4-p1?rev=1737476037&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Exercise 3.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-4-p1?rev=1737476037&amp;do=diff</link>
        <description>Question 1 Exercise 3.4

Solutions of Question 1 of Exercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)

Find the cross product $\hat{j} \times(2 \hat{j}+3 \hat{k})$\begin{align}\vec{a}=\hat{j}&amp;=0 \hat{i}+\hat{j}+0 \hat{k}\\
\vec{b}&amp;=0 \hat{i}+2 \hat{j}-3 \hat{k}\\
 \vec{a} \times \vec{b}&amp;=\hat{j} \times(2 \hat{j}+3 \hat{k})\\
&amp;=\left|\begin{array}{lll}\hat{i}…</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 Exercise 3.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-4-p2?rev=1737476037&amp;do=diff</link>
        <description>Question 2 Exercise 3.4

Solutions of Question 2 of Exercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 2(i)

Show in two different ways that the vectors $\vec{a}$$\vec{b}$$\vec{a}=-\hat{i}+2 \hat{j}-3 \hat{k}, \quad \vec{b}=2 \hat{i}-4 \hat{j}+$$6 \hat{k}$\begin{align}\vec{a} \times \vec{b}&amp;=\left|\begin{array}{ccc}
\hat{i} &amp; \hat{j} &amp; \hat{k} \\
-1 &amp; 2 &amp; -3 \\
2 …</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3 Exercise 3.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-4-p3?rev=1737476037&amp;do=diff</link>
        <description>Question 3 Exercise 3.4

Solutions of Question 3 of Exercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 3(i)

Find a unit vector that is orthogonal to
the given vector $\vec{a}=\hat{i}- 2 \hat{j}+3 \hat{k}, \quad \vec{b}=2 \hat{i}+\hat{j}-\hat{k}$$\hat{n}$$\vec{a}$$\vec{b}$\begin{align}\hat{n}&amp;=\dfrac{\vec{a} \times \vec{b}}{\mid \vec{a} \times \vec{b}} \\
\text { …</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 4 Exercise 3.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-4-p4?rev=1737476037&amp;do=diff</link>
        <description>Question 4 Exercise 3.4

Solutions of Question 4 of Exercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 4(i)

If $\vec{a}=3 \hat{i}-6 \hat{j}+5 \hat{k},\quad\vec{b}=2\hat{i}-\hat{j}+4 \hat{k} \quad$ and $\quad \vec{c}=\hat{i}+\hat{j} \quad \hat{k},\quad$$\vec{a} \times \vec{b}$\begin{align}\vec{a} \times \vec{b}&amp;=\left|\begin{array}{ccc}
\hat{i} &amp; \hat{j} &amp; \hat{k}…</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 Exercise 3.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-4-p5?rev=1737476037&amp;do=diff</link>
        <description>Question 5 Exercise 3.4

Solutions of Question 5 of Exercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 5(i)

Use the vector product to compute the area of the triangle with the given vertices $P(-2,-3), \quad Q(3,2)\quad$$\quad R(-1,-8)$$P Q$$\bar{P} R$\begin{align}\text{Area of triangle}&amp;=\dfrac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}| \\
\text { S…</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6 Exercise 3.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-4-p6?rev=1737476037&amp;do=diff</link>
        <description>Question 6 Exercise 3.4

Solutions of Question 6 of Exercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 6(i)

A force $\vec{F}=3 \hat{i}-2 \hat{j}+5 \hat{k}$$(1,-2,2)$$\vec{r}$$P(1,-2.2)$$O(0,0,0)$\begin{align}\vec{r}&amp;=\overrightarrow{O P}\\
&amp;=(1,-2,2)-(0,0,0) \\
\Rightarrow \vec{r}&amp;=(1,-2,2).\\
\text { Hence } \vec{M}-\vec{r} \times \vec{F}&amp;=\left|\begin{array}{cc…</description>
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        <dc:date>2025-01-21T16:13:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7 &amp; 8 Exercise 3.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-4-p7?rev=1737476037&amp;do=diff</link>
        <description>Question 7 &amp; 8 Exercise 3.4

Solutions of Question 7 &amp; 8 of Exercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 7

If $\vec{A}+\vec{B}+\vec{C}=\vec{O}$$$\vec{A} \times \vec{B}=\vec{B} \times \vec{C}=\vec{C} \times \vec{A}.$$$$\vec{A}+\vec{B}+\vec{C}=\vec{O} \text {. }$$$\vec{A}$$$\vec{A} \times(\vec{A}+\vec{B}+\vec{C})=0$$\begin{align}\Rightarrow \vec{A} \times \ve…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 9 Exercise 3.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-4-p8?rev=1737476038&amp;do=diff</link>
        <description>Question 9 Exercise 3.4

Solutions of Question 9 of Exercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 9(i)

Find the area of parallelogram whose diagonals are $\vec{a}=4 \hat{i}+\hat{j}-2 \hat{k}\quad$$\quad\vec{b}=-2 \hat{i}+3 \hat{j}+4 \hat{k}$$\vec{c}$$\vec{d}$$E$$E$\begin{align}\overrightarrow{A E}&amp;=\overrightarrow{E C}\\
&amp;=\dfrac{1}{2} \vec{a}\\
&amp;=2 \hat{i}+…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3 &amp; 4 Exercise 3.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-5-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 3 &amp; 4 Exercise 3.5

Solutions of Question 3 &amp; 4 of Exercise 3.5 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 3

For the vectors $\vec{a}=3 \hat{i}+2 \hat{k}$$\vec{b}=\hat{i}+2 \hat{j}+\hat{k}\quad$$\quad\vec{c}=-\hat{j}+4 \hat{k}$$\vec{a} \cdot \vec{b} \times \vec{c}=\vec{b} \cdot \vec{c} \times \vec{a}=\vec{c} \cdot \vec{a} \times \vec{b}$$\vec{a} \cdot \vec{b}…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7 Exercise 3.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-5-p6?rev=1737476038&amp;do=diff</link>
        <description>Question 7 Exercise 3.5

Solutions of Question 7 of Exercise 3.5 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 7(i)

For what value of $c$$\vec{u}=\hat{i}+2 \hat{j}+3 \hat{k}$$\vec{v}=2 \hat{i}-3 \hat{j}+4 \hat{k} \cdot \vec{w}=3 \hat{i}+\hat{j}+c \hat{k}$\begin{align}\vec{u} \cdot \vec{v} \times \vec{w}&amp;=0\\
\vec{u} \cdot \vec{v} \times \vec{w}&amp;=0\\
\Rightarrow\left|\beg…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 8 Exercise 3.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-5-p7?rev=1737476038&amp;do=diff</link>
        <description>Question 8 Exercise 3.5

Solutions of Question 8 of Exercise 3.5 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 8(i)

Find the volume of tetrahedron with the Vectors as coterminous edges
\begin{align}\vec{a}&amp;=\hat{i}+2 \hat{j}+3 \hat{k},\\ 
\vec{b}&amp;=4 \hat{i}+5 \hat{j}+6 \hat{k}, \\
\vec{c}&amp;=7 \hat{j}+8 \hat{k}\end{align}\begin{align}V&amp;=\dfrac{1}{6}[\vec{u} \cdot \vec{v} \…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 9 Exercise 3.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/ex3-5-p8?rev=1737476038&amp;do=diff</link>
        <description>Question 9 Exercise 3.5

Solutions of Question 9 of Exercise 3.5 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 9 (i)

Write the value of $(\hat{i} \times \hat{j}). \hat{k}+\hat{i}. \hat{j}$\begin{align}
(\hat{i} \times \hat{j}) \cdot \hat{k}&amp;=\left|\begin{array}{ccc}
1 &amp; 0 &amp; 0 \\
0 &amp; 1 &amp; 0 \\
0 &amp; 0 &amp; 1
\end{array}\right|&amp;=1 ....(1)\\
\text { and } \hat{i} \cdot \hat{j}&amp;=0…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Review Exercise 3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/review-ex3-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Review Exercise 3

Solutions of Question 1 of Review Exercise 3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1

$\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})$$0$$1$$1$$3$$3 \hat{i}+5 \hat{j}+2 \hat{k}$$2 \hat{i}-3 \hat{j}-5 \hat{k}$$5 \hat{i}+2 \hat{j}-3 \hat{k}$$\hat{i}-2 \hat{i}+\hat{j}+3 \…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 10 Review Exercise 3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit03/review-ex3-p6?rev=1737476038&amp;do=diff</link>
        <description>Question 10 Review Exercise 3

Solutions of Question 10 of Review Exercise 3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 10(i)
$A B C$$|\vec{a}|^2=|\vec{b}|^2+|\vec{c}|^2 -2|\vec{b}|| \vec{c}| \cos A$$A B C$$\vec{a}, \vec{b}$$\vec{c}$\begin{align}
\vec{b}&amp;=\vec{a}+\vec{c} \\
\Rightarrow \vec{a}&amp;=\vec{b}-\vec{c} \\
\Rightarrow \vec{a} \cdot \vec{a}&amp;=(\vec{b}-\vec{c}) \cd…</description>
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        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 Exercise 4.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-1-p3?rev=1737476038&amp;do=diff</link>
        <description>Question 5 Exercise 4.1

Solutions of Question 5 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 5(i)
$\sum_{j=1}^6(2 j-3)$\begin{align}\sum_{j=1}^6(2 j-3)&amp;=(2.1-3)+(2.2-3)+(2.3-3)+(2.4-3)\\&amp;+(2.5-3)+(2.6-3) \\
\implies \sum_{j=1}^6(2 j-3)&amp;=-1+1+3+5+7+9 .\end{align}$\sum_{k=1}^5(-1)^k 2^{k-1}$\begin{align}\sum_{k=1}^5(-1)^k 2^{k-1}&amp; =(-1)^1 2^{1-…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6 Exercise 4.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-1-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 6 Exercise 4.1

Solutions of Question 6 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Note

The general recursive definition formula defined for Pascal sequences is
$$P_0=1, P_{r+1}=\dfrac{n-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$$n=5$$n=5$$$P_0=1, P_{r+1}=\dfrac{5-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$$r=0$\begin{align}&amp;P_{0+1…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 12 &amp; 13 Exercise 4.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-2-p9?rev=1737476038&amp;do=diff</link>
        <description>Question 12 &amp; 13 Exercise 4.2

Solutions of Question 12 &amp; 13 of Exercise 4.2 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$a_1$$$a_1=3500.$$$=d=750$$a_{21}$\begin{align}
a_{21}&amp;=a_1+20d\\
&amp;=3500+20(750) \\
&amp;=18500. \end{align}$12$$18$$a=12, b=18$$A$\begin{align}A&amp;=\dfrac{a+b}{2}\\&amp;=\dfrac{12+18}{2}\\&amp;=\dfrac{30}{2}=15.\end{align}$\dfrac{1}{3}$$\dfrac{1}{4}$$a=\dfrac{1}{…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 14 Exercise 4.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-2-p10?rev=1737476038&amp;do=diff</link>
        <description>Question 14 Exercise 4.2

Solutions of Question 14 of Exercise 4.2 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 14(i)
$A_1, A_2, A_3$$6, A_1, A_2, A_3, 41$$$a_1=6 \text{ and } a_6=41.$$\begin{align}&amp; a_5=11\\
\Rightarrow &amp;a_1+4 d=41 \\
\Rightarrow &amp;6+4 d=41 \\
\Rightarrow &amp;d=\dfrac{41-6}{4}\\
&amp;=\dfrac{35}{4}.\end{align}\begin{align} A_1&amp;=a+d=6+\dfrac{35}{4} \…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-3-p1?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Exercise 4.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-3-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Exercise 4.3

Solutions of Question 1 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)
$9,7,5,3, \ldots$$a_1$$d$\begin{align}&amp;a_1=9 \\ 
&amp;d=7-9=-2 \\
&amp;n=20.
\end{align}\begin{align}&amp;a_n=a_1+(n-1)d \\
\implies &amp;a_20=9+(20-1)(-2)=-29.
\end{align}$S_n$$n$\begin{align}
S_n&amp;=\dfrac{n}{2}[a_1+a_n], \\
\implies S_{20}&amp;=\dfrac{20}{2}[9-29] …</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 Exercise 4.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-3-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 2 Exercise 4.3

Solutions of Question 2 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 2(i)
$a_1, a_n, n, d$$S_n$$a_1=2, n=17, d=3$$a_1=2, n=17, d=3$$a_{17}$$S_{17}$$$a_{n}=a_1+(n-1)d.$$$$a_{17}=2+(17-1)(3)=50.$$$$S_n=\dfrac{n}{2}[a_1+a_n]$$\begin{align}S_{17}&amp;=\dfrac{17}{2}(a_1+a_17) \\
&amp;=\dfrac{17}{2}(2+50)=442.\end{align}$a_{17}=50$$…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-3-p5?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7 &amp; 8 Exercise 4.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-3-p5?rev=1737476038&amp;do=diff</link>
        <description>Question 7 &amp; 8 Exercise 4.3

Solutions of Question 7 &amp; 8 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 7
$1+3-5+7+9-11+13+15-$$17+\ldots$$3 n$\begin{align}&amp;(1+7+13+\ldots)+(3+9+15+\ldots)- \\
&amp; (5+11+17+\ldots) \ldots \ldots \ldots . . .(1)\end{align}$\mathrm{n}$$n$$3 n$$$1+7+13+\ldots$$$$a_1=1, d=7-1=6$$$n$\begin{align}S_n&amp;=\dfrac{n}{2}[2 a_1+…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-4-p1?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Exercise 4.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-4-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Exercise 4.4

Solutions of Question 1 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question(i)
$a_1=5, \quad r=3$$a_1, a_1 r, a_1 r^2, a_1 r^3, a_1 r^4, \ldots$$a_1=5 ; r=3$\begin{align}&amp;5,5.3,5.3^2, 5.3^3, 5.3^4, \ldots\\
\Rightarrow &amp;5,15,45,135,405, \ldots\end{align}$a_1=8, \quad r=-\dfrac{1}{2}$$a_1, a_1 r, a_1 r^2, a_1 r^3, a_1 r^4, \ld…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-4-p5?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 8 Exercise 4.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-4-p5?rev=1737476038&amp;do=diff</link>
        <description>Question 8 Exercise 4.4

Solutions of Question 8 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 8(i)
$3.14$$2.71$$a=3.14$$b=2.71$$$G= \pm \sqrt{(3.14)(2.71)}= \pm 2.94$$$$G=2.94 \quad \text{or} \quad -2.94$$$-6$$-216$$a=-6$$b=-216$\begin{align}G&amp;= \pm \sqrt{(-6)(-216)}= \pm \sqrt{1296} \\
\Rightarrow G&amp;= \pm 36\end{align}$$G=36 \quad \text{or} \…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-4-p6?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 9 Exercise 4.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-4-p6?rev=1737476038&amp;do=diff</link>
        <description>Question 9 Exercise 4.4

Solutions of Question 9 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 9(i)
$3 \dfrac{5}{9}=\dfrac{32}{9}\quad$$\quad40 \dfrac{1}{2}=\dfrac{81}{2}$$G_1, G_2, G_3, G_4$$G_5$$\dfrac{32}{9}$$\dfrac{81}{2}$$\dfrac{32}{9}, G_1, G_2, G_3, G_4, G_5, \dfrac{81}{2}$$a_7=\dfrac{81}{2}$$a_1=\dfrac{32}{9}$\begin{align}a_1 r^6&amp;=\dfra…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-4-p7?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 10 Exercise 4.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-4-p7?rev=1737476038&amp;do=diff</link>
        <description>Question 10 Exercise 4.4

Solutions of Question 10 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 10
$48$$18$$a$$b$$1$$48$$$\quad a-b=48....(i)$$$a$$b$$$G=\sqrt{a b}$$$a$$b$$$A=\dfrac{a+b}{2}$$$2$$A \cdot M=G \cdot M+18$$A \cdot M-G \cdot M=18$$$\Rightarrow \dfrac{a+b}{2}-\sqrt{a b}=18$$$$(a+b)-2 \sqrt{a b}=36 \text {. }$$$a=b+48$\begin{align}(b…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p1?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Exercise 4.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Exercise 4.5

Solutions of Question 1 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)
$3+6+12+\ldots+3.2^9$$a_1=3, \quad r=\dfrac{6}{3}=2$$a_n=3.2^9$$n$$$a_n=a_1 r^{n-1}$$\begin{align}3.2^9&amp;=3(2)^{n-1} \text { or }(2)^{n-1}=\dfrac{3.2^9}{3} \\
\Rightarrow(2)^{n-1}&amp;=2^9 \\
\Rightarrow n-1&amp;=9 \text { or } n=10  \\
\text {. Now }\qua…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p2?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 Exercise 4.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 2 Exercise 4.5

Solutions of Question 2 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 2(i)
$a_1, a_n, n_2 r$$S_n$$a_1=1, \quad r=-2, \quad a_n=64$$n$$S_n$$a_n=a_1 r^{n-1}$\begin{align}64&amp;=(-2)^{n-1}\\
\Rightarrow(-2)^{n-1}&amp;=(-2)^6 \\
\Rightarrow n-1&amp;=6 \\
\Rightarrow n&amp;=7\\
S_7&amp;=\dfrac{a_1[r^{\prime \prime}-1]}{r-1}\\
\text{then}\\
S_7…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p9?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 13 &amp; 14 Exercise 4.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p9?rev=1737476038&amp;do=diff</link>
        <description>Question 13 &amp; 14 Exercise 4.5

Solutions of Question 13 &amp; 14 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$y=\dfrac{x}{3}+\dfrac{x^2}{3^2}+\dfrac{x^3}{3^3}+\ldots$$0&lt;x&lt;3$$x=\dfrac{3 y}{1+y}$$$1+y=1+\dfrac{x}{3}+\dfrac{x^2}{3^2}+\dfrac{x^3}{3^3}$$$a_1=1$$r=\dfrac{x}{3}$$|r|=\dfrac{x}{3}&lt;1$$0&lt;x&lt;3$$S_{\infty}=\dfrac{a_1}{1-r}$$a_1, \quad r$$$S_{\infty}=\dfr…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p1?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Exercise 5.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Exercise 5.1

Solutions of Question 1 of Exercise 5.1 of Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)
$1^2+3^2+5^2+7^2+\ldots$$n$$1+3+5+\ldots$$n^{\text {th }}$$2 n-1$$n^{t h}$$$T_j=(2 j-1)^2$$\begin{align}&amp; \sum_{j=1}^n T_j=\sum_{j=1}^n(2 j-1)^2 \\
&amp; =\sum_{j=1}^n(4 j^2-4 j+1)\\
&amp; =4 \sum_{j=1}^n j^2-4 \sum_{j=1}^n j+\sum_{j=1}^n 1 \\
&amp; =4 \dfr…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p6?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 9 Exercise 5.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p6?rev=1737476038&amp;do=diff</link>
        <description>Question 9 Exercise 5.1

Solutions of Question 9 of Exercise 5.1 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 9(i)
$n$$n$$n$\begin{align}
&amp; T_n=n^2(2 n+3)=2 n^3+3 n^2 \\
&amp; \Rightarrow T_j=2 j^3+3 j^2\end{align}\begin{align}
&amp; \sum_{j=1}^n T_j=2 \sum_{j=1}^n j^3+3 \sum_{j=1}^n j^2 \\
&amp; =2(\dfrac{n(n+1)}{2})^2+3 \dfrac{n(n+1)(2 n+1)}{6} \\
&amp; =\dfrac{n(n+1)}{2}…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-4-p1?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Exercise 5.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-4-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Exercise 5.3

Solutions of Question 1 of Exercise 5.4 of Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)
$\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\ldots$$n$$$T_n=\dfrac{1}{n(n+1)}$$$T_n$$$\dfrac{1}{n(n+1)}=\dfrac{A}{n}+\dfrac{B}{(n+1)}$$$n(n+1)$$$1=A(n+1)+B n=(A+B) n+A$$$n$$$A+B=0 \text{and} A=1$$$A=1$\begin{align}1+B&amp;=0\\
B&amp;=-1\end{align}\beg…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p1?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Review Exercise 5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Review Exercise 5

Solutions of Question 1 of Review Exercise 5 of Unit 05: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1

$t_n=6 n+5$$t_{n+1}=$$6 n-1$$6 n+11$$6 n+6$$6 n-5$$1+\dfrac{2}{3}+\dfrac{6}{3^2}+\dfrac{10}{3^3}+\dfrac{14}{3^4}+\ldots$$6$$2$$3$$4$$1+2.2+3.2^2+\cdots+100.2^{\prime \prime}$$99.2^{100}$$100.2^{100}$$99.2^{100}+1$$1000.2^{100}$$n^{t h}$$1.2+2.3+3.4+\…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p5?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7 Review Exercise</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p5?rev=1737476038&amp;do=diff</link>
        <description>Question 7 Review Exercise

Solutions of Question 7 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 7(i)
$1.2^2+3.3^2+5.4^2+\ldots$$n$$1,3,5, \ldots,(2 n-1)$$2^2, 3^2, 4^2, \ldots,(n+1)^2$\begin{align}
&amp; a_n=(2 n-1)(n+1)^2 \\
&amp; a_n=(2 n-1)(n^2+2 n+1) \\
&amp; a_n=2 n^3+3 n^2-1\end{align}\begin{align}
\sum_{r=1}^n a_r&amp;=2 \sum_{r=1}^n r^3+\sum_{r=1…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p6?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 8 Review Exercise</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p6?rev=1737476038&amp;do=diff</link>
        <description>Question 8 Review Exercise

Solutions of Question 8 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 8(i)
$n$$n^{t h}$$n^3+3^n.$$n^h$$$a_n=n^3+3^n$$\begin{align}\sum_{r=1}^n a_r&amp;=\sum_{r=1}^n r^3+\sum_{r=1}^n 3^r \\
&amp; =[\dfrac{n(n+1)}{2}]^2+\dfrac{3(3^n-1)}{3-1} \\
&amp; =\dfrac{n^2(n+1)^2}{4}+\dfrac{3}{2}(3^n-1) \end{align}$n$$$S_n=\dfrac{n^2(n+1…</description>
    </item>
    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p7?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 9 Review Exercise</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p7?rev=1737476038&amp;do=diff</link>
        <description>Question 9 Review Exercise

Solutions of Question 9 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 9(i)
$n$$3+7+13+21+31+\ldots$\begin{align}
&amp; a_2-a_1=7-3=4 \\
&amp; a_3-a_2=13-7=6 \\
&amp; a_4-a_3=21-13=8 \\
&amp; \ldots \quad \ldots \quad \ldots \\
&amp; \ldots \quad \cdots \quad \ldots \\
&amp; a_n-a_{n-1}=(n-1) \text { term of the series } \\
&amp; 4,6,8, \ldo…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-1-p3?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 Exercise 6.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-1-p3?rev=1737476038&amp;do=diff</link>
        <description>Question 5 Exercise 6.1

Solutions of Question 5 of Exercise 6.1 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\dfrac{(2 n) !}{n !}=2^n(1.3 .5 \ldots(2 n-1))$\begin{align}\dfrac{(2 n) !}{n !}&amp;=\dfrac{1}{n !}[(2 n)(2 n-1)(2 n-2) \\
&amp;=(2 n-3)(2 n-4)(2 n-5) \ldots(2 n-(2 n-4))\\
&amp;(2 n-(2 n-3))(2 n-(2 n-2))(2 n-(2 n-1))]\end{align}$2 n$\begin{align}\dfra…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p2?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3 and 4 Exercise 6.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 3 and 4 Exercise 6.2

Solutions of Question 3 and 4 of Exercise 6.2 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$^n P_r=n(^{n-1} P_{r-1})$$$^n P_r=n({ }^{n-1} P_{r-1})$$\begin{align}n(^{n-1} P_{r-1})&amp;=n \dfrac{(n-1) !}{((n-1)-(r-1)) !} \\
&amp; =\dfrac{n(n-1) !}{(n-r) !}\\
&amp;=\dfrac{n !}{(n-r) !}\\
&amp;=^n P_r\end{align}$^n P_r=^{n-1} P_r+r(^{n-1} …</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p4?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7 and 8 Exercise 6.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 7 and 8 Exercise 6.2

Solutions of Question 7 and 8 of Exercise 6.2 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$1,2,3,4$$E_1$$m_1=5$$E_2$$\cdot m_2=5$$E_3$$m_3=5$$$m_1 \cdot m_2 \cdot m_3=5.5 \cdot 5=125$$$1,2,3,4$$E_1$$m_1=5$$E_2$$m_2=4$$E_3$$m_3=3$$$m_1 \cdot m_2 \cdot m_3=5 \cdot 4 \cdot 3=60$$$8$$5$$=4$$=4$$=5$$=3$$4 ! \cdot 5 ! \cdot …</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 10 Exercise 6.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p6?rev=1737476038&amp;do=diff</link>
        <description>Question 10 Exercise 6.2

Solutions of Question 10 of Exercise 6.2 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$n=8$$r=5$\begin{align}^8 P_5&amp;=\dfrac{8 !}{(8-5) !}\\
&amp;=\dfrac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !}\\
&amp;=6720\end{align}\begin{align}^2 P_2 \times^7 P_4&amp;=2 \times \dfrac{7 !}{(7-4) !}\\
&amp;=2 \times\dfrac{7.6 .5 .4 .3 !}{3 !}\\
&amp;=…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p7?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 11 Exercise 6.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p7?rev=1737476038&amp;do=diff</link>
        <description>Question 11 Exercise 6.2

Solutions of Question 11 of Exercise 6.2 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$10$$1000$$2.3,4,0,8,9$$10$$1000$$10$$100$$E_1$$m_1=5$$E_2$$m_2=5$$10$$100$$$m_1 \cdot m_2=5.5=25$$$100$$1000$$0$$E_1$$m_1=5$$E_2$$\boldsymbol{m}_2=5$$E_3$$m_3=4$$100$$1000$$$m_1 \cdot m_2 \cdot m_3=5.5 \cdot 4=100$$$10$$1000$$$100 + 25=125…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p8?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 12 Exercise 6.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p8?rev=1737476038&amp;do=diff</link>
        <description>Question 12 Exercise 6.2

Solutions of Question 12 of Exercise 6.2 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$8.$$n=8$$\mathrm{O}$$m_1=3$\begin{align}
 \left(\begin{array}{c}
n \\
m 1
\end{array}\right)&amp;=\left(\begin{array}{l}
8 \\
3
\end{array}\right) \\
&amp; =\dfrac{8 !}{3 !}\\
&amp;=\dfrac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !}\\
&amp;=6,720 \e…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 13 Exercise 6.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-2-p9?rev=1737476038&amp;do=diff</link>
        <description>Question 13 Exercise 6.2

Solutions of Question 13 of Exercise 6.2 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\mathrm{E}$$n=10$$m_1=4$$E, m_2=2$$L$$m_3=2$$C$\begin{align}\text{total number of permutations are}
 &amp;=\left(\begin{array}{c}
n \\
m_1, m_2, m_3
\end{array}\right)\\&amp;=\left(\begin{array}{c}
10 \\
4,2,2
\end{array}\right) \\
&amp; =\dfrac{10 !}…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-3-p1?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Exercise 6.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-3-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Exercise 6.3

Solutions of Question 1 of Exercise 6.3 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$^n C_2=36$$n$\begin{align}&amp;^n C_2=36\\
&amp; \Rightarrow \dfrac{n !}{(n-2) ! 2 !}=36 \\
&amp; \Rightarrow \dfrac{n(n-1)(n-2) !}{(n-2) ! \cdot 2}=36 \\
&amp; \Rightarrow n(n-1)=72 \\
&amp; \Rightarrow n^2-n-72=0 \\
&amp; \Rightarrow n^2-9 n+8 n-72=0\\
&amp; \Rightar…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-3-p4?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 4 Exercise 6.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-3-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 4 Exercise 6.3

Solutions of Question 4 of Exercise 6.3 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.${ }^{n-1} C_r+{ }^{n-1} C_{r-1}={ }^n C_r$$${ }^n{ }^1 C_r+{ }^n{ }^1 C_{r-1}={ }^n C_s$$\begin{align}
{ }^{n-1} C_r+{ }^{n-1} C_{r-1}&amp;=\dfrac{(n-1) !}{(n-r-1) ! r !}+\dfrac{(n-1) !}{(n-1-(r-1)) !(r-1) !} \\
&amp; =\dfrac{(n-1) !}{(n-r-1) ! r(r-…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-3-p5?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 and 6 Exercise 6.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-3-p5?rev=1737476038&amp;do=diff</link>
        <description>Question 5 and 6 Exercise 6.3

Solutions of Question 5 and 6 of Exercise 6.3 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$12$$n=12$${ }^{12} C_2=66$$12$$n=12$${ }^{12} C_3=220$$${ }^6 C_2=\dfrac{6 !}{(6-2) ! 2 !}=15 $$$6$$\quad 15-6=9$</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 9 Exercise 6.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-3-p7?rev=1737476038&amp;do=diff</link>
        <description>Question 9 Exercise 6.3

Solutions of Question 9 of Exercise 6.3 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$8$$6$$7$$7$$6.$$=7+6=13$${ }^7 C_4$${ }^6 C_4$\begin{align}{ }^7 C_4 \cdot{ }^6 C_4&amp;=\dfrac{7 !}{(7-4) ! 4 !} \cdot \dfrac{6 !}{(6-4)}\\\
&amp;= 525\end{align}$8$$6$$7$$7$$6$$=7+6=13$$3,4,5,6$$6$\begin{align}{ }^7 C_2 \cdot{ }^6 C_6&amp;=\dfrac{7 !}…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 Exercise 6.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-4-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 2 Exercise 6.4

Solutions of Question 2 of Exercise 6.4 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$4$$5$$6$$3$$4+5+6=15$$${ }^{15} C_3=\dfrac{15 !}{(15-3) ! 3 !}=455 $$$${ }^6 C_4=\dfrac{6 !}{(6-4) ! 4 !}=15$$$$=\dfrac{15}{455}=\dfrac{3}{91}$$$4$$5$$6$$3$$4+5+6=15$$${ }^{15} C_3=\dfrac{15 !}{(15-3) ! 3 !}=455 $$$${ }^4 C_3=\dfrac{4 !}{(4-…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 4 Exercise 6.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-4-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 4 Exercise 6.4

Solutions of Question 4 of Exercise 6.4 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.\begin{align}S&amp;=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\
\text{then} n(S)&amp;=2^3=8\end{align}$$A=\{H H H\}$$$$n(A)=1$$$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{1}{8}$\begin{align}S&amp;=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ 
\text{then} n(S)&amp;=2^3=8\end{align}…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 Exercise 6.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-4-p5?rev=1737476038&amp;do=diff</link>
        <description>Question 5 Exercise 6.4

Solutions of Question 5 of Exercise 6.4 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$6$$4$$3$$2$$=6+4=10$$5$$10$\begin{align}{ }^{10)} C_5 &amp;=\dfrac{10 !}{(10-5) ! 5 !}\\
&amp;=252\\ 
n(S)&amp;=252\end{align}$3$$2$$3$$2$\begin{align}{ }^6 \mathrm{C}_3\cdot{ }^{4} \mathrm{C}_2&amp;=\dfrac{6 !}{(6-3) ! 3 !} \cdot \dfrac{4 !}{(4-2) ! 2 !}\\…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6 Exercise 6.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-4-p6?rev=1737476038&amp;do=diff</link>
        <description>Question 6 Exercise 6.4

Solutions of Question 6 of Exercise 6.4 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$52$$$=52$$$=4$$$=\dfrac{4}{52}=\dfrac{1}{13}$$$52$$=52$$13$$13$$$\dfrac{13}{52}+ \dfrac{13}{52}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}$$$52$$=52$$13.$$$=\dfrac{13}{52}=\dfrac{1}{4}$$$52$$=52$$12.$$$=\dfrac{12}{52}=\dfrac{3}{13}$…</description>
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        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7 Exercise 6.4</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-4-p7?rev=1737476038&amp;do=diff</link>
        <description>Question 7 Exercise 6.4

Solutions of Question 7 of Exercise 6.4 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.\begin{align}S&amp;=\{(i, j) ; i, j=1,2,3,4,5,6\}\\
&amp;=\left[\begin{array}{llllll}
(1,1) &amp; (1,2) &amp; (1,3) &amp; (1,4) &amp; (1,5) &amp; (1,6) \\
(2,1) &amp; (2,2) &amp; (2,3) &amp; (2,4) &amp; (2,5) &amp; (2,6) \\
(3,1) &amp; (3,2) &amp; (3,3) &amp; (3,4) &amp; (3,5) &amp; (3,6) \\
(4,1) &amp; (4,2) &amp; (…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-5-p1?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 and 2 Exercise 6.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-5-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 and 2 Exercise 6.5

Solutions of Question 1 and 2 of Exercise 6.5 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$A$$B$$P(A)=\dfrac{2}{5}, P(B)=\dfrac{2}{5}$$P(A \cup B)=\dfrac{1}{2}$$P(A \cap B)$\begin{align}
 P(A \cup B)&amp;=P(A)+P(B)-P(A \cap B) \\
 \Rightarrow P(A \cap B)&amp;=P(A)+P(B)-P(A \cup B)
\end{align}$P(A), P(B)$$P(A \cup B)$$$P(A \cap…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 9 Exercise 6.5</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/ex6-5-p6?rev=1737476038&amp;do=diff</link>
        <description>Question 9 Exercise 6.5

Solutions of Question 9 of Exercise 6.5 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$2$$\dfrac{1}{7}$$\dfrac{1}{5}$\begin{align}
P(\text { Ajmal scicction })&amp;=\dfrac{1}{7} \\
\Rightarrow P(\text { Ajmal not selected })&amp;=\dfrac{6}{7} \\
P(\text { Bushra selection })&amp;=\dfrac{1}{5} \\
\Rightarrow P(\text { Bushra not selected }…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Review Exercise 6</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/re-ex6-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Review Exercise 6

Solutions of Question 1 of Review Exercise 6 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$O, P, Q, R, S, T, U$$2520$$9040$$5140$$4880$$\{1,2,3,4,5,6,7\}$$14$$42$$28$$21$$\{1,2,3,4,6,7,8\}$$3$$7$$120$$180$$144$$96$$\dfrac{(n+2) !(n-2) !}{(n+1) !(n-1) !}$$(n-3)$$(\dot{n}-1)$$\dfrac{n+1}{n+2}$$\dfrac{n+2}{n-1}$$5$$768$$724…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 Review Exercise 6</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/re-ex6-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 2 Review Exercise 6

Solutions of Question 2 of Review Exercise 6 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.${ }^{2 n} C_r={ }^{2 n} C_{r+2}$$r$\begin{align}
{ }^{2 n} C_r&amp;={ }^{2 n} C_{r+2} \\
\Rightarrow \dfrac{(2 n) !}{(2 n-r) ! r !}&amp;=\dfrac{(2 n) !}{(2 n-(r+2)) !(r+2) !}\end{align}$(2 n)$\begin{align}
\Rightarrow \dfrac{1}{(2 n-r) ! r…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 &amp; 6 Review Exercise 6</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/re-ex6-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 5 &amp; 6 Review Exercise 6

Solutions of Question 5 &amp; 6 of Review Exercise 6 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$$n=6$$$$(n-1) !=(6-1) !=5 !=120$$$120-24=96$$n=6$$(n-1) !=(6-1) !=5 !=120$$$(n-1) !=(5-1) !=4 !=24$$$$(n-1) !=(6-1) !=5 !=120$$$$4 ! \cdot 2 !=48$$$(5-1) !$</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7 &amp; 8 Review Exercise 6</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit06/re-ex6-p5?rev=1737476038&amp;do=diff</link>
        <description>Question 7 &amp; 8 Review Exercise 6

Solutions of Question 7 &amp; 8 of Review Exercise 6 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$P(A)=0.8, P(B)=0.5$$P(B / A)=0.4$$P(A \cap B)$\begin{align}
P(B \mid A)&amp;=\dfrac{P(A \cap B)}{P(A)} \\
\Rightarrow P(A \cap B)&amp;=P(B \mid A) \cdot P(A)\\
&amp;=0.4 \times 0.8=0.32\end{align}$P(A)=0.8, P(B)=0.5$$P(B / A)=0.4$$P(A …</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 12 Exercise 7.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-1-p12?rev=1737476038&amp;do=diff</link>
        <description>Question 12 Exercise 7.1

Solutions of Question 12 of Exercise 7.1 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\dfrac{5^{2 n}-1}{24}$$n=1$$$\dfrac{5^{2 n}-1}{24}=\dfrac{5^{2.1}-1}{24}=\dfrac{24}{24}=1 \in \mathbb{Z}$$$n=1$$n=k&gt;1$$$\dfrac{5^{2 k}-1}{24} \in \mathbb{Z}$$$n=k+1$\begin{align}\dfrac{5^{2(k+1)}-1}{24}&amp;=\dfrac{5^{2 k+2}-1}{24} \\
&amp; =\dfra…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-1-p13?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 13 Exercise 7.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-1-p13?rev=1737476038&amp;do=diff</link>
        <description>Question 13 Exercise 7.1

Solutions of Question 13 of Exercise 7.1 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$2^n&gt;n \forall n \in \mathbf{N}$$n=1$$2^n=2^1=2$$n=1$$2&gt;1$$n=1$$n=l&gt;I$$2^k&gt;k\cdots(i)$$n=k+1$\begin{align}
&amp; 2^{k+1}=2^k \cdot 2&gt;k \cdot 2 \quad \text { by (i) } \\
&amp; \Rightarrow 2^{k+1}&gt;2 k=k+k \\
&amp;\Rightarrow 2^{k+1}&gt;k+1 \text {. as } k&gt;1…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-1-p14?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 14 Exercise 7.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-1-p14?rev=1737476038&amp;do=diff</link>
        <description>Question 14 Exercise 7.1

Solutions of Question 14 of Exercise 7.1 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$5$$3^{2 n-1}+2^{2 n-1}$$n$$n=1$$$3^{2 n-1}+2^{2 n-1}=3^{2.1-1}+2^{2.1-1}=5 \text {. }$$$5$$5$$5$$5.$$n=1$$n=k&gt;1$$54$$3^{2 k} 1+2^{2 k} \quad 1$$$3^{2 k-1}+2^{2 k-1}=5 Q$$$Q$$n=k+1$\begin{align}
3^{2(k+1)-1}+2^{2(k+1)-1} &amp; =3^{2 k+2-1}+2^{2…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-2-p1?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Exercise 7.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-2-p1?rev=1737476038&amp;do=diff</link>
        <description>Question 1 Exercise 7.2

Solutions of Question 1 of Exercise 7.2 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$(x^2-\dfrac{1}{y})^4$\begin{align}(x^2-\dfrac{1}{y})^4&amp;=(x^2)^4+{ }^4 C_1(x^2)^3(-\dfrac{1}{y})+ \\
&amp; { }^4 C_2(x^2)^2(-\dfrac{1}{y})^2+{ }^4 C_3(x^2)(-\dfrac{1}{y})^3 + { }^4 C_4(-\dfrac{1}{y})^4 \\
&amp; =x^8- \dfrac{4x^6}{y}+\dfrac{6x^4}{y^2}…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-2-p2?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 Exercise 7.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-2-p2?rev=1737476038&amp;do=diff</link>
        <description>Question 2 Exercise 7.2

Solutions of Question 2 of Exercise 7.2 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$4^{th}$$(2+a)^7$$\ln$$n=7$$a=2$$b=a$$$T_{r+1}=\frac{7 !}{(7-r) ! r !}(2)^{7-r } a^r $$$4^{\text {th }}$$r=3$\begin{align}
&amp; T_{3+1}=\dfrac{7 !}{(7-3) ! 3 !} 2^{7-3} a^3 \\
&amp; \Rightarrow T_4=\dfrac{7 !}{4 ! 3 !} \cdot 2^4 a^3 \\
&amp; \Rightarrow…</description>
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3 Exercise 7.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-2-p3?rev=1737476038&amp;do=diff</link>
        <description>Question 3 Exercise 7.2

Solutions of Question 3 of Exercise 7.2 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$x$$(\dfrac{4 x^2}{3}-\dfrac{3}{2 x})$$n=9, \quad a=\dfrac{4 x^2}{3}$$b=-\dfrac{3}{2 x}$$T_{r+1}$$x$$T_{r+1}$\begin{align}T_{r+1}&amp;=\dfrac{9 !}{(9-r) ! r !}(\dfrac{4 x^2}{3})^{9-r}(-\dfrac{3}{2 x})^r \\
&amp; =\dfrac{9 !}{(9-r) ! r !} \cdot \dfrac…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-2-p4?rev=1737476038&amp;do=diff">
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        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 4 Exercise 7.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-2-p4?rev=1737476038&amp;do=diff</link>
        <description>Question 4 Exercise 7.2

Solutions of Question 4 of Exercise 7.2 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$x^{23}$$(x^2-x)^{20}$$n=20, \quad a=x^2$$b=-x$$T_{r, 1}$$x^{23}$\begin{align}T_{r-1}&amp;=\dfrac{20 !}{(20-r) ! r !}(x^2)^{20 r}(-x)^r \\
&amp; =\dfrac{20 !}{(20-r) ! r !}(-1)^r \cdot x^{40-2 r+r} \\
&amp; =\dfrac{20 !}{(20-r) ! r !}(-1)^r x^{40-r}\end{…</description>
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        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5 Exercise 7.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-2-p5?rev=1737476038&amp;do=diff</link>
        <description>Question 5 Exercise 7.2

Solutions of Question 5 of Exercise 7.2 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$(\dfrac{a}{x}+b x)^8$$a=\dfrac{a}{x}$$b=b x$$n=8$$n-8$$8+1=9$$$(\dfrac{8+2}{2})^{t h}=5^{t h}$$T_{r+1}$$$T_{r+1}=\dfrac{8 !}{(8-r) ! r !}(\dfrac{a}{x})^{8-r}(b x)^r$$$T_5$$r=4$\begin{align}T_5&amp;=\dfrac{8 !}{(8-4) ! 4 !}(\dfrac{a}{x})^{8-4}(b …</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-2-p7?rev=1737476038&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2025-01-21T16:13:58+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7 Exercise 7.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-2-p7?rev=1737476038&amp;do=diff</link>
        <description>Question 7 Exercise 7.2

Solutions of Question 7 of Exercise 7.2 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$(2+\sqrt{3})^5+(2-\sqrt{3})^5$\begin{align}(2+\sqrt{3})^5+(2 \cdot \sqrt{3})^5&amp; =[(2)^5+{ }^5 C_1 \cdot 2^4 \cdot \sqrt{3}+{ }^5 C_2 \cdot 2^3 \cdot(\sqrt{3})^2 \\
&amp; +^5 C_3 \cdot 2^2 \cdot(\sqrt{3})^4+{ }^5 C_4 \cdot 2 \cdot(\sqrt{3})^4 \\
…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1 Exercise 7.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-3-p1?rev=1737476039&amp;do=diff</link>
        <description>Question 1 Exercise 7.3

Solutions of Question 1 of Exercise 7.3 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\frac{1}{2}$$$
\begin{aligned}
&amp; (1-x)^{\frac{1}{2}}=1+\frac{1}{2} x+ \\
&amp; \frac{\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2 !}(-x)^2
\end{aligned}
$$$$
\begin{aligned}
&amp; +\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2 Exercise 7.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-3-p2?rev=1737476039&amp;do=diff</link>
        <description>Question 2 Exercise 7.3

Solutions of Question 2 of Exercise 7.3 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\sqrt{26}$$$
\begin{aligned}
&amp; \sqrt{26}=\sqrt{25+1} \\
&amp; =\sqrt{25} \sqrt{1+\frac{1}{25}}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}}
\end{aligned}
$$$$
\begin{aligned}
&amp; \sqrt{26}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \\
&amp; =5\left[1+\f…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7 and 8 Exercise 7.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-3-p6?rev=1737476039&amp;do=diff</link>
        <description>Question 7 and 8 Exercise 7.3

Solutions of Question 7 and 8 of Exercise 7.3 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$x^4$$(1-x)^{\frac{1}{4}}+(1-x)^{\frac{1}{4}}=a-b x^2$$a$$b$$$
\begin{aligned}
&amp; (1+x)^{\frac{1}{4}}+(1-x)^{\frac{1}{4}} \\
&amp; =\left[1+\frac{x}{4}+\frac{\frac{1}{4}\left(\frac{1}{4}-1\right)}{2 !} x^2+\right. \\
&amp; \left.\frac{\fra…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 10 Exercise 7.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit07/ex7-3-p8?rev=1737476039&amp;do=diff</link>
        <description>Question 10 Exercise 7.3

Solutions of Question 10 of Exercise 7.3 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$1-\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\ldots$$(1+x)^n$$$
\begin{aligned}
&amp; 1+n x+\frac{n(n-1)}{2 !} x^2 \\
&amp; +\frac{n(n-1(n-2))}{3 !} x^3+\ldots
\end{aligned}
$$$n x=-\frac{1}{4}$$\frac{n(n-1)}{2 !} x^2=\frac{1.3}{2 !} \cdot …</description>
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        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Exercise 10.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p1?rev=1737476039&amp;do=diff</link>
        <description>Question 1, Exercise 10.1

Solutions of Question 1 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. There are four parts in Question 1.$\sin {{37}^{\circ }}\cos {{22}^{\circ }}+\cos {{37}^{\circ }}\sin {{22}^{\circ }}$\begin{align} \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta, \end{align}\begin{a…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2, Exercise 10.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p2?rev=1737476039&amp;do=diff</link>
        <description>Question 2, Exercise 10.1

Solutions of Question 2 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\sin \dfrac{\pi }{12}$$\dfrac{\pi }{12}$$\dfrac{\pi }{3}-\dfrac{\pi }{4}$\begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin.\end{align}\begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\f…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3, Exercise 10.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p3?rev=1737476039&amp;do=diff</link>
        <description>Question 3, Exercise 10.1

Solutions of Question 3 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\sin u=\dfrac{3}{5}$$\sin v=\dfrac{4}{5}$$u$$v$$0$$\dfrac{\pi }{2}$$\cos \left( u+v \right)$$\sin u=\dfrac{3}{5},$$0\le u\le \dfrac{\pi }{2}.$$\sin v=\dfrac{4}{5},$$0\le v\le \dfrac{\pi }{2}.$$\cos u=\pm \sqrt{1-{{\sin }^…</description>
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        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5, Exercise 10.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p5?rev=1737476039&amp;do=diff</link>
        <description>Question 5, Exercise 10.1

Solutions of Question 5 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\tan \alpha =\dfrac{3}{4}$$\sec \beta =\dfrac{13}{5}$$\alpha$$\beta$$\sin \left( \alpha +\beta  \right)$$\tan\alpha =\dfrac{3}{4}$$\tan\alpha$$\alpha$\begin{align}{{\sec}^{2}}\alpha &amp;=1+{{\tan}^{2}}\alpha\\
\Rightarrow \q…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6, Exercise 10.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p6?rev=1737476039&amp;do=diff</link>
        <description>Question 6, Exercise 10.1

Solutions of Question 1 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\cos \alpha =2{{\cos }^{2}}\dfrac{\alpha }{2}-1=1-2{{\sin }^{2}}\dfrac{\alpha }{2}$\begin{align}\cos \alpha &amp;=\cos 2\dfrac{\alpha }{2}\\
&amp;={{\cos }^{2}}\dfrac{\alpha }{2}-{{\sin }^{2}}\dfrac{\alpha }{2}\\ 
&amp;={{\cos }^{2}}…</description>
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        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7, Exercise 10.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p7?rev=1737476039&amp;do=diff</link>
        <description>Question 7, Exercise 10.1

Solutions of Question 1 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\cot \left( \alpha +\beta  \right)=\dfrac{\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }$\begin{align}L.H.S.&amp;=\cot (\alpha +\beta )\\
&amp;=\dfrac{1}{\tan (\alpha +\beta )}\\
&amp;=\dfrac{1}{\,\dfrac{\tan \alpha +\tan \beta…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 8, Exercise 10.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p8?rev=1737476039&amp;do=diff</link>
        <description>Question 8, Exercise 10.1

Solutions of Question 8 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\tan \left( \dfrac{\pi }{4}+\theta  \right)=\dfrac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$\begin{align}L.H.S.&amp;=\tan \left( \dfrac{\pi }{4}+\theta  \right)\\ 
&amp;=\dfrac{\sin \left( \dfrac{\pi }{4}+\theta  \ri…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p11?rev=1737476039&amp;do=diff">
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 13, Exercise 10.1</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p11?rev=1737476039&amp;do=diff</link>
        <description>Question 13, Exercise 10.1

Solutions of Question 13 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$r\,\,\sin \left( \theta +\phi  \right)$$\theta$$\phi$$4\sin \theta +3\cos \theta .$$4\sin \theta +3\cos \theta$$r\sin(\theta + \varphi)$$$4\sin \theta +3\cos \theta=r\cos\varphi\sin\theta+r\sin\varphi\cos\theta --- (1)$…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Exercise 10.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-2-p1?rev=1737476039&amp;do=diff</link>
        <description>Question 1, Exercise 10.2

Solutions of Question 1 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. There are four parts in Question 1.$\sin 2\theta ,\,\,\cos 2\theta$$\tan 2\theta$$\tan \theta =-\dfrac{1}{5}$$\theta$$\sin \theta =\dfrac{1}{\sqrt{26}}$$\cos \theta =\dfrac{-5}{\sqrt{26}}$\begin{align}\sin 2\theta &amp;=2\sin…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2, Exercise 10.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-2-p2?rev=1737476039&amp;do=diff</link>
        <description>Question 2, Exercise 10.2

Solutions of Question 2 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\sin \theta =\dfrac{5}{13}$$\theta $$\sin 2\theta $$\sin \theta =\dfrac{5}{13}$$$\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }.$$$\theta$$\cos$\begin{align}\cos\theta &amp;=-\sqrt{1-{{\sin }^{2}}\theta }\\
&amp;=-\sqrt{1-\left(\…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3, Exercise 10.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-2-p3?rev=1737476039&amp;do=diff</link>
        <description>Question 3, Exercise 10.2

Solutions of Question 3 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\sin \theta =\dfrac{4}{5}$$\theta$$\sin2\theta$$\sin \theta =\dfrac{4}{5}$$\theta$$\cos \theta =-\dfrac{3}{5}$\begin{align}\sin 2\theta &amp;=2\sin \theta \cos \theta \\
&amp;=2\left( \dfrac{4}{5} \right)\left( -\dfrac{3}{5} \rig…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 4 and 5, Exercise 10.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-2-p4?rev=1737476039&amp;do=diff</link>
        <description>Question 4 and 5, Exercise 10.2

Solutions of Question 4 and 5 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\cos \theta =-\dfrac{3}{7}$$\theta $$\sin \dfrac{\theta }{2}$$\cos \theta =-\dfrac{3}{7}$$\theta$\begin{align}&amp;\pi &lt; \theta &lt; \dfrac{3\pi}{2} \\
\implies &amp;\frac{\pi}{2} &lt; \frac{\theta}{2} &lt; \dfrac{3\pi}{4}\end…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-2-p5?rev=1737476039&amp;do=diff">
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 6, Exercise 10.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-2-p5?rev=1737476039&amp;do=diff</link>
        <description>Question 6, Exercise 10.2

Solutions of Question 6 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\cos {{15}^{\circ }}$${{15}^{\circ }}=\dfrac{{{30}^{\circ }}}{2}$$\dfrac{\theta }{2}=\dfrac{{{30}^{\circ }}}{2}$$\cos {{15}^{\circ }}$\begin{align}\cos {{15}^{\circ }}&amp;=\cos \dfrac{{{30}^{\circ }}}{2}=\sqrt{\dfrac{1+\cos …</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-2-p6?rev=1737476039&amp;do=diff">
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 7, Exercise 10.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-2-p6?rev=1737476039&amp;do=diff</link>
        <description>Question 7, Exercise 10.2

Solutions of Question 7 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.${{\cos }^{4}}\theta -{{\sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$\begin{align}L.H.S&amp;={{\cos }^{4}}\theta -{{\sin }^{4}}\theta \\ 
&amp;=\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta  \right)\left( {{\cos }^{2}}\theta +{{\…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-2-p7?rev=1737476039&amp;do=diff">
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 8 and 9, Exercise 10.2</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-2-p7?rev=1737476039&amp;do=diff</link>
        <description>Question 8 and 9, Exercise 10.2

Solutions of Question 8 and 9 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.${{\cos }^{4}}\theta $\begin{align}{{\cos}^{4}}\theta &amp;={{\left( {{\cos }^{2}}\theta  \right)}^{2}}\\
&amp;={{\left( \dfrac{1+\cos 2\theta }{2} \right)}^{2}}\\ 
&amp;=\dfrac{1+2\cos 2\theta +{{\cos }^{2}}2\theta }{4}\\…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-3-p1?rev=1737476039&amp;do=diff">
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Exercise 10.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-3-p1?rev=1737476039&amp;do=diff</link>
        <description>Question 1, Exercise 10.3

Solutions of Question 1 of Exercise 10.3 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. There are four parts in Question 1.$2\sin 6x\sin x$$$-2\sin \alpha \sin \beta =\cos (\alpha +\beta )-\cos (\alpha -\beta ).$$$\alpha =6x$$\beta =x$\begin{align}-\,2\sin 6x\sin x&amp;=\cos (6x+x)-\cos (6x-x)\\
&amp;=\cos 7x-\cos x…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-3-p2?rev=1737476039&amp;do=diff">
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 2, Exercise 10.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-3-p2?rev=1737476039&amp;do=diff</link>
        <description>Question 2, Exercise 10.3

Solutions of Question 2 of Exercise 10.3 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$$\sin {{37}^{\circ }}+\sin {{43}^{\circ }}.$$$$\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right).$$$\alpha ={{37}^{\circ }}$$\beta ={{43}^{\circ }}$\begin…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-3-p3?rev=1737476039&amp;do=diff">
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 3, Exercise 10.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-3-p3?rev=1737476039&amp;do=diff</link>
        <description>Question 3, Exercise 10.3

Solutions of Question 3 of Exercise 10.3 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$$\dfrac{\cos {{75}^{\circ }}+\cos {{15}^{\circ }}}{\sin {{75}^{\circ }}-\sin {{15}^{\circ }}}=\sqrt{3}.$$$$\cos \alpha +\cos \beta =2\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \righ…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-3-p4?rev=1737476039&amp;do=diff">
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5, Exercise 10.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-3-p4?rev=1737476039&amp;do=diff</link>
        <description>Question 5, Exercise 10.3

Solutions of Question 5 of Exercise 10.3 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}.$$$2\cos \alpha \cos \beta =\cos \left( \alpha +\beta  \right)+\cos \left( \alpha -\beta  \right)$\begin{align}L.H.S.&amp;=\cos {…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-3-p5?rev=1737476039&amp;do=diff">
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 5, Exercise 10.3</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/ex10-3-p5?rev=1737476039&amp;do=diff</link>
        <description>Question 5, Exercise 10.3

Solutions of Question 5 of Exercise 10.3 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}$$2\cos \alpha \cos \beta =\cos \left( \alpha +\beta  \right)+\cos \left( \alpha -\beta  \right)$\begin{align}L.H.S.&amp;=\cos {{20…</description>
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    <item rdf:about="https://beta.mathcity.org/math-11-kpk/sol/unit10/re-ex10-p1?rev=1737476039&amp;do=diff">
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 1, Review Exercise 10</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/re-ex10-p1?rev=1737476039&amp;do=diff</link>
        <description>Question 1, Review Exercise 10

Solutions of Question 1 of Review Exercise 10 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\cos {{50}^{\circ }}5{0}&#039;\cos {{9}^{\circ }}1{0}&#039;-\sin {{50}^{\circ }}5{0}&#039;\sin {{9}^{\circ }}1{0}&#039;=$$0$$\dfrac{1}{2}$$1$$\dfrac{\sqrt{3}}{2}$$\tan {{15}^{\circ }}=2-\sqrt{3}$${{\cot }^{2}}{{75}^{\circ }}$$7+\sq…</description>
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        <dc:date>2025-01-21T16:13:59+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>Question 8 &amp; 9, Review Exercise 10</title>
        <link>https://beta.mathcity.org/math-11-kpk/sol/unit10/re-ex10-p5?rev=1737476039&amp;do=diff</link>
        <description>Question 8 &amp; 9, Review Exercise 10

Solutions of Question 8 &amp; 9 of Review Exercise 10 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\sin \left( \dfrac{\pi }{4}-\theta  \right)\sin \left( \dfrac{\pi }{4}+\theta  \right)=\dfrac{1}{2}\cos 2\theta $$2\sin \alpha \sin \beta =\cos \left( \alpha -\beta  \right)-\cos \left( \alpha +\beta  \r…</description>
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