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- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... Question 2(i)===== Find the inverse of the matrix by using elementary row operation. $$\begin{bmatrix}... \\ 0 & 0 & 1 \end{matrix} \right. \right]\text{ by }R_1+3R_3\text{ and }R_2+2R_3 \\ \underset{\sim}{... \\ 1 & 0 & 4 \end{matrix} \right. \right]\text{ by }R_3+R_1\\ \underset{\sim}{R}&\left[\begin{matrix
- Question 1, Exercise 2.3 @math-11-kpk:sol:unit02
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... \ 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_2-2R_1 \text{ and } R_3-3R_1 \\ \underset{\si... 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_3-R_2 \\ \underset{\sim}{R}&\begin{bmatrix} 1... \\ 3 & \quad 1 & 3 & \quad 2 \end{bmatrix}\text{ by } R_1\leftrightarrow R_2\\ \underset{\sim}{R}& \b
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... & c\end{bmatrix}\\ &=\begin{bmatrix}ax+hy+gz & hx+by+fz & gx+fy+cz \end{bmatrix} \\ &=\begin{bmatrix}ax+hy+gz & hx+by+fz & gx+fy+cz \end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\\ &=\left[ x( ax+hy+gz )+y( hx+by+fz )+z( gx+fy+cz ) \right] \\ &=\left[ a{{x}^{2}}
- Question 12 Exercise 7.1 @math-11-kpk:sol:unit07
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... eshawar, Pakistan. =====Question 12(i)===== Show by mathematical induction that $\dfrac{5^{2 n}-1}{24... thbf{N}$, $\dfrac{5^{2 k}-1}{24} \in \mathbb{Z}$ by (a). Thus the given statement is true for $n=k+1$. Hence by mathematical induction it is true for all $n \in
- Question 13 Exercise 7.1 @math-11-kpk:sol:unit07
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... gn} & 2^{k+1}=2^k \cdot 2>k \cdot 2 \quad \text { by (i) } \\ & \Rightarrow 2^{k+1}>2 k=k+k \\ &\Right... the form of proposition taken when $n$ is replace by $k+1$, hence true for $n=k+1$. Thus by mathematical induction it is true for all $n \in \mathbf{N}$
- Question 6, Exercise 10.2 @math-11-kpk:sol:unit10
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... {\circ }}}{2}$, we can find $\cos {{15}^{\circ }}$by using half angle identity as, \begin{align}\cos {... c }}}{2}$, so we can find $\tan {{67.5}^{\circ }}$by using half angle identity as, \begin{align}\tan {... rc }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half angle identity as, \begin{align} sin{
- Question 5, Exercise 2.2 @math-11-kpk:sol:unit02
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... b+3b & 3-3c+3c \\ 4 & 5 & 6 \end{vmatrix} \text{ by } R_2-3R_1 \\ &=\begin{vmatrix} a & b & c \\1 & ... & c\\ a+b+c & b+c+a & c+a+b \end{vmatrix} \text{ by } R_3+R_2 \\ &=(a+b+c)\begin{vmatrix} 1 & 1 & 1\\... text{taking common from }R_3 \\ &=(a+b+c)0 \text{ by } R_1\simeq R_3 \\ &=0=R.H.S.\end{align} =====
- Question 3, Exercise 2.3 @math-11-kpk:sol:unit02
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... \\ 0 & 2 & 5 \\ 0 & 2 & 1 \end{bmatrix} \text{ by }R_2-2R_1 \text{ and } R_1-2R_3\\ \underset{\sim}... 2 \\ 0 & 0 & 4 \\ 0 & 2 & 1 \end{bmatrix}\text{ by }R_2-R_3\end{align} The last matrix is the echelo... \\ 0 & 2 & 1 \\ 3 & 1 & -4 \end{bmatrix} \text{ by }R_1\leftrightarrow R_3\\ \underset{\sim}{R}&\beg
- Question 1 and 2 Exercise 6.5 @math-11-kpk:sol:unit06
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... $.\\ Find $P(A \cap B)$. ====Solution==== We know by addition law of probability \begin{align} P(A \c... =\dfrac{5}{8}, P(A \cup B)=\dfrac{3}{4}$$ We know by complementary events $$P(B)=1-P(\bar{B})$$ Puttin... r{B})&=1-\dfrac{1}{2}=\dfrac{1}{2}\end{align} Now by addition law of probability, we know that: \begi
- Question 7 & 8 Review Exercise 7 @math-11-kpk:sol:unit07
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... tive intege: n. prove that $7^n-3^n$ is divisible by 4 . Solution: We using mathematical induction to ... rrow 7^{k+1}-3^{k+1}=4.7^k+3.4 Q \end{aligned} $$ by induction hypothesis $$ \begin{aligned} & \Righta... \Rightarrow 7^{k-1}-3^{k+1} \text { is divisible by } 4 . \end{aligned} $$ Thus the given statement
- Question 2, Exercise 1.3 @math-11-kpk:sol:unit01
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... == Given: $$p\left( z \right)={{z}^{3}}+6z+20$$ By factor theorem, $(z-a)$ is a factor of $P(z)$ iff... } Thus $z+2$ is a factor of ${{z}^{3}}+6z+20$.\\ By using synthetic division, we have $$\begin{array}... $$P\left( z \right)={{z}^{3}}-2{{z}^{2}}+z-2$$ \\ By factor theorem $(z-a)$ is a factor of $()$ iff $P
- Question 4, Exercise 2.3 @math-11-kpk:sol:unit02
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix} \text{by}R_2-R_1\\ \underset{\sim}{R}&\begin{bmatrix} 1 & ... & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix} \text{by}R_1\leftrightarrow R_2\\ \underset{\sim}{R}&\begi... & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \end{bmatrix} \text{by}R_2-2R_1\text{,}R_3-4R_1 \text{and} R_4-9R_1\\ \u
- Question 3 & 4 Exercise 4.3 @math-11-kpk:sol:unit04
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... tion 3===== Find sum of all the numbers divisible by $5$ from $25$ to $350$. GOOD ====Solution==== The numbers divisible by $5$ from $25$ tò $350$ are\\ $$25,30,35, \ldots, ... pose the three numbers are $a-d, a, a+d$\\. then by first condition their sum is equal to $36$\\ \beg
- Question 1 Exercise 5.3 @math-11-kpk:sol:unit05
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... c{A}{n}+\dfrac{B}{(n+1)}$$ Multiplying both sides by $n(n+1)$ $$1=A(n+1)+B n=(A+B) n+A$$ Comparing the... n-1)}+\dfrac{B}{(2 n+1)}$$ Multiplying both sides by $(2 n-1)(2 n+1)$. we get, \begin{align} & \mathrm... {3 n-1}-\dfrac{B}{3 n+2}$$ Multiplying both sides by $(3 n-1)(3 n+2)$ we get, \begin{align} 1&=A(3 n+2
- Question 7 and 8 Exercise 6.2 @math-11-kpk:sol:unit06
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... git, ten digit and unit digit place) to be filled by five digits, Moreover repetition is allowed. H... t m_2=5$ ways $E_3$ occurs in $m_3=5$ ways Thus by fundamental principle of counting the total numbe... $m_2=4$ ways $E_3$ occurs in $m_3=3$ ways. Thus by fundamental principle of 'counting the total numb