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- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- 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. ===... 1 & 2 \\ 0 & 0 & 1 \end{matrix} \right. \right]\text{ by }R_1+3R_3\text{ and }R_2+2R_3 \\ \underset{\sim}{R}&\left[\begin{matrix} 1 & 4 & 14 \\ 0 & 5 & 6
- Question 1, Exercise 2.3 @math-11-kpk:sol:unit02
- 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. ==... -1 \\ 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_2-2R_1 \text{ and } R_3-3R_1 \\ \underset{\sim}{R}&\begin{bmatrix} 1 & 3 & -1 \\ 0 & -5 & 6 \
- Question 9 Exercise 6.5 @math-11-kpk:sol:unit06
- on, 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. ==... ll be selected. ====Solution==== \begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7} \\ \Rightarrow P(\text { Ajmal not selected })&=\dfrac{6}{7} \\ P(\text
- Question 9 & 10 Exercise 4.3 @math-11-kpk:sol:unit04
- f 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. ==... ots, 693$$.\\ Here, $a=306$,\\ $$d=(315-306) = 9 \text { and } a_n=693 .$$\\ Let the number of terms be $n$. Then\\ \begin{align}a_n&=a_1+(n-1) d \text { becomes } \\ \Rightarrow a_1+(n-1) d&=693 \\ \R
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- on, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... $ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} &=\left(\begin... L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{Number of permulations are} &=\left(\begin{array}
- Question 5 Exercise 3.4 @math-11-kpk:sol:unit03
- ercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... ea of the parallelogram, that is:\\ \begin{align}\text{Area of triangle}&=\dfrac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}| \\ \text { Since } \overrightarrow{P Q}&=(3,2)-(-2 ,-3) \\
- Question 2 & 3 Exercise 4.4 @math-11-kpk:sol:unit04
- f Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... he sequence. ====Solution==== Here $$a_3=27 \quad\text{and}\quad a_5=243$$ and we know\\ \begin{align}a_... _1 r^2}&=\dfrac{243}{27}=9 \\ \Rightarrow r^2&=9 \text { or } r= \pm 3 .\end{align} Putting this in (i),
- Question 2 and 3 Exercise 3.3 @math-11-kpk:sol:unit03
- ercise 3.3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... gle hetween $\vec{a}$ and $\vec{b}$ \begin{align}\text { then } \cos \theta&=\dfrac{\vec{a} \cdot \vec{b... gle between $\vec{a}$ and $\vec{b}$ \begin{align}\text { then } \quad \cos \theta &=\dfrac{\vec{a} \cdot
- Question 5 & 6 Exercise 4.3 @math-11-kpk:sol:unit04
- f Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... \\ \Rightarrow 20 d^2&=20 \\ \Rightarrow d^2&=1 \text { or } d= \pm 1\end{align} When $a=5$ and $d=1$ t... align} a-3d&=5-3=2, \\ a-d&=5-1=4, \\ a+d&=5+1=6 \text { and } \\ a+3 d&=5+3=8.\end{align} When $a=5$ an
- Question 5 & 6 Exercise 4.5 @math-11-kpk:sol:unit04
- f Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... then\\ $$S_{10}=\dfrac{a_1(r^{10}-1)}{r-1} \quad \text{and}\quad S_5=\dfrac{a_1(r^5-1)}{r-1}$$\\ Putting... \\ \Rightarrow(r^5-1)(r^5-243)&=0 \\ \Rightarrow \text { Either } r^5-10&=0 \text { or } r^5-243=0 \\ \R
- Question 5, Exercise 2.2 @math-11-kpk:sol:unit02
- 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... & 2-3b+3b & 3-3c+3c \\ 4 & 5 & 6 \end{vmatrix} \text{ by } R_2-3R_1 \\ &=\begin{vmatrix} a & b & c \\... a & b & c\\ a+b+c & b+c+a & c+a+b \end{vmatrix} \text{ by } R_3+R_2 \\ &=(a+b+c)\begin{vmatrix} 1 & 1 &
- Question 4 Exercise 4.5 @math-11-kpk:sol:unit04
- f Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... verline{8}&=0.8+0.08+0.008 \div 0.0008+ \ldots\\ \text { or } 0 . \overline{8}&=0.8+(0.1)(0.8) +(0.1)^2(... \overline{63}&=1+0.63+0.0063+0.000063 +\ldots \\ \text { or } 1 . \overline{63}&=1+[0.63+ (0.01)(0.63)-(
- Question 3, Exercise 2.3 @math-11-kpk:sol:unit02
- 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... & -2 \\ 0 & 2 & 5 \\ 0 & 2 & 1 \end{bmatrix} \text{ by }R_2-2R_1 \text{ and } R_1-2R_3\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & -2 \\ 0 & 0 & 4 \\ 0
- Question 1 Exercise 4.5 @math-11-kpk:sol:unit04
- f Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... =a_1 r^{n-1}$$,\\ \begin{align}3.2^9&=3(2)^{n-1} \text { or }(2)^{n-1}=\dfrac{3.2^9}{3} \\ \Rightarrow(2)^{n-1}&=2^9 \\ \Rightarrow n-1&=9 \text { or } n=10 \\ \text {. Now }\quad S_n&=\dfrac{a
- Question 7 & 8 Exercise 4.5 @math-11-kpk:sol:unit04
- f 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. ==... ightarrow a^3&=1728\\ \Rightarrow \quad a&=12,\\ \text{putting}\text{in} (1)\\ \dfrac{12}{r}+12+12 r&=38\\ \Rightarrow \dfrac{1}{r}+1+r&=\dfrac{38}{12}=\dfrac