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- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- 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 ... 1 & 2 \\ 1 & 0 & 4 \end{matrix} \right. \right]\text{ by }R_3+R_1\\ \underset{\sim}{R}&\left[\begin{ma... -1 & 2 \\ 1 & 0 & 4 \end{matrix} \right. \right]\text{by}-R_2+R_3\\ \underset{\sim}{R}&\left[\begin{ma
- Question 1, Exercise 2.3 @math-11-kpk:sol:unit02
- -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 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_3-R_2 \\ \underset{\sim}{R}&\begin{bmatri... 1 & 9 \\ 3 & \quad 1 & 3 & \quad 2 \end{bmatrix}\text{ by } R_1\leftrightarrow R_2\\ \underset{\sim}{R}
- Question 9 Exercise 6.5 @math-11-kpk:sol:unit06
- ll be selected. ====Solution==== \begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7} \\ \Rightarrow P(\text { Ajmal not selected })&=\dfrac{6}{7} \\ P(\text { Bushra selection })&=\dfrac{1}{5} \\ \Rightarrow P(\text { Bushra not selected })&=\dfrac{4}{5}\end{align}
- Question 9 & 10 Exercise 4.3 @math-11-kpk:sol:unit04
- ots, 693$$.\\ Here, $a=306$,\\ $$d=(315-306) = 9 \text { and } a_n=693 .$$\\ Let the number of terms be $n$. Then\\ \begin{align}a_n&=a_1+(n-1) d \text { becomes } \\ \Rightarrow a_1+(n-1) d&=693 \\ \R... 1000$, \\ therefore we have $n=4$\\ \begin{align}\text{Let the first person receives}&=R s . a\\ \text{Then the seçond receives}&=\operatorname{Rs}(a-20)\\ \
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- $ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} &=\left(\begin... L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{Number of permulations are} &=\left(\begin{array}... L$ $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} & =\left(\begin... and $m_3=1$ are $C$. $\therefore$ \begin{align}\text{Number of permutations are}& =\left(\begin{array}
- Question 5 Exercise 3.4 @math-11-kpk:sol:unit03
- ea of the parallelogram, that is:\\ \begin{align}\text{Area of triangle}&=\dfrac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}| \\ \text { Since } \overrightarrow{P Q}&=(3,2)-(-2 ,-3) \\... } \times \overrightarrow{P R}|&=30 \\ \therefore \text { Area of triangle }& =\dfrac{1}{2}|\overrightarr... times \overrightarrow{P R}|\\ &=\dfrac{30}{2}=15 \text { units square. }\end{align} =====Queswtion 5(ii
- Question 2 & 3 Exercise 4.4 @math-11-kpk:sol:unit04
- he sequence. ====Solution==== Here $$a_3=27 \quad\text{and}\quad a_5=243$$ and we know\\ \begin{align}a_... _1 r^2}&=\dfrac{243}{27}=9 \\ \Rightarrow r^2&=9 \text { or } r= \pm 3 .\end{align} Putting this in (i), then\\ $$a_1(9)=27 \quad \text{then} \quad a_1=3$$ $$\text{Hence}\quad a_1=3, r= \pm 3$$ =====Question 3===== Find the seventh term o
- Question 2 and 3 Exercise 3.3 @math-11-kpk:sol:unit03
- gle hetween $\vec{a}$ and $\vec{b}$ \begin{align}\text { then } \cos \theta&=\dfrac{\vec{a} \cdot \vec{b... gle between $\vec{a}$ and $\vec{b}$ \begin{align}\text { then } \quad \cos \theta &=\dfrac{\vec{a} \cdot... \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=8 \\ \text { Also }|\vec{a}|&=\sqrt{(3)^2+(4)^2}=\sqrt{9+19} \\ \Rightarrow \quad|\vec{a}|&=\sqrt{25}=5\\ \text { and }|\vec{b}|&=\sqrt{(2)^2+(-5)^2}\\ \Rightarr
- Question 5 & 6 Exercise 4.3 @math-11-kpk:sol:unit04
- \\ \Rightarrow 20 d^2&=20 \\ \Rightarrow d^2&=1 \text { or } d= \pm 1\end{align} When $a=5$ and $d=1$ t... align} a-3d&=5-3=2, \\ a-d&=5-1=4, \\ a+d&=5+1=6 \text { and } \\ a+3 d&=5+3=8.\end{align} When $a=5$ an... =5-3(-1)=8, \\ a-d&=5-(-1)=6, \\ a+d&=5+(-1)=, 4 \text { and } \\ a+3 d&=5-3=2\\ & 2,4,6,8; 8,6,4,2\end{... are in A.P. If\\ \begin{align}x_1+x_7+x_{10}&=-6 \text { and } \\ x_3+x_8+x_{12}&=-11, \text { find } \\
- Question 5 & 6 Exercise 4.5 @math-11-kpk:sol:unit04
- then\\ $$S_{10}=\dfrac{a_1(r^{10}-1)}{r-1} \quad \text{and}\quad S_5=\dfrac{a_1(r^5-1)}{r-1}$$\\ Putting... \\ \Rightarrow(r^5-1)(r^5-243)&=0 \\ \Rightarrow \text { Either } r^5-10&=0 \text { or } r^5-243=0 \\ \Rightarrow r^5&=1 \text { or } r^5=3^5 \\ \Rightarrow r&=1 \text { or } r=3 . \\ \
- Question 5, Exercise 2.2 @math-11-kpk:sol:unit02
- & 2-3b+3b & 3-3c+3c \\ 4 & 5 & 6 \end{vmatrix} \text{ by } R_2-3R_1 \\ &=\begin{vmatrix} a & b & c \\... a & b & c\\ a+b+c & b+c+a & c+a+b \end{vmatrix} \text{ by } R_3+R_2 \\ &=(a+b+c)\begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ 1 & 1 & 1 \end{vmatrix} \text{taking common from }R_3 \\ &=(a+b+c)0 \text{ by } R_1\simeq R_3 \\ &=0=R.H.S.\end{align} =====Questi
- Question 4 Exercise 4.5 @math-11-kpk:sol:unit04
- verline{8}&=0.8+0.08+0.008 \div 0.0008+ \ldots\\ \text { or } 0 . \overline{8}&=0.8+(0.1)(0.8) +(0.1)^2(... \overline{63}&=1+0.63+0.0063+0.000063 +\ldots \\ \text { or } 1 . \overline{63}&=1+[0.63+ (0.01)(0.63)-(0.01)^2 0.63+\ldots \ldots \text { (i) }\end{align} The serics in braces is infini... =\dfrac{7}{11} \ldots \ldots \ldots \ldots . . . \text { (ii) }\end{align} Putting (ii) in (i), we get\\
- Question 3, Exercise 2.3 @math-11-kpk:sol:unit02
- & -2 \\ 0 & 2 & 5 \\ 0 & 2 & 1 \end{bmatrix} \text{ by }R_2-2R_1 \text{ and } R_1-2R_3\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & -2 \\ 0 & 0 & 4 \\ 0 & 2 & 1 \end{bmatrix}\text{ by }R_2-R_3\end{align} The last matrix is the ec... & -2 \\ 0 & 2 & 1 \\ 3 & 1 & -4 \end{bmatrix} \text{ by }R_1\leftrightarrow R_3\\ \underset{\sim}{R}&
- Question 1 Exercise 4.5 @math-11-kpk:sol:unit04
- =a_1 r^{n-1}$$,\\ \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_1(r^n-1)}{r-1},\end{align} becomes in the given case\... es. We know that\\ \begin{align}a_n&=a_1 r^{n-1} \text {, }\\ \therefore \dfrac{1}{16}&=8(\dfrac{1}{2})^
- Question 7 & 8 Exercise 4.5 @math-11-kpk:sol:unit04
- 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... htarrow(3 r-2)(2 r-3)&=0 \\ \Rightarrow 3 r-2&=0 \text { or } 2 r-3=0 \\ \Rightarrow r=\dfrac{2}{3} \text { or } r=\dfrac{3}{2}\end{align} When $a=12$ and $r=