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- Question 9 Exercise 6.3 @math-11-kpk:sol:unit06
- 5,6$ or 7 men. If committee contains two men and then it will obviously contain $6$ women. so in this ... \\ &=21\end{align} If committee contains $3$ men then it will contain $5$ women, in this case the tota... \\\ &=210\end{align} If committee contain $4$ men then it will contain $4$ women, in this case total nu... \\ &=525\end{align} If committee contains $5$ men then it will contain $3$ women, in this case total num
- Question 7 Exercise 6.4 @math-11-kpk:sol:unit06
- s than $6$ Let $B=\{$ a number less than 6$\}$, then from sample space, we see that $n(B)=10$. Thus t... more than $7$ Let $C=\{$ a sum mure than 7$\}$, then from sample space, we see that $n(C)=5$. Thus th... than $10$ Let $D=\{$ a sum greater than 10$\}$, then from sample space, we get $n(D)=3$. Thus the pro... at least $10$ Let $E=\{a$ sum at least 10$\}$, then from sample space, we see that $n(E)=6$. Thus th
- Question 8, Exercise 1.2 @math-11-kpk:sol:unit01
- t( z \right)$. ====Solution==== Assume $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z+\overline{z}&... \right)$. ====Solution==== Assume that $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z-\overline{z}&... ight]}^{2}}$. ====Solution==== Suppose $z=a+ib$, then $\overline{z}=a-ib$. Then \begin{align}z\overline{z}&=\left( a+ib \right)\cdot \left( a-ib \right)\\ &
- Question 1, Exercise 3.2 @math-11-kpk:sol:unit03
- at{i}-5\hat{j}$ and $\vec{b}=-2\hat{i}+3\hat{j}$, then find $\vec{a}+2\vec{b}$. ====Solution==== \begi... at{i}-5\hat{j}$ and $\vec{b}=-2\hat{i}+3\hat{j}$, then find $3\vec{a}-2\vec{b}$. ====Solution==== \beg... at{i}-5\hat{j}$ and $\vec{b}=-2\hat{i}+3\hat{j}$, then find $2(\vec{a}-\vec{b})$. ====Solution==== Fir... at{i}-5\hat{j}$ and $\vec{b}=-2\hat{i}+3\hat{j}$, then find $|\vec{a}+\vec{b}|$. ====Solution==== We h
- Question 7 & 8 Exercise 4.3 @math-11-kpk:sol:unit04
- find the sum of $\mathrm{n}$ terms of each, and\\ then adding the $n$ terms sum of each one will give us... .\\ For $$1+7+13+\ldots$$ here $$a_1=1, d=7-1=6$$ then sum of the $n$ terms\\ \begin{align}S_n&=\dfrac{n... \\ Now for $3+9+15+\ldots$, with $a_1=3, d=9-3=6$ then sum of the $\mathrm{n}$ terms, let denote\\ \begi... n}S_n^{\prime}&=\dfrac{n}{2}[2 a_1+(n-1) d] \text{then}\\ \cdot S_n^{\prime}&=\dfrac{n}{2}[2.3+(n-1) 6]\
- Question 9 Exercise 6.2 @math-11-kpk:sol:unit06
- e six flags. If each signal consist of one color then total number of signals $=^6 P_1=6$. If each signal consist of two color then total number of signal $s=^6 P_2=30$. If each signal consist of three color then total number of signals $=^6 P_3=120$. If each signal consist of four color then total number of signals $=^6 P_4=360$. If each s
- Question 4 Exercise 6.4 @math-11-kpk:sol:unit06
- n}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When all heads. Let $$A=\{H H H\}$$ then $$n(A)=1$$ Hence the probability of obtaining all... }S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When two heads Let $$B=\{H T$ T.THT.TTH $\}$$ then $$n(B)=3$$ Hence the probability of gelling two h
- Question 12 Exercise 7.1 @math-11-kpk:sol:unit07
- n}-1}{24}$ is an integer. Solution: 1. For $n=1$, then $$\dfrac{5^{2 n}-1}{24}=\dfrac{5^{2.1}-1}{24}=\df... is true for $n=1$ 2. Let it be true for $n=k>1$ then $$\dfrac{5^{2 k}-1}{24} \in \mathbb{Z}$$ 3. For $n=k+1$ then consider \begin{align}\dfrac{5^{2(k+1)}-1}{24}&=\... 81}$ is an integer. ====Solution==== 1. For $n=1$ then \begin{align}\dfrac{10^{n+1}-9 n-10}{81}&=\dfrac{
- Question 13 Exercise 7.1 @math-11-kpk:sol:unit07
- n \in \mathbf{N}$. ====Solution==== 1. For $n=1$ then $2^n=2^1=2$ and $n=1$. Clearly $2>1$. hence the ... is true for $n=1$. 2. Let it be true for $n=l>I$ then $2^k>k\cdots(i)$ 3. For $n=k+1$ then we consider \begin{align} & 2^{k+1}=2^k \cdot 2>k \cdot 2 \quad \... integer $n \geq 4$ ====Solution==== 1. For $n=4$ then $n !=4 !=24$ and $n^2=4^2=16$. Clearly $24>16$,
- Question 14 Exercise 7.1 @math-11-kpk:sol:unit07
- y positive integer. ====Solution==== 1. For $n=1$ then $$3^{2 n-1}+2^{2 n-1}=3^{2.1-1}+2^{2.1-1}=5 \text... true for $n=1$ 2. Let it be true for $n=k>1$, then $54$ divides $3^{2 k} 1+2^{2 k} \quad 1$ which im... 1}=5 Q$$ where $Q$ is a quotient. 3. For $n=k+1$ then considering \begin{align} 3^{2(k+1)-1}+2^{2(k+1)-... ll natural numbers. ====Solution==== 1. For $n=1$ then $$2^{2 n}-1=2^{2.1}-1=4-1=3 $$ $3$ divides $3$, h
- Question 7 & 8 Review Exercise 7 @math-11-kpk:sol:unit07
- tion to prove the given statement. (1.) For $n=1$ then $7^k-3^k=7-4=4$. Thus 4 divides 4. Hence given is true for $n=1$. (2.) Let it be true for $n=k>1$ then $7^n-3^n=4 Q$ where $Q$ is the quotient in the induction hypothesis. (3.) For $n=k+1$ then we have $$ \begin{aligned} & 7^{k+1}-3^{k+1}=7.7^... this using mathernatical induction. 1. For $n=1$ then $$ (1+x)^1=1+x=1+1 x $$ Thus it is true for $n=1
- Question 12 & 13 Exercise 4.2 @math-11-kpk:sol:unit04
- $a_1$ represents salary of worker at first year. Then $$a_1=3500.$$ Increase in salary in each year $=d... $a=12, b=18$.\\ Let say $A$ be arithmetic means. Then \\ \begin{align}A&=\dfrac{a+b}{2}\\&=\dfrac{12+18... }, b=\dfrac{1}{4}$,\\ Let $A$ be arithmetic mean. Then\\ \begin{align}A&=\dfrac{a+b}{2}\\&=\dfrac{\dfrac... re $a=-6, b=-216$.\\ Let $A$ be arithmetic mean. Then\\ $$A=\dfrac{a+b}{2}=\dfrac{-6-216}{2}=-111$$ GOO
- Question 9, Exercise 1.2 @math-11-kpk:sol:unit01
- Pakistan. =====Question 9(i)===== If $z=3+2i,$ then verify that $-|z|\leq \operatorname{Re}\left( z \... ight)\leq |z|$ ====Solution==== Given $z=3+2i$. Then $|z|=\sqrt{9+4}=\sqrt{13}$ and ${\rm Re}z=3=\sqrt... nd{align} =====Question 9(ii)===== If $z=3+2i,$ then verify that $-|z|\leq \operatorname{Im}\left( z \... ight)\leq |z|$ ====Solution==== Given $z=3+2i$. Then $|z|=\sqrt{9+4}=\sqrt{13}$ and ${\rm Im}z=2=\sqrt
- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}.$$ Then \begin{align}|A|&=\begin{vmatrix}4 & -2 & 5 \\ 2... & 6 \\ 1 & 3 & 4 \\ -1 & 5 & 1 \end{bmatrix}.$$ Then \begin{align}|A|&=\left| \begin{matrix} 3 & -1 & ... 3 \\ 0 & -2 & 0 \\ -2 & -2 & 2 \end{bmatrix}.$$ Then \begin{align}|A|&=\left| \begin{matrix} 1 & 2 & -... -1 \\ 0 & -1 & 3 \\ 1 & 0 & 2 \end{bmatrix}.$$ Then \begin{align}|A|&=\left| \begin{matrix} 1 & 2 & -
- Question 2, Exercise 3.2 @math-11-kpk:sol:unit03
- {\scriptscriptstyle\rightharpoonup}{a}=3\hat{i}$$ Then $$|\overset{\scriptscriptstyle\rightharpoonup}{a}... criptstyle\rightharpoonup}{a}=3\hat{i}-4\hat{j}$$ Then $$|\overset{\scriptscriptstyle\rightharpoonup}{a}... \rightharpoonup}{a}=\hat{i}+\hat{j}-2\hat{k}$$ \\ Then \\ $$|\overset{\scriptscriptstyle\rightharpoonup}... frac{\sqrt{3}}{2}\hat{i}-\dfrac{1}{2}\hat{j}$$ \\ Then \\ $$|\overset{\scriptscriptstyle\rightharpoonup}