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- Question 6, Exercise 1.3 @math-11-kpk:sol:unit01
- $ $$(z^2+z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{-1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{-1\p... $ $$(z^2-z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{1\pm ... }i}{2}$$ The value of $z$ from both equations, we have $$z=\pm \dfrac{1}{2}\pm \dfrac{\sqrt{3}}{2}i$$ ... 2-2z+4=0$$ According to the quadratic formula, we have $a=1$, $b=-2$ and $c=4$ Thus, we have \begin{alig
- Question 2 Exercise 4.3 @math-11-kpk:sol:unit04
- on==== Given: $a_1=-40$ and $S_{21}=210$.\\ So we have $n=21$ and we have to find $a_{21}$ and $d$. As \begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \impl... ===Solution==== Given: $a_1=-7, d=8, S_n=225$, we have to find $n$ and $a_n$. We know that $$S_n=\dfrac{n}{2}[2 a_1+(n-1) d].$$ Thus, we have \begin{align} & 225=\dfrac{n}{2}[2 \cdot(-7)+(n-1
- Question 5, Exercise 1.3 @math-11-kpk:sol:unit01
- }+z+3=0.$$ According to the quadratic formula, we have\\ $a=1$, $b=1$ and $c=3$.\\ Thus we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\... }}-z-1=0$$ According to the quadratic formula, we have\\ $a=1$, $b=-1$ and $c=-1$\\ So we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\d
- Question 9 Exercise 3.4 @math-11-kpk:sol:unit03
- }+2 \hat{k}\end{align} From $\triangle A E B$, we have\\ \begin{align}\vec{c}&=\overrightarrow{A E}+\ove... s \ldots(1)\end{align} From $\triangle A E D$. we have\\ \begin{align}\vec{d}&=\overrightarrow{A E}+\ove... {. }....(2) \end{align} By using (1) and (2), we have,\\ \begin{align}\vec{c} \times \vec{d}&=\left| \b... at{k}\end{align} Which from $\triangle A E B$, we have\\ \begin{align}\vec{c} & =\overrightarrow{A E}+\v
- Question 2, Exercise 10.3 @math-11-kpk:sol:unit10
- c }}+\sin {{43}^{\circ }}.$$ ====Solution==== We have an identity: $$\sin \alpha +\sin \beta =2\sin \le... ht)\end{align} Since $cos(-\theta)=cos\theta$, we have $$\sin {{37}^{\circ }}+\sin {{43}^{\circ }}=2\sin... rc }}-\cos {{82}^{\circ }}$. ====Solution==== We have an identity: $$\cos \alpha -\cos \beta =-2\sin \l... &=-2\sin(59^\circ)\sin(-23^\circ) \end{align} We have $\sin(-\theta)=-\sin\theta$, therefore $$\cos {{3
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- and $C=\begin{bmatrix}x\\y\\z\end{bmatrix}$.\\ We have to prove that $$(AB)C=A(BC)$$ First, we take \beg... right] \ldots (2)\end{align} From (1) and (2), we have $$(AB)C=A(BC).$$ =====Question 3(ii)(a)===== If... \right] ... (2) \end{align} From (1) and (2), we have $$A(B+C)=AB+AC.$$ =====Question 3(ii)(b)===== I... x}\right]... (2) \end{align} From (1) and (2), we have $$A(B-C)=AB-BC.$$ ====Go To==== <text align="l
- Question 1, Exercise 3.2 @math-11-kpk:sol:unit03
- $2(\vec{a}-\vec{b})$. ====Solution==== First we have, \begin{align}\vec{a}-\vec{b}&=3\hat{i}-5\hat{j}-... \hat{j}\end{align} Multiply both sides by $2$. We have, $$2(\vec{a}-\vec{b})=10\hat{i}-16\hat{j}$$ ===... n find $|\vec{a}+\vec{b}|$. ====Solution==== We have, \begin{align}\vec{a}+\vec{b}&=3\hat{i}-5\hat{j}+... at{j}\end{align} Taking modulus of both sides. We have, $$|\vec{a}+\vec{b}|=\sqrt{(1)^2+(-2)^2}=\sqrt{5}
- Question 3 & 4, Exercise 3.2 @math-11-kpk:sol:unit03
- ing the coeffients of $\hat{i}$ and $\hat{j}$, we have, $$p+5q=1…(i)$$ $$2p-q=-9 …(ii)$$ Multiply $2$ by (i) and subtract (ii) from (i). We have \[\begin{array}{ccc} 2p&+10q&=2 \\ \mathop+\lim... \implies q=1$$ \\ Put the value of $q$ in (i). We have, $$p+5(1)=1 \quad \implies p=-4$$ Hence we have $p=-4$ and $q=1$. =====Question 4===== If $\vec{p}=2\h
- Question 12, 13 & 14, Exercise 3.2 @math-11-kpk:sol:unit03
- 2}&=3.\end{align} Taking square on both sides, we have, \begin{align}&{\alpha ^2+(\alpha +1)^2}+4=9\\ \i... 13}} By head to tail rule of vectors addition, we have\\ \begin{align}\vec{u}+\vec{v}&=\vec{w}\\ (2\hat... By comparison $\hat{i},\hat{j}$ and $\hat{k}.$ we have,\\ $$3=-z$$ $$-z=3$$ $$\Rightarrow \,\,\,z=-3$$ ... ghtarrow{OD}=-\hat{i}-2\hat{j}+\hat{k}$. Now, we have \begin{align}\overrightarrow{AB}&=\overrightarrow
- Question 12 & 13, Exercise 3.3 @math-11-kpk:sol:unit03
- ering a triangle inside a semicircle as shown. We have to show $\overrightarrow{B A} \cdot \overrightarr... opposile in direction. From $\triangle A B O$, we have \begin{align}\overrightarrow{O B}+\overrightarrow... ...(1)\end{align} Also from $\triangle A C O$, we have \begin{align}\overrightarrow{O A}+\overrightarrow... $ intersect at point $O$ as shown in figure.\\ We have to show that side bisector of $A B$ also passes t
- Question 15 Exercise 4.2 @math-11-kpk:sol:unit04
- $, then $$ A=\dfrac{a+b}{2}. --- (1) $$ Also, we have given $$ A=\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}. --- (2) $$ Comparing (1) and (2), we have \begin{align}&\dfrac{a+b}{2}=\dfrac{a^{n+1}+b^{n+... ^n(a-b)=a^n(a-b)\end{align} If $a\neq b$, then we have \begin{align} &b^n =a^n \\ \implies &\dfrac{b^... lies &n=0.\end{align} If $a=b$, then from (3), we have \begin{align}&\dfrac{a+a}{2}=\dfrac{a^{n+1}+a^{n+
- Question 1 Exercise 4.4 @math-11-kpk:sol:unit04
- 1 r^3, a_1 r^4, \ldots$, so for $a_1=5 ; r=3$, we have \begin{align}&5,5.3,5.3^2, 5.3^3, 5.3^4, \ldots\\... \ldots$,\\ so for $$a_1=8 ; r=-\dfrac{1}{2}$$ we have\\ \begin{align}&8,8(-\dfrac{1}{2}), 8(-\dfrac{1}{... for $$a_1=-\dfrac{9}{16} ; r=-\dfrac{2}{3}$$ we have\\ \begin{align}&-\dfrac{9}{16} ,-\dfrac{9}{16}(-\... \\ so for $$a_1=\dfrac{x}{y} r=-\dfrac{y}{x}$$ we have,\\ \begin{align}&\dfrac{x}{y}, \dfrac{x}{y} \cdot
- Question 3 Exercise 6.4 @math-11-kpk:sol:unit06
- that $8$ answers are correct. ====Solution==== We have $8$ questions, each question has two options. Th... that $7$ answers are correct. ====Solution==== We have $8$ questions, each question has two options. Th... that $6$ answers are correct. ====Solution==== We have $8$ questions, each question has two options. Th... east $6$ answers are correct. ====Solution==== We have $8$ questions, each question has two options. Th
- Question 10 Exercise 7.3 @math-11-kpk:sol:unit07
- s \end{aligned} $$ Comparing both the series, we have $n x=-\frac{1}{4}$ (I) and $\frac{n(n-1)}{2 !} x^... \cdot \frac{1}{2^4}$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq.(3),... ts \end{aligned} $$ Comparing both the series, we have $$ \begin{aligned} & n x=\frac{5}{8} \\ & \frac{n... } . \end{aligned} $$ Taking square of Eq.(1), we have $$ n^2 x^2=\frac{25}{64} $$ Dividing Eq.(2) by E
- Question 13, Exercise 10.1 @math-11-kpk:sol:unit10
- rphi =\dfrac{3}{5}.\end{align} Thus, from (1), we have \begin{align}&4\sin\theta +3\cos\theta \\ &=5 \le... phi =\dfrac{8}{17}.\end{align} Thus, from (1), we have \begin{align}&15\sin\theta +8\cos\theta \\ &=17 \... rac{-5}{\sqrt{29}}.\end{align} Thus, from (1), we have \begin{align}&2\sin\theta -5\cos\theta \\ &=\sqrt... dfrac{1}{\sqrt{2}}.\end{align} Thus, from (1), we have \begin{align}&\sin\theta+\cos\theta \\ &=\sqrt{2}