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- Question 13, Exercise 10.1 @math-11-kpk:sol:unit10
- &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4^2+3^2\\ \implies &r^2\left(\cos^2\varphi+\sin^2\varphi\right) = 16+9\\ \implies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implies 5\cos\varphi = 4 \implies \cos\varphi =\dfrac{4}{
- Question 6 Exercise 4.1 @math-11-kpk:sol:unit04
- 0$ \begin{align}&P_{0+1}=\dfrac{5-0}{0+1} P_0\\ \implies &P_1=5.\end{align} For $r=1$ \begin{align}&P_{1+1}=\dfrac{5-1}{1+1} P_1\\ \implies &P_2=2\cdot 5=10.\end{align} For $r=2$ \begin{align}&P_{2+1}=\dfrac{5-2}{2+1} P_2\\ \implies &P_3=1\cdot 10=10\end{align} For $r=3$ \begin{align}&P_{3+1}=\dfrac{5-3}{3+1} P_3\\ \implies &P_4=\dfrac{10}{2}=5.\end{align} For $r=4$ \begin
- Question 2 Exercise 4.3 @math-11-kpk:sol:unit04
- begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \implies &210=\dfrac{21}{2}(-40+a_{21}) \\ \implies &-40+a_{21}=\dfrac{2 \times 210}{21}=20 \\ \implies &a_{21}=20+40=60.\end{align} Also $a_{21}=a_1+20 d$, then\\ \begin{align}&20d=60-(-40)=100 \\ \implies &d=\dfrac{100}{20}=5 \end{align} Hence $a_{21}=60
- Question11 and 12, Exercise 10.1 @math-11-kpk:sol:unit10
- \begin{align}&\alpha+\beta +\gamma =180^\circ\\ \implies &\alpha +\beta =180^\circ-\gamma \\ \implies &\dfrac{\alpha +\beta }{2}=\dfrac{180^\circ-\gamma }{2}\\ \implies &\dfrac{\alpha}{2}+\dfrac{\beta}{2}=90^\circ-\dfr... ght)=\tan \left( 90-\dfrac{\gamma }{2} \right)\\ \implies &\dfrac{\tan\dfrac{\alpha}{2}+\tan\dfrac{\beta}{2
- Question 8, Exercise 1.2 @math-11-kpk:sol:unit01
- &={{a}^{2}}-{{bi}^{2}}\\ &={{a}^{2}}-b^2 (-1)\\ \implies z\overline{z}&={{a}^{2}}+b^2. \ldots (1) \end{al... . We have given \begin{align}&z=\overline{z} \\ \implies &a+bi=a-bi \\ \implies &bi=bi \\ \implies &2bi=0 \\ \implies &b=0 \end{align} Using it in (1), we get $z=a+0i=a$, that is, $z$ is
- Question 17 Exercise 4.2 @math-11-kpk:sol:unit04
- gn}a_{n+2}&=a_1+(n+2-1) d \\ & =a_1+(n+1) d \\ \implies 32&=5+(n+1)d \\ \implies (n+1)d&=32-5\\ \implies n+1&=\dfrac{27}{d}\\ \implies n&=\dfrac{27}{d}-1 ---(i)\end{align} Also, we have given $A_3:A_7=7:13$, wh
- Question 15 Exercise 4.2 @math-11-kpk:sol:unit04
- 2}=\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}, --- (3) \\ \implies &(a^n+b^n)(a+b)=2(a^{n+1}+b^{n+1}) \\ \implies &a^{n+1}+a b^n+a^n b+b^{n+1}=2 a^{n+1}+2 b^{n+1} \\ \implies &a b^n+b^{n+1}-2 b^{n+1}=2 a^{n+1}-a^{n+1}-a^n b \\ \implies &a b^n-b^{n+1}=a^{n+1}-a^n b \\ \implies & b^n(a
- Question 1, Exercise 1.3 @math-11-kpk:sol:unit01
- 1\\ \end{array} \] \begin{align}-11w&=11i-11\\ \implies w&=\dfrac{11-11i}{11}\\ \implies w&=1-i\end{align} Put value of $w$ in (i). \begin{align}z&-4(1-i)=3i\\ \implies z &=4(1-i)+3i\\ &=4-4i+3i\\ &=4-i\end{align} He... lue of $w$ in (i).\\ \begin{align}z&+(2-6i)=3i\\ \implies z&=3i-2+6i\\ &=-2+9i\end{align} Hence $$z=-2+9i
- Question 1, Exercise 2.2 @math-11-kpk:sol:unit02
- 2 & 0 \\ 0 & -2 \\ \end{matrix} \right|$$ $$\implies{{A}_{11}}=-4$$ $${{A}_{21}}={{\left( -1 \right)}^... 3 & 0 \\ 1 & -2 \\ \end{matrix} \right|$$ $$\implies{{A}_{21}}=6$$ $${{A}_{23}}={{\left( -1 \right)}^{... 1 & 3 \\ 2 & 0 \\ \end{matrix} \right|$$ $$\implies{{A}_{23}}=6$$ $${{A}_{31}}={{\left( -1 \right)}^{... 3 & 1 \\ 2 & 0 \\ \end{matrix} \right|$$ $$\implies{{A}_{31}}=-2$$ $${{A}_{32}}={{\left( -1 \right)}^
- Question 12, 13 & 14, Exercise 3.2 @math-11-kpk:sol:unit03
- e, \begin{align}&{\alpha ^2+(\alpha +1)^2}+4=9\\ \implies & {\alpha ^2+\alpha ^2}+2\alpha +1+4=9\\ \implies & 2{\alpha ^2}+2\alpha +5-9=0\\ \implies & 2{\alpha ^2}+2\alpha -4=0\\ \implies & {\alpha ^2}+\alpha -2=0.\end{align} This is quadratic equation in
- Question 7 Exercise 4.2 @math-11-kpk:sol:unit04
- ce of A.P. As given \begin{align} &a_6+a_4=6 \\ \implies & a_1+5d+a_1+3d=6\\ \implies & 2a_1+8d=6\\ \implies & a_1+4d=3 --- (1) \end{align} Also, we have given \begin{align} &a_6-a_4=\dfrac{2}{3} \\ \implies & a_1+5d-a_1-3d=\dfrac{2}{3}\\ \implies & 2d=\dfr
- Question 6, Exercise 1.3 @math-11-kpk:sol:unit01
- z^3=-8.$$ This gives \begin{align} & z^3+2^3=0\\ \implies &(z+2)\left(z^2-2z+4 \right)=0 \\ & \quad \becaus... \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\\ \implies & z+2=0 \quad \text{or} \quad z^2-2z+4=0.\end{ali... efore, we have \begin{align} &z^3-1-3z(z-1)=-1\\ \implies &{{z}^{3}}-1-3{{z}^{2}}+3z+1=0\\ \implies &{{z}^{3}}-3{{z}^{2}}+3z=0\\ \implies &z\left(z^2-3z+3 \right)=0
- Question 3 & 4, Exercise 3.2 @math-11-kpk:sol:unit03
- t{j}=p(\hat{i}+2\hat{j})+q(5\hat{i}-\hat{j})$$ $$\implies \hat{i}-9\hat{j}=(p+5q)\hat{i}+(2p-q)\hat{j}.$$ B... mits_{+}9 \\ \hline &11q&=11\\ \end{array} \] $$\implies q=1$$ \\ Put the value of $q$ in (i). We have, $$p+5(1)=1 \quad \implies p=-4$$ Hence we have $p=-4$ and $q=1$. =====Ques... given that \begin{align}&|\vec{p}+\vec{q}|=5 \\ \implies & \sqrt{{{x}^{2}}+4x+8}=5 \\ \implies & x^2+4x+8=
- Question 1 and 2 Exercise 4.2 @math-11-kpk:sol:unit04
- $ This gives \begin{align} &a_{21}=8+(21-1) d \\ \implies &108=8+20 d\\ \implies &20 d=108-8=100 \\ \implies &d=\dfrac{100}{20}\\ \implies &d=5. \end{align} Now \begin{align}&a_7=a_1+(n-1) d \\ \implies &a_7=8+(7-1
- Question 5, Exercise 10.1 @math-11-kpk:sol:unit10
- dfrac{1}{\sec\beta} = \dfrac{1}{\tfrac{13}{5}}\\ \implies \cos\beta &=\dfrac{5}{13}\end{align} As $\cos\be... ht)\\ &=-\frac{3}{13}+\frac{48}{65}\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin(\alpha +\b... dfrac{1}{\sec\beta} = \dfrac{1}{\tfrac{13}{5}}\\ \implies \cos\beta &=\dfrac{5}{13}\end{align} As $\cos\be... t)\\ &=-\frac{20}{65}+\frac{36}{65}\end{align} $$\implies \bbox[4px,border:2px solid black]{\cos(\alpha +\b