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- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- ta<\dfrac{\pi}{2}$, i.e. $\theta$ lies in QI. We have $$\sin\theta = \pm \sqrt{1-\cos^2}.$$ Since $\the... }. \end{align*} (d) $\sin \dfrac{\theta}{2}$ We have $$\sin\left(\frac{\theta}{2} \right) = \pm \sqrt{... }} \end{align*} (e) $\cos \dfrac{\theta}{2}$ We have $$\cos\left(\frac{\theta}{2} \right) = \pm \sqrt{... ac{3\pi}{2}\), i.e., \(\theta\) lies in QIII. We have: \begin{align*} \sec \theta &= \pm \sqrt{1+tan^2
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- {53}i\end{align} Put value of $\omega$ in (1), we have \begin{align} &(1-i) z+(1+i)\left(\dfrac{2}{53}-\... c{155}{106}+\dfrac{145}{106}i\end{align} Thus, we have $$z=\dfrac{155}{106}+\dfrac{145}{106}i, \omega=\d... -12+10i \quad \cdots(4) \end{align} $(3)-(4)$,we have \begin{align} &(-1-8i+4-6i)\omega=3-i+12-10i\\ \i... c{36}{205}+\dfrac{373}{205}i \end{align} Thus, we have: $$z = \dfrac{36}{205} + \dfrac{373}{205}i;\quad
- Question 2, Exercise 2.6 @math-11-nbf:sol:unit02
- ich the system of homogeneous linear equation may have non-trivial solution. Also solve the system for v... }+4 x_{3}=0\cdots(vi)\\ \end{align*} (iv)-(v), we have\\ \begin{align*} &\begin{array}{cccc} 2x_1&+\frac... \end{align*} Put the value of $x_2$ in (vi), we have \begin{align*} &3 x_{1}-2(\frac{11}{13})x_{3}+4 x... ich the system of homogeneous linear equation may have non-trivial solution. Also solve the system for v
- Question 5 Exercise 8.2 @math-11-nbf:sol:unit08
- in 2\theta=\dfrac{24}{25}$, $2\theta$ in QII. We have $$\cos 2\theta = \pm \sqrt{1-\sin^2 2\theta}$$ S... c{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta = \pm \sqrt{\frac{1-\cos 2\theta}{2}... dfrac{7}{25}\) and \(2\theta\) lies in QIII. We have: \[\sin 2\theta = \pm \sqrt{1 - \cos^2 2\theta}\... 76}{625}} = -\frac{24}{25}. \end{align*} Also, we have: \[ \sin\theta = \pm \sqrt{\frac{1 - \cos 2\the
- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- -- (4) \end{align} From (1), (2), (3) and (4), we have: $$\left| z_1 \right| = \left| -z_1 \right| = \le... + \frac{7}{10}i. \,\, -- (i) \end{align} Now, we have \begin{align} \overline{z_1} = -3 - 2i, \quad \ov... frac{7}{10}i.\,\, -- (ii)$$ From (i) and (ii), we have \[ \overline{\left( \frac{z_1}{z_2} \right)} = \f... 11i. -- (ii) \end{align} From (i) and (ii), we have \[ \overline{z_1 z_2} = \overline{z_1} \overline
- Question 6 Exercise 8.2 @math-11-nbf:sol:unit08
- 15^{\circ} \cos 15^{\circ}$ ** Solution. ** We have double-angle identity: $$\sin 2 \theta = 2\sin\th... \circ}-\sin ^{2} 15^{\circ}$ ** Solution. ** We have double-angle identity: $$\cos^2\theta -\sin^2\the... 2}\left(\frac{\pi}{8}\right)$ ** Solution. ** We have a double-angle identity: $$\cos 2\alpha = 1-2\sin... =\cos 2\alpha.$$ Put $\alpha= \dfrac{\pi}{8}$, we have \begin{align*} 1-2\sin^2 \left(\frac{\pi}{8}\righ
- Question 8, Exercise 1.2 @math-11-nbf:sol:unit01
- ution.** Given: $$|2z-i|=4.$$ Put $z=x+i y$, we have \begin{align} & |2(x+iy)-i|=4 \\ \implies & |2x+i... * Given: $$|z-1|=|\bar{z}+i|.$$ Put $z=x+iy$, we have \begin{align} & |(x+iy)-1| = |(x-iy)+i| \\ \impli... n: $$|z-4i| + |z+4i| = 10.$$ Put $z = x + iy$, we have \begin{align} & |(x + iy) - 4i| + |(x + iy) + 4i|... Put $z = x + i y$, then $\bar{z} = x - i y$. We have \begin{align} & \dfrac{1}{2}Re(i(x-iy)) = 4 \\ \i
- Question 20 and 21, Exercise 4.4 @math-11-nbf:sol:unit04
- _\_ , \_\_\_ , \_\_\_ , 48$$ ** Solution. ** We have given $a_1=3$ and $a_5=48$. Assume $r$ be common... ference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*} &a_... ="true"> **The good solution is as follows:** We have given $a_1=3$ and $a_5=48$. Assume $r$ be common... ference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*} &a_
- Question 3, Exercise 1.4 @math-11-nbf:sol:unit01
- d z=|z|e^{i\theta} \,\,-- (1) \end{align*} Now we have given \begin{align*} & \left(x_{1}+i y_{1}\right)... \cdots z_n = z. \end{align*} By using $(1)$, we have \begin{align*} &|z_1| e^{i\theta_1}\cdot |z_2| e^... right)=a^{2}+b^{2}\,\,-- (4)$$ Now from $(3)$, we have $$ \sum_{r=1}^{n} \theta_r = \theta + 2k\pi, \qua... results. **Alternative Method for Part (i)** We have given \begin{align*} & \left(x_{1}+i y_{1}\right)
- Question 28 and 29, Exercise 4.4 @math-11-nbf:sol:unit04
- tance fallen. How far (up and down) will the ball have travelled when it hits the ground for the 6th tim... are in staff at high school? ** Solution. ** We have given \\ first person $=a_1= 1$ principal \\ Peop... \\ ... \\ People in 6ht round $= a_7$.\\ Thus we have the series $$ 1+2+4+...+a_7 $$ We have to find sum of geometric series with $a_1=1$, $r=2$, $n=7$. As
- Question 11 and 12, Exercise 4.8 @math-11-nbf:sol:unit04
- + Bk \ldots (2) \end{align*} Put $k=0$ in (2), we have \begin{align*} &1=2A + 0 \\ \implies & A = \frac{... \end{align*} Put $k+2=0 \implies k=-2$ in (2), we have \begin{align*} &1=0-2B\\ \implies &B = -\frac{1}{... {1}{k+2} \right). \end{align*} Taking the sum, we have \begin{align*} S_n &= \sum_{k=1}^n T_k = \frac{1}... Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have \begin{align*} &1 = \left(3\times\frac{2}{3}+1 \r
- Question 2, Exercise 1.2 @math-11-nbf:sol:unit01
- plicative assocative law} \end{align} That is, we have proved $$(z_1 z_2)(z_3 z_4)=(z_1 z_3) (z_2 z_4) .... licative associative law} \end{align} That is, we have proved $$(z_1 z_3) (z_2 z_4)=z_3 (z_1 z_2) z_4 ... (ii)$$ From (i) and (ii), we have the required result. **Remark:** For any three complex numbers $z_1$, $z_2$ and $z_3$, we have $$z_1 (z_2 z_3) = (z_1 z_2)z_3 = z_1 z_2 z_3.$$ L
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- \] For \(z_1 = 7 + 2i\) and \(z_2 = 3 - i\), we have: \[x_1 = 7, \quad y_1 = 2, \quad x_2 = 3, \quad... Given \(z_1 = 4 + 2i\) and \(z_2 = 2 + 5i\), we have: \[x_1 = 4, \quad y_1 = 2, \quad x_2 = 2, \quad... ] For \(z_1 = 5 - 4i\) and \(z_2 = 5 + 4i\), we have: \[x_1 = 5, \quad y_1 = -4, \quad x_2 = 5, \qua... Given \(z_1 = 3 - 7i\) and \(z_2 = 2 + 5i\), we have: \[x_1 = 3, \quad y_1 = -7, \quad x_2 = 2, \quad
- Question 1, Exercise 2.6 @math-11-nbf:sol:unit02
- n-trivial solution. \text{By}\quad(i)-2(ii), we have \begin{align*} &\begin{array}{cccc} 2x_1&-3 x_{2}... \end{align*} Put the value of $x_3$ in (iii), we have \begin{align*} &4 x_{1}+2x_{3}-6 x_{3}=0\\ &4 x_{... ii)} \end{align*} For the system of equations, we have: \begin{align*} A &= \left[ \begin{array}{ccc} 2 ... \end{align*} Put the value of $x_2$ in (iii), we have \begin{align*} &x_1 - 4\left(\frac{2}{5}x_3\right
- Question 1, Exercise 4.2 @math-11-nbf:sol:unit04
- ** Solution. ** Given: $a_1= 16$, $d=-2$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&... ** Solution. ** Given: $a_1= 38$, $d=-4$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&... Given: $a_1=\frac{3}{4}$, $d=\frac{1}{4}$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&... Given: $a_1=\frac{3}{8}$, $d=\frac{5}{8}$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2