Problem-1

Let, $\quad\frac{-\pi}{6}<\theta<\frac{-\pi}{12}$, suppose $\alpha$ and $\beta$ are the roots of the equation $x^2-2 x \sec \theta+1=0$ , and $\alpha_2$ and $\beta_2$ are the root of the equation $x^2+2 x \tan \theta-1=0.\quad \text{If} \quad \alpha_ 1>\beta_ 1$, and $\alpha_ 2>\beta_ 2$ then, $\alpha_ 1+\beta_2$ equals?

Solution

$x^2-2 x \sec \theta+1=0$

$\left(\alpha_1, \beta_1\right) =\frac{2 \sec \theta \pm \sqrt{4 \sec ^2 \theta-4}}{2}$

$=\sec \theta \pm \sqrt{\sec ^2 \theta-1}$

$\sec ^2 \theta =1+\tan ^2 \theta$

$\left(\alpha_1, \beta_1\right)=\sec \theta \pm \tan \theta$

$=\left(\frac{1 \pm \sin \theta}{\cos \theta}\right)$

$\theta \in\left(-\frac{\pi}{6},-\frac{\pi}{12}\right) \Rightarrow \quad\sin \theta<0, \cos \theta>0$

$\frac{\alpha_1=(1-\sin \theta)}{\cos \theta}$

$x^2+2 x \tan \theta-1=0$

$\left(\alpha_2, \beta_2\right) =\frac{-2 \tan \theta \pm \sqrt{4 \tan ^2 \theta+4}}{2}$

$=-\tan \theta \pm \sqrt{1+\tan^2 \theta } = \sec^{2}\theta$ $\quad$ $[\because 1+ \tan^2 \theta = {\sec ^2 \theta}]$

Problem-2

The number of possible values of $\theta$ where $0<\theta<\pi$, for which the system of equation

$(y+z) \cos 3 \theta= (x y z) \sin 3 \theta$

$x \sin 3 \theta= \frac{2\cos 3 \theta}{y} +\frac{2 \sin 3 \theta}{z}$

$x y z \sin 3 \theta =(y+2 z) \cos 3 \theta +y \sin 3 \theta$

have a solution $(x_0,y_0,z_0)$ with $\quad y_0 z_0\neq 0$ is ?

Solution

$(y+z) \cos 3 \theta= (xyz) \sin 3 \theta$

$xyz \sin 3 \theta =2 z \cos 3 \theta + 2y \sin 3 \theta$

$xyz \sin 3 \theta = (y+2z)\cos 3 \theta + 2y \sin 3 \theta$

$(y+3) \cos 3 \theta= 2 z \cos 3\theta+2 y \sin 3 \theta$

$(y+3) \cos 3 \theta= (y+2 z) \cos 3 \theta+y \sin 3 \theta$

$= (y\cos 3 \theta)+2 z \cos 3 \theta +y \sin 3 \theta$

${2 z \cos 3} \theta+2 y \sin 3 \theta =y \cos 3 \theta+2 {z\cos 3} \theta+y \sin 3 \theta$

$y \sin 3 \theta=y \cos 3 \theta$

$y(\sin 3 \theta-\cos 3 \theta)=0$

$\rightarrow \sin 3 \theta-\cos 3 \theta=0, \quad y \neq 0$

$\rightarrow(y+z) \cos 3 \theta =(2 z+2 y) \cos 3 \theta$

$(y+z)=0 \quad \quad(y+z) \cos 3 \theta=0$

$\sin 3 \theta-\cos 3 \theta=0$

$\frac{1}{\sqrt{2}} \sin 3 \theta-\frac{1}{\sqrt{2}} \cos 3 \theta=0$

$\sin 3 \theta \cos \frac{\pi}{4}-\sin \frac{\pi}{4} \cos 3 \theta=0$

$\sin (3 \theta-\frac{\pi}{4})=0$

$3 \theta-\frac{\pi}{4}=n \pi \quad$ for some integer n

$\theta=\frac{n \pi}{3}+\frac{\pi}{12} \in(0, \pi)$

$\theta=\left(\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{3 \pi}{4}\right)$

Problem-3

The number of distinct solutions of the equation

$\frac{5}{4} \cos ^2 2 x+\cos ^4 x+\sin ^4 x+\cos ^6 x + \sin ^6 x=2$

$\text{in the interval} [0,2 \pi]$ is?

Solution

$\left(\sin ^2 x+\cos ^2 x\right)^3=1$

$[\because (A+B)^3= A^3+B^3+3AB^2+3A^2B]$

$\sin ^6 x+\cos ^6 x +3 \sin ^2 x \cos ^4 x +3 \cos ^2 x \sin ^4 x=1$

$\sin ^6 x+\cos ^6 x=1 - 3 \sin ^2 x \cos ^4 x -3 \cos^2 x \sin^4 x$

$\rightarrow \sin ^6 x+\cos ^6 x=1-3 \sin ^2 x \cos ^4 x -3\cos^2 x \sin^4 x$

$\left(\sin ^2 x+\cos ^2 x\right)^2=1$

$\sin ^4 x+\cos ^4 x +2 \sin^2 x \cos^2 x =1$

$\rightarrow \sin ^4 x+\cos ^4 x=1 - 2 \sin ^2 x \cos ^2 x$

$\cos 2 x=\left(\cos ^2 x - \sin ^2 x \right)$

$\rightarrow \cos^2 2 x = \cos^4 x+\sin^4 x -2 \sin^2 x \cos^2 x$

$\frac{5}{4} \cos ^2 2x +\left(1-2 \sin ^2 x \cos ^2 x\right) +1-3 \sin ^2 x \cos ^4 x$

$-3 \cos^2 x \sin^4 x = 2$

$\frac{5}{4} \cos ^2 2 x= 2 \sin ^2 x \cos ^2 x+3 \sin ^2 x \cos ^2 x (\sin ^2 x + \cos^2 x)$

$\frac{5}{4} \cos ^2 x =5 \sin ^2 x \cos ^2 x$

$\cos ^2 2 x= 4 \sin ^2 x \cos ^2 x = \left(2 \sin x \cos x\right)^2$

$\cos ^2 2x =\sin ^2 2 x$

$\cos ^2 2 x=\sin ^2 2 x$

${\cos ^2 2 x-\sin ^2 2 x}=0$ $\quad {\cos 4 x=0}$

$4 x=(2 n+1) \frac{\pi}{2},\quad$ n is an integer

$x=(2n+1) \frac{\pi}{8}, \quad \quad x \in[0,2 \pi]$

$x=\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}, \frac{9 \pi}{8}, \frac{11 \pi}{8}, \frac{13 \pi}{8}, \frac{15 \pi}{8}, \frac{17}{8}$

Problem-4

Prove that the values of the function $\left(\frac{\sin x \cos 3 x}{\sin 3 x \cos x}\right)$ do not lie between $\frac{1}{3}$ and 3 for any real $x$.

Solution

$\frac{\tan x}{\tan 3 x}$

$\tan 3 x =\frac{3 \tan x-\tan ^3 x}{1-3 \tan ^2 x}$

$=\frac{1-3 \tan ^2 x}{3-\tan ^2 x} =\frac{1-3 a}{3-a}$

$a = \tan ^2 x \geqslant 0 \quad \quad a \in[0, \infty]$

$\frac{1-3 a}{3-a} \notin\left(\frac{1}{3}, 3\right) \quad a \in[0, \infty]$

$\frac{1-3 a}{3-a}=\frac{(9-3 a)-8}{3-a}=3-\frac{8}{3-a}=3+\frac{8}{a-3}$

i) $a \in[0,3]$

$3 \geqslant 3-a>0$

$-3 \leqslant a-3 < 0$

$-\infty<\frac{8}{a-3} \leqslant \frac{-8}{3}$

$-\infty< 3+ \frac{8}{a-3} \leqslant (3\frac{-8}{3})$

If $\quad a \in[0,3]$ then,

$\frac{1-3 a}{3-a} \in\left(-\infty, \frac{1}{3}\right]$

ii) if $\quad a \in(3, \infty] \Rightarrow a-3>0$

$\frac{1-3 a}{3-a}=3+\left(\frac{8}{a-3}\right), \quad 0<\frac{8}{a-3}<\infty$

$\frac{1-3 a}{3-a} \in(3, \infty)$

Problem-5

$\sum_{k=1}^{13} \frac{2}{2 \sin \left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right) 2 \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$

Solution

$2 \sin A \sin B=\cos (A-B)-\cos (A+B)$

$=\cos \frac{\pi}{6}-\cos \left(\frac{\pi}{2}+\frac{(2 k-1) \pi}{6}\right)$

$=\cos \frac{\pi}{6}+\frac{\sin (2 k-1) \pi}{6} \quad [\because \cos (\frac{\pi}{2}+\theta) \ =-\sin \theta]$

$\sum_{k=1}^{13} \frac{2}{[\frac{\sqrt{3}{}}2 +\sin \frac{(2 k-1) \pi}{6}]}$

$(k+6)^{th}$term: $\sin (\frac{2(k+6)-1) \pi}{6}$

$=\sin \left[\frac{(2 k-1) \pi}{6}+\frac{12 \pi}{6}\right]$

$=\sin \left[\frac{(2 k-1) \pi}{6}+2 \pi\right]$

$=\sin \frac{(2 k-1) \pi}{6} \qquad [\sin \frac{25 \pi}{6}]$

$=\sin (4 \pi + \frac{\pi}{6})=\sin (\frac{\pi}{6})$

$=k^{th}$ term

$\left(\sum_{k=1}^{6} \frac{4}{[\frac{\sqrt{3}{}}2 +\sin \frac{(2 k-1) \pi}{6}]}\right) + \frac{2}{[\frac{\sqrt{3}{}}2 +\sin \frac{\pi}{6}]}$

$=\frac{2}{\left[\frac{\sqrt{3}}{2}+\frac{1}{2}\right]}=\frac{4}{[\sqrt{3}+1]}$

$=\frac{24 (\sqrt{3}-1)}{2}=2(\sqrt{3}-1)$

$\frac{4}{\left[\frac{\sqrt{3}}{2}+\sin \frac{(2 k-1) \pi}{6}\right]}$

$k=1 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}+\frac{1}{2}\right]}$

$k=2 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}+1\right]}$

$k=3 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}+\frac{1}{2}\right]}$ $\quad$ $[\because \sin \frac{5 \pi}{6} =\sin \frac{\pi}{6} =\frac{1}2]$

$k=4 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\right]}$

$k=5 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}-1\right]}$

$k=6 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\right]}$ $\quad$ $[\because \sin \frac{11 \pi}{6} =-\sin \frac{\pi}{6}]$