Question

Find the range of values of $t$ for which $2\sin t=\frac{1-2x+5x^2}{3x^2-2x-1},tϵ[-\frac{\pi }{2},\frac{\pi }{2}].$

• A. $[-\frac{\pi }{2},-\frac{\pi }{10}]\cup [\frac{3\pi }{10},\frac{\pi }{2}]$
• B. $[-\frac{\pi }{2},-\frac{\pi }{5}]\cup [\frac{3\pi }{10},\frac{\pi }{2}]$
• C. $[-\frac{\pi }{4},-\frac{\pi }{10}]\cup [\frac{3\pi }{10},\frac{\pi }{2}]$
• D. $[-\frac{\pi }{3},-\frac{\pi }{10}]\cup [\frac{3\pi }{10},\frac{\pi }{2}]$

Answer 1

Answer: (A)

$[-\frac{\pi }{2},-\frac{\pi }{10}]\cup [\frac{3\pi }{10},\frac{\pi }{2}]$

Given, $2\sin t=\frac{1-2x+{5x}^2}{{3x}^2-2x-1},t\in [-\frac{\pi }{2},\frac{\pi }{2}]$

Put $2\sin t=y,$ where $-2\le y\le 2$

$\therefore y=\frac{1-2x+{5x}^2}{{3x}^2-2x-1}$

$\Rightarrow (3y-5)x^2-2x(y-1)-(y+1)=0$

since $x\in$ R $-\{1,-\frac{1}{3}\}$ $(as,{3x}^2-2x-1\ne 0\Rightarrow (x-1)(x+\frac{1}{3})\ne 0)$
$\therefore D\ge 0$

$\Rightarrow 4{(y-1)}^2+4(3y-5)(y+1)\ge 0\Rightarrow y^2-y-1\ge 0$

$\Rightarrow {(y-\frac{1}{2})}^2-\frac{5}{4}\ge 0\Rightarrow (y-\frac{1}{2}-\frac{\sqrt{5}}{2})(y-\frac{1}{2}+\frac{\sqrt{5}}{2})\ge 0$

$\Rightarrow y\le \frac{1-\sqrt{5}}{2}$ or $y\ge \frac{1+\sqrt{5}}{2}$

$\Rightarrow 2\sin t\le \frac{1-\sqrt{5}}{2}$ or $\sin t\ge \frac{1+\sqrt{5}}{2}$

$\Rightarrow \sin t\le \sin (-\frac{\pi }{10})$ or $\sin t\ge \sin \left(\frac{3\pi }{10}\right)$

$\Rightarrow t\le -\frac{\pi }{10}$ or $t\ge \frac{3\pi }{10}$

Hence, range of $t$ is $[-\frac{\pi }{2},-\frac{\pi }{10}]\cup [\frac{3\pi }{10},\frac{\pi }{2}]$.