Question

The range of $t$ such that $2\sin t=\frac{1-2x+5x^2}{3x^2-2x-1}$ has a solution, where $t\in [-\frac{\pi }{2},\frac{\pi }{2}]$is

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

Answer 1

The correct option is A $[-\frac{\pi }{2},-\frac{\pi }{10}]\cup [\frac{3\pi }{10},\frac{\pi }{2}]$

$2\sin t=\frac{1-2x+5x^2}{3x^2-2x-1}\Rightarrow 2\sin t=\frac{1-2x+5x^2}{(3x+1)(x-1)}\Rightarrow x^2(6\sin t-5)+x(2-4\sin t)-(1+2\sin t)=0\because x\in R\Rightarrow \Delta \ge 0\Rightarrow (2-4\sin t{)}^2+4(6\sin t-5\left)\right(1+2\sin t)\ge 0\Rightarrow 4(4{\sin }^{2}t-2\sin t+1)+4(12{\sin }^{2}t-4\sin t-5)\ge 0\Rightarrow 16{\sin }^{2}t-8\sin t-4\ge 0\Rightarrow 4{\sin }^{2}t-2\sin t-1\ge 0\Rightarrow [\sin t-\left(\frac{1-\sqrt{5}}{4}\right)][\sin t-\left(\frac{1+\sqrt{5}}{4}\right)]\ge 0\Rightarrow \sin t\le \frac{1-\sqrt{5}}{4}\text{or}\sin t\ge \frac{1+\sqrt{5}}{4}$
Now using the graph, we get
$\therefore t\in [-\frac{\pi }{2},-\frac{\pi }{10}]\cup [\frac{3\pi }{10},\frac{\pi }{2}]$