Question

The set of values of $x$ such that $({\cos }^{-1}x{)}^2-3{\cos }^{-1}x+2>0$, is/are

• A. $[\cos 2,\cos 1]$
• B. $[0,1]$
• C. $[-1,\cos 2)$
• D. $(\cos 1,1]$

Answer 1

Answer: (C), (D)

$(\cos 1,1]$

Given : $({\cos }^{-1}x{)}^2-3{\cos }^{-1}x+2>0$
For inequality to be satisfied : $-1\le x\le 1$
Let $t={\cos }^{-1}x\in [0,\pi ]$, then inequality becomes, $t^2-3t+2>0$
$\Rightarrow (t-1)(t-2)>0$
$\Rightarrow t<1\text{or}t>2$
$\Rightarrow 0\le {\cos }^{-1}x<1\text{or}\pi \ge {\cos }^{-1}x>2$
Since, ${\cos }^{-1}x$ is decreasing function,
$\Rightarrow \cos 1<x\le 1\text{or}-1\le x<\cos 2$
$\therefore x\in [-1,\cos 2)\cup (\cos 1,1]$