If ^ -1 x> ^ -1 x, then: - Vidyadip

MCQ

If $\cos^{-1}\text{x}>\sin^{-1}\text{x},$ then:

A
$\frac{1}{\sqrt2}<\text{x}\leq1$
B
$0\leq\text{x}\leq\frac{1}{\sqrt2}$
✓
$-1\leq\text{x}<\frac{1}{\sqrt2}$
D
$\text{x}>0$

✓

Answer

Correct option: C.

$-1\leq\text{x}<\frac{1}{\sqrt2}$

The correct option is $C$
$
-1 \leq x < \frac{1}{\sqrt{2}}
$
Explanation for the correct options:
Step $1$.
Find the intervals in which $x$ lies:
We have given $\cos ^{-1} x > \sin ^{-1} x$, and we know that,
$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
$\Rightarrow \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x$
$\text { But } \frac{\pi}{2}-\sin ^{-1} x > \sin ^{-1} x$
$\Rightarrow \frac{\pi}{2} > 2 \sin ^{-1} x$
$\Rightarrow \frac{\pi}{4} > \sin ^{-1} x \ldots \ldots \ldots(1)$
Also $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2} \ldots \ldots(2)$

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