First slide

Properties of ITF

Question

If $\sin^{- 1} (2 x \sqrt{1 - x^{2}}) - 2 \sin^{- 1} x = 0,$ then $x$ belongs to the interval

Moderate

A

$[- 1, 1]$
B

$[- 1 / \sqrt{2}, 1 / \sqrt{2}]$
C

$[- 1, - 1 / \sqrt{2}]$
D

$[- 1 / \sqrt{2}, 1]$

Solution

Let $\sin^{- 1} x = θ .$ Then, $x = \sin θ$ and $\sqrt{1 - x^{2}} = \cos θ$
Now,
$\sin^{- 1} (2 x \sqrt{1 - x^{2}})$
$= \sin^{- 1} (\sin 2 θ) = 2 θ,$ if $- \frac{π}{2} \leq 20 \leq \frac{π}{2}$
$= 2 \sin^{- 1} x,$ if $- \frac{π}{4} \leq θ \leq \frac{π}{4}$ i.e. if $- \frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$
$∴ \sin^{- 1} (2 x \sqrt{1 - x^{2}}) - 2 \sin^{- 1} x = 0,$ if $- \frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$

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