First slide

Properties of ITF

Question

If $- \frac{1}{2} \leq x \leq \frac{1}{2},$ then $\sin^{- 1} (3 x - 4 x^{3})$ equals

Moderate

A

$3 \sin^{- 1} x$
B

$π - 3 \sin^{- 1} x$
C

$- π - 3 \sin^{- 1} x$
D

none of these

Solution

Let $\sin^{- 1} x = θ .$ Then, $x = \sin θ$
Also,
$- \frac{1}{2} \leq x \leq \frac{1}{2} \Rightarrow - \frac{1}{2} \leq \sin θ \leq \frac{1}{2} \Rightarrow - \frac{π}{6} \leq θ \leq \frac{π}{6} .$
Now,
$\begin{array}{l} \sin^{- 1} (3 x - 4 x^{3}) = \sin^{- 1} (\sin 3 θ) \\ \Rightarrow \sin^{- 1} (3 x - 4 x^{3}) = 3 θ [∵ - \frac{π}{6} \leq 0 \leq \frac{π}{6} \Rightarrow - \frac{π}{2} \leq 3 θ \leq \frac{π}{2}] \\ \Rightarrow \sin^{- 1} (3 x - 4 x^{3}) = 3 \sin^{- 1} x \end{array}$

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