Considering the principal values of the inverse trigonometric functions, $\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^{2}}\right),-\frac{1}{2}<x<\frac{1}{\sqrt{2}}$, is equal to

(B) $\frac{\pi}{6}+\sin ^{-1} x$

(A) $\frac{\pi}{4}+\sin ^{-1} x$

(C) $\frac{-5 \pi}{6}-\sin ^{-1} x$

(D) $\frac{5 \pi}{6}-\sin ^{-1} x$

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Inverse Trigonometric Functions

Considering the principal values of the inverse trigonometric functions,

sin^{- 1} (\frac{3}{2} x + \frac{1}{2} 1 - x^{2}), - \frac{1}{2} < x < \frac{1}{2}

, is equal to

Answer: (B)

Setting

x = sin θ

with

θ = sin^{- 1} x

makes

1 - x^{2} = cos θ

, so the argument is

\frac{3}{2} sin θ + \frac{1}{2} cos θ = sin (θ + \frac{π}{6})

. The range

- \frac{1}{2} < x < \frac{1}{2}

gives

- \frac{π}{6} < θ < \frac{π}{4}

, so

θ + \frac{π}{6}

lies in

(0, \frac{5 π}{12})

, inside the principal branch of the inverse sine. Hence the expression equals

θ + \frac{π}{6} = \frac{π}{6} + sin^{- 1} x

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option