The domain of the function $f(x)=\sin ^{-1}\left(\frac{3 x^{2}+x-1}{(x-1)^{2}}\right)+\cos ^{-1}\left(\frac{x-1}{x+1}\right)$ is :

(A) $\left[\frac{1}{4}, \frac{1}{2}\right] \cup\{0\}$

(B) $\left[0, \frac{1}{2}\right]$

(C) $\left[0, \frac{1}{4}\right]$

(D) $[-2,0] \cup\left[\frac{1}{4}, \frac{1}{2}\right]$

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Inverse Trigonometric Functions

The domain of the function

f (x) = sin^{- 1} (\frac{3 x ^{2} + x - 1}{( x - 1 ) ^{2}}) + cos^{- 1} (\frac{x - 1}{x + 1})

is :

Answer: (A)

The inverse sine requires

- 1 \leq \frac{3 x ^{2} + x - 1}{( x - 1 ) ^{2}} \leq 1

with

x \neq = 1

. The upper bound gives

2 x^{2} + 3 x - 2 \leq 0

, i.e.

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

; the lower bound gives

4 x^{2} - x \geq 0

, i.e.

x \leq 0

x \geq \frac{1}{4}

. Together these give

[- 2, 0] \cup [\frac{1}{4}, \frac{1}{2}]

. The inverse cosine requires

- 1 \leq \frac{x - 1}{x + 1} \leq 1

, which holds for

x \geq 0

. Intersecting the two conditions gives

[\frac{1}{4}, \frac{1}{2}] \cup {0}

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option