Let $f(x)=\sin ^{-1} x$ and $g(x)=\frac{x^{2}-x-2}{2 x^{2}-x-6}$. If $g(2)=\lim _{x \rightarrow 2} g(x)$, then the domain of the function fog is:

(B) $(-\infty,-2] \cup\left[-\frac{4}{3}, \infty\right)$

(A) $(-\infty,-1] \cup[2, \infty)$

(C) $(-\infty,-2] \cup[-1, \infty)$

(D) $(-\infty,-2] \cup\left[-\frac{3}{2}, \infty\right)$

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Sets Relations and Functions

Let

f (x) = sin^{- 1} x

and

g (x) = \frac{x ^{2} - x - 2}{2 x ^{2} - x - 6}

. If

g (2) = lim_{x \to 2} g (x)

, then the domain of the function fog is:

Answer: (B)

For

x \neq = 2

g (x) = \frac{( x - 2 ) ( x + 1 )}{( x - 2 ) ( 2 x + 3 )} = \frac{x + 1}{2 x + 3}

and the stipulation

g (2) = lim_{x \to 2} g (x) = \frac{3}{7}

makes this reduced formula valid at

x = 2

too, so

g

is defined for every

x \neq = - \frac{3}{2}

.
Since

sin^{- 1} u

requires

- 1 \leq u \leq 1

, the domain of

f \circ g

consists of those

x

with

\frac{x + 1}{2 x + 3} \leq 1

\frac{x + 1}{2 x + 3} \leq 1 ⟺ \frac{x + 2}{2 x + 3} \geq 0 ⟺ x \leq - 2 or x > - \frac{3}{2}

\frac{x + 1}{2 x + 3} \geq - 1 ⟺ \frac{3 x + 4}{2 x + 3} \geq 0 ⟺ x < - \frac{3}{2} or x \geq - \frac{4}{3}

The two conditions hold together exactly when

x \leq - 2

x \geq - \frac{4}{3}

, so the domain of

f \circ g

(- \infty, - 2] \cup [- \frac{4}{3}, \infty)

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option