Let fx=sin^(-1)x and gx=(x^2-x-2)/(2x^2-x-6). If g2=limx→2gx, then the domain of the function fog is

Correct answer: (C) (-∞,-2]∪-4/3,∞. Domain of fogx=sin^(-1)gx ⇒g(x)≤1,g2=3/7 (x^2-x-2)/(2x^2-x-6)≤1 (x+1x-2)/(2x+3x-2)≤1 (x+1)/(2x+3)≤1 and (x+1)/(2x+3)≥-1 (x+1-2x-3)/(2x+3)≤0 and (x+1+2x+3)/(2x+3)≥0 (x+2)/(2x+3)≥0 and (3x+4)/(2x+3)≥0 x∈(-∞,-2]∪-4/3,∞

Q45 / 90 · JEE Main 2021 (26 Feb Shift 2) · Mathematics

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 $f o g$ is

A
$(- \infty, - 1] \cup [2, \infty)$
B
$(- \infty, - 2] \cup [- \frac{3}{2}, \infty)$
Correct answer
C
$(- \infty, - 2] \cup [- \frac{4}{3}, \infty)$
D
$(- \infty, - 2] \cup [- 1, \infty)$

Solution

Domain of $f o g (x) = \sin^{- 1} (g (x))$

$\Rightarrow |g (x)| \leq 1, g (2) = \frac{3}{7}$

$|\frac{x^{2} - x - 2}{2 x^{2} - x - 6}| \leq 1$

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

$\frac{x + 1}{2 x + 3} \leq 1$ and $\frac{x + 1}{2 x + 3} \geq - 1$

$\frac{x + 1 - 2 x - 3}{2 x + 3} \leq 0$ and $\frac{x + 1 + 2 x + 3}{2 x + 3} \geq 0$

$\frac{x + 2}{2 x + 3} \geq 0$ and $\frac{3 x + 4}{2 x + 3} \geq 0$

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

Hence, the correct answer is option: 'C'.

Let fx=sin-1x and gx=x2-x-22x2-x-6. If g2=limx→2gx, then the domain of the function fog is

Solution

Relevant Questions

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 $f o g$ is