Let f(x) = ^ - 1 x and g(x) = x ^ 2 - x - 2 2 x ^ 2 - x - 6 . If g(2) = _ x → 2 g(x) , then the domain of the function fog is

Correct answer: C. ( - ∞ , - 2 ] [- (4)/(3), ∞) Domain of fog (x) = ^ - 1 (g (x) ) ⇒ |g ( x )| ≤ 1, g(2) = (3)/(7) | x ^ 2 - x - 2 2 x ^ 2 - x - 6 | ≤ 1 |((x + 1)(x - 2) )/((2x + 3)(x - 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), ∞)

JEE Main 202126 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 26 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2021 · 26 February · Evening Shift · Mathematics · Functions

Let f(x) = ^ - 1 x and g(x) = x ^ 2 - x - 2 2 x ^ 2 - x - 6 . If...

Let

f\left(x\right) = \sin^{- 1}x

and

g\left(x\right) = \dfrac{{x}^{2} - x - 2}{2{x}^{2} - x - 6}

. If

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

, then the domain of the function

fog

Options

A $( - \infty , - 1 ] \cup [ 2, \infty )$
B $( - \infty , - 2 ] \cup \left[- \dfrac{3}{2}, \infty\right)$
C $( - \infty , - 2 ] \cup \left[- \dfrac{4}{3}, \infty\right)$
D $( - \infty , - 2 ] \cup [ - 1, \infty )$

Correct answer

C. $( - \infty , - 2 ] \cup \left[- \dfrac{4}{3}, \infty\right)$

Step-by-step solution

Domain of $fog \left(x\right) =$ \sin^{- 1}\left(g \left(x\right) $\right)$

$⇒ \left|g ( x )\right| \le 1, g\left(2\right) = \dfrac{3}{7}\left|\dfrac{{x}^{2} - x - 2}{2{x}^{2} - x - 6}\right| \le 1$

\left|\dfrac{\left(x + 1\right)\left(x - 2\right) $}{\left(2x + 3\right)\left(x - 2\right)}\right| \le 1$

$\dfrac{x + 1}{2x + 3} \le 1$ and $\dfrac{x + 1}{2x + 3}$ \ge - 1 $\dfrac{x + 1 - 2x - 3}{2x + 3}$ \le 0 and $\dfrac{x + 1 + 2x + 3}{2x + 3} \ge 0$

$\dfrac{x + 2}{2x + 3} \ge 0$ and $\dfrac{3x + 4}{2x + 3}$ \ge 0 $$ x \in ( - \infty , - 2 ] \cup \left[- \dfrac{4}{3}, $\infty\right)$

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