JEE PYQ: Functions Question 12
Question 12 - 2021 (26 Feb Shift 2)
Let $f(x) = \sin^{-1} x$ and $g(x) = \frac{x^2 - x - 2}{2x^2 + x - 6}$. If $g(2) = \lim_{x \to 2} g(x)$, then the domain of the function $f \circ g$ is:
(1) $(-\infty, -2] \cup \left[-\frac{4}{3}, \infty\right)$
(2) $(-\infty, -1] \cup [2, \infty)$
(3) $(-\infty, -2] \cup [-1, \infty)$
(4) $(-\infty, -2] \cup \left[-\frac{4}{3}, \infty\right)$
Type: MCQ
Show Answer
Answer: (1) $(-\infty, -2] \cup \left[-\frac{4}{3}, \infty\right)$
Solution
$g(2) = \lim_{x \to 2} \frac{(x-2)(x+1)}{(2x-3)(x+2)} = \frac{3}{7}$
For domain of $f \circ g$: $(3x + 4)(x + 2) \ge 0$
$x \in (-\infty, -2] \cup \left[-\frac{4}{3}, \infty\right)$
BETA
