JEE PYQ: Functions Question 13
Question 13 - 2020 (02 Sep Shift 1)
The domain of the function $f(x) = \sin^{-1}\left(\frac{|x| + 5}{x^2 + 1}\right)$ is $(-\infty, -a] \cup [a, \infty)$. Then $a$ is equal to:
(1) $\frac{\sqrt{17}}{2}$
(2) $\frac{\sqrt{17} - 1}{2}$
(3) $\frac{1 + \sqrt{17}}{2}$
(4) $\frac{\sqrt{17}}{2} + 1$
Type: MCQ
Show Answer
Answer: (3) $\frac{1 + \sqrt{17}}{2}$
Solution
$x^2 - |x| - 4 \ge 0$
$\therefore a = \frac{1 + \sqrt{17}}{2}$
BETA
