JEE PYQ: Inverse Trigonometric Functions Question 5
Question 5 - 2021 (17 Mar Shift 2)
The number of solutions of the equation $\sin^{-1}\left[x^2 + \frac{1}{3}\right] + \cos^{-1}\left[x^2 - \frac{2}{3}\right] = x^2$ for $x \in [-1, 1]$, and $[x]$ denotes the greatest integer less than or equal to $x$, is:
(1) 2
(2) 0
(3) 4
(4) Infinite
Show Answer
Answer: (2)
Solution
$\sin^{-1}\left[x^2 + \frac{1}{3}\right]$ is defined if $-1 \le x^2 + \frac{1}{3} < 2$, so $0 \le x^2 < \frac{5}{3}$. And $\cos^{-1}\left[x^2 - \frac{2}{3}\right]$ is defined if $-1 \le x^2 - \frac{2}{3} < 2$, so $0 \le x^2 < \frac{8}{3}$. Combined: $0 \le x^2 < \frac{5}{3}$. Case I: $0 \le x^2 < \frac{2}{3}$: $\sin^{-1}(0) + \cos^{-1}(-1) = x^2 \Rightarrow \pi = x^2$, but $\pi \notin [0, \frac{2}{3})$. Case II: $\frac{2}{3} \le x^2 < \frac{5}{3}$: $\sin^{-1}(1) + \cos^{-1}(0) = x^2 \Rightarrow \pi = x^2$, but $\pi \notin [\frac{2}{3}, \frac{5}{3})$. No value of $x$ works. Number of solutions = 0.
BETA
