Sets, Relations & Functions
Functions
star_batch_jee_advanced_2025
Grade 11
Question:
The range of $f(x) = \tan x + \frac{1}{2}\sin^{-1} x$ is:
(-\pi, \pi)
[\left(-\frac{3\pi}{4}, \frac{3\pi}{4}\right)]
(\left(-\frac{3\pi}{4}, \frac{3\pi}{4}\right))
[\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)]
Step-by-Step Solution
Key Concept: The range of inverse trigonometric functions must be matched with the domain constraint and monotonicity of the original function.
Given domain of $f(x)$ is $[-1,1]$ and $f(x)$ is continuous and increasing, we find that $\tan^{-1}x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$ and $\frac{1}{2}\sin^{-1}x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$, both matching the required range.
Correct Answer: 4