Question:medium

The range of  the function  \( f(x) = \sin^{-1}(x - \sqrt{x}) \) is equal to?

Show Hint

For functions involving inverse trigonometric functions, first determine the range of the inner function, and then map it to the range of the inverse trigonometric function.
Updated On: Mar 28, 2026
  • \( \left[ \sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right] \)
  • \( \left[ \sin^{-1}\left(\frac{1}{2}\right), \frac{\pi}{2} \right] \)
  • \( \left[ -\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right] \)
  • \( \left[ -\sin^{-1}\left(\frac{1}{2}\right), \frac{\pi}{2} \right] \)
Show Solution

The Correct Option is C

Solution and Explanation

Given the function: \[f(x) = \sin^{-1}(x - \sqrt{x}).\]
Step 1: Analyze \( g(x) = x - \sqrt{x} \) The domain for \( x \) is limited to \( [0, 1] \). Let: \[g(x) = x - \sqrt{x}.\]
Compute the derivative of \( g(x) \): \[g'(x) = 1 - \frac{1}{2\sqrt{x}}.\]
Find critical points by setting \( g'(x) = 0 \): \[1 - \frac{1}{2\sqrt{x}} = 0 \implies \sqrt{x} = \frac{1}{2} \implies x = \frac{1}{4}.\]
Evaluate \( g(x) \) at critical points and endpoints: \[g(0) = 0, \quad g(1) = 1, \quad g\left(\frac{1}{4}\right) = -\frac{1}{4}.\]
The range of \( g(x) \) is: \[\left[-\frac{1}{4}, 1\right].\] Step 2: Apply \( \sin^{-1} \) The function \( \sin^{-1}(g(x)) \) maps the range of \( g(x) \) to: \[\left[ \sin^{-1}\left(-\frac{1}{4}\right), \sin^{-1}(1) \right].\]
Substitute the values: \[f(x) \in \left[ -\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right].\]
Thus, the range of \( f(x) \) is: \[\left[ -\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right].\] Final Answer: \[\boxed{\left[ -\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right]}\]
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