Question

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

• A. \( \left[ \sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right] \)
• B. \( \left[ \sin^{-1}\left(\frac{1}{2}\right), \frac{\pi}{2} \right] \)
• C. \( \left[ -\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right] \)
• D. \( \left[ -\sin^{-1}\left(\frac{1}{2}\right), \frac{\pi}{2} \right] \)

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.

Answer 1

The Correct Option is C

From the question given function is: \[ f(x) = \sin^{-1}(x - \sqrt{x}). \]
Step 1: Analyze \( g(x) = x - \sqrt{x} \) The domain of \( x \) is restricted to \( [0, 1] \).
Define:
\[ g(x) = x - \sqrt{x}. \]
Now, compute the derivative of \( g(x) \): \[ g'(x) = 1 - \frac{1}{2\sqrt{x}}. \]
To find the critical points, set \( g'(x) = 0 \): \[ 1 - \frac{1}{2\sqrt{x}} = 0 \implies \sqrt{x} = \frac{1}{2} \implies x = \frac{1}{4}. \]
Evaluate \( g(x) \) at the critical points: \[ g(0) = 0, \quad g(1) = 1, \quad g\left(\frac{1}{4}\right) = -\frac{1}{4}. \]
Thus, 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: \[ \left[ \sin^{-1}\left(-\frac{1}{4}\right), \sin^{-1}(1) \right]. \]
Substituting the values: \[ f(x) \in \left[ -\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right]. \]
Therefore, the range of \( f(x) \) is: \[ \left[ -\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right]. \] Final Answer: \[ \boxed{\left[ -\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right]} \]