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Question #40410

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \left[\pi, \frac{3\pi}{2}\right] \to [-1, 1]$ be defined as $f(x) = \sqrt{\sin^2 x} + \cos x$ . If $g(x)$ is the inverse function of $f(x)$ , then $g(x)$ is equal to:

Options

A.
$\frac{7\pi}{4} - \arccos\left(\frac{x}{\sqrt{2}}\right)$
B.
$\frac{5\pi}{4} - \arcsin\left(\frac{x}{\sqrt{2}}\right)$
C.
$\pi + \frac{1}{2}\arcsin(1 - x^2)$
D.
$\frac{3\pi}{4} + \arccos\left(\frac{x}{\sqrt{2}}\right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Square Root and Modulus Logic|Extraneous Solution Rejection|The Modulus Trap|Bijectivity and Inverse

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous handling of the modulus function $\sqrt{\sin^2 x} = |\sin x|$ and the selection of the correct inverse trigonometric branch.

A lethal trap occurs if students square $y = \cos x - \sin x$ to obtain $\sin 2x = 1 - y^2$ , which destroys sign information and yields the extraneous branch $\pi + \frac{1}{2}\arcsin(1 - x^2)$ .

Additionally, mapping the domain $[\pi, 3\pi/2]$ requires applying the symmetry $\cos(2\pi - \theta) = \cos \theta$ to correctly shift the range of the standard arccosine function, penalizing students who rely on naive algebraic manipulation.

Notably, all distractors yield the correct value at $x=0$ , forcing students to verify the endpoints or derive the function rigorously.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Square Root and Modulus Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free