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Question #41220

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: (0, 2\pi) \to \mathbb{R}$ be defined as $f(x) = \sin(2x) + \cos^{-1}(\cos x) + \sin^{-1}(\sin x)$ Let $C$ be the number of critical points of $f(x)$ and $S$ be the number of stationary points of $f(x)$ . Find the ordered pair $(C, S)$ .

Options

A.
$(5, 2)$
B.
$(5, 4)$
C.
$(4, 4)$
D.
$(2, 2)$

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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

Inverse Function Periodicity|Critical Point vs Stationary Point|Differentiability vs Continuity

Why This Question Is Tricky

Common Mistakes

This question tests the distinction between critical points and stationary points in the context of piecewise periodic inverse trigonometric functions.

A common trap is confusing $\cos^{-1}(\cos x) + \sin^{-1}(\sin x)$ with the identity $\cos^{-1}x + \sin^{-1}x = \pi/2$ , which incorrectly yields 4 stationary points.

Furthermore, students often blindly set $f'(x)=0$ and find that one-sided derivatives vanish at the boundary points (e.g., $f'(\pi/2^-)=0$ and $f'(\pi^+)=0$ ).

Without checking both the LHD and RHD, they falsely classify these non-differentiable cusps as stationary points.

A stationary point strictly requires the two-sided derivative to exist and equal zero, whereas a critical point also includes points of non-differentiability.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Inverse Function Periodicity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free