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Question #41115

Difficulty: 8/10 | 🟢 Free

The Question

Let $X = [0, 2\pi]$ . A relation $R$ is defined on $X$ by $x R y \iff f(x) = f(y)$ , where $f(x) = \cos^{-1}(\cos x) - \left| \sin^{-1}(\sin x) + \frac{\pi}{6} \right|$ Let $E_x$ denote the equivalence class of $x$ under $R$ . Let $S$ be the set of all $c \in \mathbb{R}$ such that the equation $f(x) = c$ has exactly $2$ solutions in $X$ (i.e., $|E_x| = 2$ for any $x$ with $f(x) = c$ ). If $S$ can be expressed as the union of two disjoint open intervals $(a, b) \cup (c, d)$ with $a < b \le c < d$ , then the value of $a+b+c+d$ is:

Options

A.
$\pi$
B.
$\frac{5\pi}{6}$
C.
$\frac{4\pi}{3}$
D.
$\frac{7\pi}{6}$

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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

Equivalence Class Partitioning|Domain Constraint Intersections|Modulus Inequality Logic

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to partition a domain based on the intersection of inverse trigonometric constraints and modulus inequalities, framing it within the context of equivalence classes.

The primary trap lies in the modulus sign change: students often mechanically apply the piecewise definitions of $\sin^{-1}(\sin x)$ and $\cos^{-1}(\cos x)$ at multiples of $\pi/2$ , but fail to split the domain at the critical points of the modulus argument ( $7\pi/6$ and $11\pi/6$ ).

Ignoring these critical points leads to incorrect linear segments instead of the constant intervals, which drastically alters the range analysis and the cardinality of the equivalence classes.

By requiring the identification of values that yield exactly two solutions, the problem forces a rigorous, full-domain piecewise construction.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Equivalence Class Partitioning
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free