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Question #41177

Difficulty: 8/10 | 🟢 Free

The Question

Let $P(x)$ be a polynomial of degree 3 satisfying $P(x) + P(-x) = 0$ for all $x \in \mathbb{R}$ , $P(2) = 0$ , and $\lim_{x \to 0} \frac{P(x)}{x} = -4$ . Consider the function $f(x) = \sin^{-1} \left( \frac{P(|x|) + |P(x)|}{2x} \right)$ The domain of $f(x)$ is:

Options

A.
$[-\sqrt{5}, \sqrt{5}] \setminus \{0\}$
B.
$[-\sqrt{5}, -2] \cup [2, \sqrt{5}]$
C.
$[-\sqrt{3}, \sqrt{3}] \setminus \{0\}$
D.
$[-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Domain Constraint Intersections|Polynomial Function Parity|Modulus Sign Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle Modulus Sign Dependency and Parity constraints within composite functions.

A common trap is ignoring the piecewise nature of the modulus and parity, incorrectly assuming $P(|x|) = P(x)$ or $|P(x)| = P(x)$ globally.

Students falling for this will solve $-1 \le x^2 - 4 \le 1$ across the entire domain, leading directly to the trap option D.

The problem forces a strict bifurcation of the domain based on the sign of $x$ and the roots of $P(x)$ , filtering out those who rely on algebraic shortcuts instead of rigorous interval analysis.

Quick Facts About This Question

Question Information

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