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Question #40966

Difficulty: 8/10 | 🟢 Free

The Question

Let $\omega = e^{i 2\pi/5}$ and define the polynomial $P(x) = \prod_{k=1}^4 (x - \omega^k)$ for $x \in \mathbb{R}$ . Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = \sqrt{\max(0, 1 - P(x))}$ and $g: \mathbb{R} \to \mathbb{R}$ be defined as $g(x) = x^2$ . If $m$ and $n$ are the number of points of non-differentiability of the composite functions $f(g(x))$ and $g(f(x))$ respectively, find the value of $m + n$ .

Options

A.
0
B.
1
C.
2
D.
4

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

Polynomial Continuity/Differentiability Link, Roots of Unity Properties, Geometry of Complex Inequalities, Composition Non-Commutativity

Why This Question Is Tricky

Common Mistakes

This question tests the non-commutativity of composition in the context of differentiability and domain restrictions arising from complex roots.

Students often blindly apply the chain rule $H'(x) = f'(g(x))g'(x)$ , assuming $f(g(x))$ inherits singularities from $f(x)$ .

However, the range of $g(x) = x^2$ completely avoids the interval $(-1, 0)$ where $1-P(x) > 0$ , making $f(g(x))$ a constant zero function (hence everywhere differentiable).

Conversely, for $g(f(x)) = f(x)^2$ , students might assume $g'(f(x)) = 2f(x) = 0$ at the boundaries cancels the infinite derivative of $f(x)$ .

This is a $0 \cdot \infty$ trap; evaluating the piecewise polynomial directly reveals non-zero one-sided derivatives, proving non-differentiability at the boundaries.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Limits and Derivatives - Class 11
Concept	Polynomial Continuity/Differentiability Link
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free