Question

\(f(x)\) is an \(n^{th}\) degree polynomial and \(\alpha_1,\alpha_2,\ldots,\alpha_n\) are distinct zeros of \(f(x)\). \(g(x)\) is a polynomial having three zeros common with the zeros of \(f(x)\). Assertion (A): \(|f(x)|g(x)\) is continuous and differentiable at all \(\alpha_i\).
Reason (R): \[ \lim_{x\to a}\frac{|x-a|}{x-a} \] does not exist and \[ \lim_{x\to a}|x-a|=0. \] Choose the correct option.}

• A. Both (A) and (R) are correct, (R) is the correct explanation of (A)
• B. Both (A) and (R) are correct, (R) is not the correct explanation of (A)
• C. (A) is correct, but (R) is not correct
• D. (A) is not correct, but (R) is correct

Hint:

A function involving modulus is always continuous, but it may not be differentiable where the expression inside the modulus changes sign.

Answer 1

The Correct Option is D

Concept: The modulus function can destroy differentiability at zeros of a function. A polynomial is continuous everywhere, but \(|f(x)|\) may fail to be differentiable where \(f(x)=0\).

Step 1: Examine the assertion.
Given \[ |f(x)|g(x). \] Since \(f(x)\) and \(g(x)\) are polynomials, \[ |f(x)|g(x) \] is continuous everywhere. However, differentiability at each zero \(\alpha_i\) is not guaranteed. If \(f(x)\) changes sign at a simple root, \[ |f(x)| \] develops a sharp corner at that point. Therefore differentiability may fail. Hence Assertion (A) is false.

Step 2: Examine the reason statement.
Consider \[ \lim_{x\to a} \frac{|x-a|}{x-a}. \] For \(x>a\), \[ \frac{|x-a|}{x-a}=1. \] For \(x<a\), \[ \frac{|x-a|}{x-a}=-1. \] Since LHL and RHL are different, \[ \lim_{x\to a} \frac{|x-a|}{x-a} \] does not exist. Also, \[ \lim_{x\to a}|x-a|=0. \] Thus Reason (R) is true.

Step 3: Choose the correct conclusion.
Assertion is false. Reason is true. Therefore, \[ \boxed{\text{(A) is not correct, but (R) is correct}}. \]