The number of points, where the function f: R → R, f(x) = |x - 1| |x - 2| |x - 1| + (x - 3)| x ^ 2 - 5x + 4| , is NOT differentiable, is

JEE Main 202229 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 29 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2022 · 29 July · Morning Shift · Mathematics · Continuity and Differentiability

The number of points, where the function f: R → R, f(x) = |x - 1|...

The number of points, where the function

f: R \to R, f\left(x\right) = \left|x - 1\right|\cos \left|x - 2\right|\sin \left|x - 1\right| + \left(x - 3\right)\left|{x}^{2} - 5x + 4\right|

, is NOT differentiable, is

Options

A $1$
B $2$
C $3$
D $4$

Correct answer

B. $2$

Step-by-step solution

Given, $f\left(x\right) = \left|x - 1\right|\cos \left|x - 2\right|\sin \left|x - 1\right| + \left(x - 3\right)\left|{x}^{2} - 5x + 4\right|$
$= \left|x - 1\right|\cos \left|x - 2\right|\sin \left|x - 1\right| + \left(x - 3\right)\left|x - 1\right|\left|x - 4\right|$
$= \left|x - 1\right|\left[\cos \left|x - 2\right|\sin \left|x - 1\right| + \left(x - 3\right)\left|x - 4\right|\right]$
We know that modulus function is not differentiable at values where modulus function is zero
So, we can say that function is non-differentiable at $x = 1$ and $x = 4$ .

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