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JEE Main 202125 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2021 · 25 February · Morning Shift · Mathematics · Continuity and Differentiability

The number of points, at which the function f(x) = |2x + 1| - 3|x +...

The number of points, at which the function f(x)=∣2x+1∣−3∣x+2∣+∣x2+x−2∣,x∈Rf\left(x\right) = \left|2x + 1\right| - 3\left|x + 2\right| + \left|{x}^{2} + x - 2\right|, x \in R is not differentiable, is

Correct answer

2

Step-by-step solution

f(x)=∣2x+1∣−3∣x+2∣+∣x2+x−2∣f\left(x\right) = \left|2x + 1\right| - 3\left|x + 2\right| + \left|{x}^{2} + x - 2\right|

=∣2x+1∣−3∣x+2∣+∣x+2∣∣x−1∣= \left|2x + 1\right| - 3\left|x + 2\right| + \left|x + 2\right|\left|x - 1\right|

=∣2x+1∣+∣x+2∣(∣x−1∣−3)= \left|2x + 1\right| + \left|x + 2\right|\left(\left|x - 1\right| - 3\right)

Critical points are x=−12,−2,−1x = \dfrac{- 1}{2}, - 2, - 1

but x=−2x = - 2 is making a zero twice in product so, points of non differentiability are x=−12x = \dfrac{- 1}{2} and x=−1x = - 1

∴∴ Number of points of non-differentiability =2= 2

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