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Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 1 February, Shift 2 — Question 1

Let f(x)=∣2x2+5∣x∣−3∣,x∈Rf(x)=\left|2 x^{2}+5\right| x|-3|, x \in R. If mm and nn denote the number of points where ff is not continuous and not differentiable respectively, then m+nm+n is equal to :

  1. Option A:

    5

  2. Option B:

    2

  3. Option C:

    0

  4. Option D:

    3

    Correct

Answer: D

Step-by-step solution

f(x)=∣2x2+5∣x∣−3∣f(x)=\left|2 x^{2}+5\right| x|-3|Graph of y=∣2x2+5x−3∣y=\left|2 x^{2}+5 x-3\right|

Number of points of discontinuity =0=m=0=m

Number of points of non-differentiability =3=n=3=\mathrm{n}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 2
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Differentiability

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