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Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 31 January, Shift 2 — Question 14

Consider the function f:(0,∞)→Rf:(0, \infty) \rightarrow \mathrm{R} defined by f(x)=e−∣log⁡ex∣f(x)=e^{-\left|\log _{e} x\right|}. If mm and nn be respectively the number of points at which ff is not continuous and ff is not differentiable, then m+n\mathrm{m}+\mathrm{n} is

  1. Option A:

    0

  2. Option B:

    3

  3. Option C:

    1

    Correct
  4. Option D:

    2

Answer: C

Step-by-step solution

Given

f:(0,∞)→R,f(x)=e−∣ln⁡x∣f:(0,\infty)\to \mathbb{R}, \qquad f(x)=e^{-|\ln x|}

First remove modulus by cases.

Case 1: x≥1 x\ge 1

Then ln⁡x≥0\ln x \ge 0, so

f(x)=e−ln⁡x=1xf(x)=e^{-\ln x}=\frac{1}{x}

Case 2: 0<x<10<x<1

Then ln⁡x<0\ln x<0, so

f(x)=eln⁡x=xf(x)=e^{\ln x}=x

Thus

f(x)={x,0<x<1,1x,x≥1.f(x)= \begin{cases} x, & 0<x<1,\\[4pt] \dfrac{1}{x}, & x\ge 1. \end{cases}

Continuity at x=1:x=1:

lim⁡x→1−f(x)=1\lim_{x\to1^-} f(x)=1 lim⁡x→1+f(x)=1\lim_{x\to1^+} f(x)=1 f(1)=1f(1)=1

Hence continuous at x=1x=1.

So,

m=0.m=0.

Differentiability at x=1: x=1:

Left derivative:

ddx(x)=1\frac{d}{dx}(x)=1

Right derivative:

ddx(1x)=−1\frac{d}{dx}\left(\frac1x\right)=-1

Since

1≠−1,1\ne -1,

not differentiable at x=1x=1.

Elsewhere both pieces are smooth.

So,

n=1.n=1. m+n=1\boxed{m+n=1}

Answer key and solution verified before publishing.

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

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