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Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 30 January, Shift 1 — Question 7

Let g:R→R\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R} be a non constant twice differentiable such that g′(12)=g′(32)g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right). If a real valued function f is defined as f(x)=12[g(x)+g(2−x)]f(x)=\frac{1}{2}[g(x)+g(2-x)], then :

  1. Option A:

    f′(x)=0f^{\prime}(x)=0 for atleast two xx in (0,2)(0,2)

    Correct
  2. Option B:

    f′′(x)=0f^{\prime \prime}(x)=0 for exactly one xx in (0,1)(0,1)

  3. Option C:

    f′(x)=0\mathrm{f}^{\prime}(\mathrm{x})=0 for no x in (0,1)(0,1)

  4. Option D:

    f′(32)+f′(12)=1f^{\prime}\left(\frac{3}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)=1

Answer: A

Step-by-step solution

Given f(x)=12[g(x)+g(2−x)]f(x) = \frac{1}{2}[g(x) + g(2-x)]. Differentiate: f′(x)=12[g′(x)−g′(2−x)]f'(x) = \frac{1}{2}[g'(x) - g'(2-x)]. Given g′(12)=g′(32)g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right). Then f′(12)=12[g′(1/2)−g′(3/2)]=0f'\left(\frac{1}{2}\right) = \frac{1}{2}[g'(1/2) - g'(3/2)] = 0

and f′(32)=12[g′(3/2)−g′(1/2)]=0f'\left(\frac{3}{2}\right) = \frac{1}{2}[g'(3/2) - g'(1/2)] = 0. Also f′(1)=12[g′(1)−g′(1)]=0f'(1) = \frac{1}{2}[g'(1) - g'(1)] = 0. Thus f′(x)=0f'(x) = 0 at x=12,1,32x = \frac{1}{2}, 1, \frac{3}{2},

so at least two zeros in (0,2)(0,2). Hence option A is correct.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
30 January, Shift 1
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Intermediate value theorem, Root Location Theorem

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