Mathematics · Methods of Differentiation

JEE Main 2024 — 6 April, Shift 1 — Question 20

Let f:(−∞,∞)−{0}→R\mathrm{f}:(-\infty, \infty)-\{0\} \rightarrow \mathrm{R} be a differentiable function such that f′(1)=lim⁡a→∞a2f(1a)f^{\prime}(1)=\lim _{a \rightarrow \infty} a^{2} f\left(\frac{1}{a}\right). Then lim⁡a→∞a(a+1)2tan⁡−1(1a)+a2−2log⁡ea\lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^{2}-2 \log _{e} a is equal to

  1. Option A:

    32+π4\frac{3}{2}+\frac{\pi}{4}

  2. Option B:

    38+π4\frac{3}{8}+\frac{\pi}{4}

  3. Option C:

    52+π8\frac{5}{2}+\frac{\pi}{8}

    Correct
  4. Option D:

    34+π8\frac{3}{4}+\frac{\pi}{8}

Answer: C

Step-by-step solution

f : (−∞,∞)−{0}→R(-\infty, \infty)-\{0\} \rightarrow R

f′(1)=lim⁡a→∞a2f(1a)f^{\prime}(1)=\lim _{a \rightarrow \infty} a^{2} f\left(\frac{1}{a}\right)

lim⁡a→∞a(a+1)2tan⁡−1(1a)+a2−2ln⁡(a)\lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^{2}-2 \ln (a)

lim⁡a→∞a2((1+1a)2tan⁡−1(1a)+1−2a2ln⁡(a))\lim _{a \rightarrow \infty} a^{2}\left(\frac{\left(1+\frac{1}{a}\right)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+1-\frac{2}{a^{2}} \ln (a)\right)

f(x)=12(1+x)tan⁡−1(x)+1−2x2ln⁡(x)f(x)=\frac{1}{2}(1+x) \tan ^{-1}(x)+1-2 x^{2} \ln (x)

f′(x)=12(1+x1+x2+tan⁡−1(x)+4xln⁡(x))+2x\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2}\left(\frac{1+\mathrm{x}}{1+\mathrm{x}^{2}}+\tan ^{-1}(\mathrm{x})+4 \mathrm{x} \ln (\mathrm{x})\right)+2 \mathrm{x}

f′(1)=12(1+π4)+2\mathrm{f}^{\prime}(1)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)+2

f′(1)=52+π8\mathrm{f}^{\prime}(1)=\frac{5}{2}+\frac{\pi}{8}

Answer key and solution verified before publishing.

Practise Methods of Differentiation

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2024
Subject
Mathematics
Chapter
Methods of Differentiation
Topic
Introduction to Differentiation
Let f :(-∞, ∞)-\ 0\ rightarrow R be a differentiable function such… | JEE Main 2024 PYQ with Solution · DhiX AI