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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Limits

Let f:(-∞, ∞)-\ 0\ arrow ℝ be a differentiable function such that...

Let f:(−∞,∞)−{0}→Rf:(-\infty, \infty)-\{0\} \rightarrow \mathbb{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

Options

  1. A32+π4\frac{3}{2}+\frac{\pi}{4}
  2. B34+π8\frac{3}{4}+\frac{\pi}{8}
  3. C38+π4\frac{3}{8}+\frac{\pi}{4}
  4. D52+π8\frac{5}{2}+\frac{\pi}{8}

Correct answer

D. 52+π8\frac{5}{2}+\frac{\pi}{8}

Step-by-step solution

f:(−∞,∞)−{0}→Rf′(1)=lim⁡a→∞a2f(1a)lim⁡a→∞a(a+1)2tan⁡−1(1a)+a2−2ln⁡(a)lim⁡a→∞a2((1+1a)2tan⁡−1(1a)+1−2a2ln⁡(a))f(x)=12(1+x)tan⁡−1(x)+1−2x2ln⁡(x)\begin{aligned} & f:(-\infty, \infty)-\{0\} \rightarrow R \\ & f^{\prime}(1)=\lim _{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right) \\ & \lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \ln (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)=\frac{1}{2}(1+x) \tan ^{-1}(x)+1-2 x^2 \ln (x)\end{aligned}
f′(x)=12(1+x1+x2+tan⁡−1(x)+4xℓ(x))+2xf′(1)=12(1+π4)+2f′(1)=52+π8\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2}\left(\frac{1+\mathrm{x}}{1+\mathrm{x}^2}+\tan ^{-1}(\mathrm{x})+4 \mathrm{x} \ell(\mathrm{x})\right)+2 \mathrm{x} \\ & \mathrm{f}^{\prime}(1)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)+2 \\ & \mathrm{f}^{\prime}(1)=\frac{5}{2}+\frac{\pi}{8} \end{aligned}
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