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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
Let $\quad f:(-\infty, \infty)-\{0\} \rightarrow R$ be a differentiable function such that $f^{\prime}(1)=\lim _{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right)$.

Then $\lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \log _e a$ is equal to

  • A
     $\frac{3}{2}+\frac{\pi}{4}$
  • B
     $\frac{3}{8}+\frac{\pi}{4}$
  • ✓
     $\frac{5}{2}+\frac{\pi}{8}$
  • D
     $\frac{3}{4}+\frac{\pi}{8}$
✓

Answer

Correct option: C.
 $\frac{5}{2}+\frac{\pi}{8}$
c
$ 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) $

$ f^{\prime}(x)=\frac{1}{2}\left(\frac{1+x}{1+x^2}+\tan ^{-1}(x)+4 x \ln (x)\right)+2 x $

$ f^{\prime}(1)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)+2 $

$ f^{\prime}(1)=\frac{5}{2}+\frac{\pi}{8}$

Ans. ($3$)

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