Let f : (-∞, ∞) - {0} be a differentiable function such that f'(1) = lim(a → ∞) a^2 f(1/a).

Question

Let \(\mathrm{f}:(-\infty, \infty)-\{0\} \rightarrow \mathrm{R}\) 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 

(1) \(\frac{3}{2}+\frac{\pi}{4}\)

(2) \(\frac{3}{8}+\frac{\pi}{4}\)

(3) \(\frac{5}{2}+\frac{\pi}{8}\)

(4) \(\frac{3}{4}+\frac{\pi}{8}\)

Answer 1

Correct option is (3) \(\frac{5}{2}+\frac{\pi}{8}\)

\(\mathrm{f}:(-\infty, \infty)-\{0\} \rightarrow \mathrm{R}\)

\(f^{\prime}(1)=\lim _{a \rightarrow \infty} a^{2} f\left(\frac{1}{a}\right)\)

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

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

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

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

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

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