Q92 / 155 · WBJEE 2024 · Mathematics
If $y=\tan ^{-1}\left[\frac{\log _e\left(\frac{e}{x^2}\right)}{\log _e\left(e x^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log _e x}{1-6 \cdot \log _e x}\right]$, then $\frac{d^2 y}{d x^2}=$
- A2
- B1
- Correct answerC0
- D-1
Solution
\(\begin{aligned} & f(x)=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(e x^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right] \\ & f(x)=\tan ^{-1}\left[\frac{1-2 \log x}{1+2 \log x}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-3 \cdot 2 \log x}\right] \\ & f(x)=\tan ^{-1}(1)-\tan ^{-1}(2 \log x)+\tan ^{-1}(3)+\tan ^{-1}(2 \log x) \\ & f(x)=\tan ^{-1}(1)+\tan ^{-1}(3) \\ & \therefore f^{\prime \prime}(x)=0\end{aligned}\)
Hence, the correct answer is option: 'C'.