Question

If $y=Ta{n}^{-1}\left(\frac{\log \left(e/x^2\right)}{\log e{x}^2}\right)+Ta{n}^{-1}\left(\frac{3+2\log x}{1-6\log x}\right)$ then ${\left(\frac{dy}{dx}\right)}_{x=2}+{\left(\frac{dy}{dx}\right)}_{x=3}$.

• A. $-6$
• B. $2$
• C. $0$
• D. $-2$

Answer 1

Answer: (C)

$0$

$y={\tan }^{-1}\left(\frac{\log \frac{e}{x^2}}{\log e{x}^2}\right)+{\tan }^{-1}\left(\frac{3+2\log x}{1-6\log x}\right)={\tan }^{-1}\left(\frac{\log e-\log x^2}{\log e+\log x^2}\right)+{\tan }^{-1}\left[\frac{3+2\log x}{1-3-2\log x}\right]={\tan }^{-1}\left[\frac{1-\log x^2}{1+\log x^2}\right]+{\tan }^{-1}3+{\tan }^{-1}2\log x={\tan }^{-1}1-{\tan }^{-1}\log {\left(x\right)}^2+{\tan }^{-1}3+{\tan }^{-1}\log {\left(x\right)}^2={\tan }^{-1}1+{\tan }^{-1}3y={\tan }^{-1}1+{\tan }^{-1}3$ is constant

So, $\frac{dy}{dx}=0$

Answer C