Question

If$\mathbf{f(x) =}\mathbf{\tan}^{\mathbf{- 1}}\left\{ \frac{\mathbf{\log}\left( \frac{\mathbf{e}}{\mathbf{x}^{\mathbf{2}}} \right)}{\mathbf{\log}\mathbf{(}\mathbf{e}\mathbf{x}^{\mathbf{2}}\mathbf{)}} \right\}\mathbf{+}\mathbf{\tan}^{\mathbf{- 1}}\left( \frac{\mathbf{3 + 2}\mathbf{\log}\mathbf{x}}{\mathbf{1 - 6}\mathbf{\log}\mathbf{x}} \right)$, then $\frac{d^{n}y}{dx^{n}}$ is

$(n \geq 1)$

• A. $\tan^{- 1}\{(\log x)^{n}\}$
• B. 0
• C. 1/2
• D. None of these

Answer 1

Answer: (B)

0

Answer 2

We have

$$$= \tan^{- 1}\left( \frac{1 - 2\log x}{1 + 2\log x} \right) + \tan^{- 1}\left( \frac{3 + 2\log x}{1 - 6\log x} \right) = \tan^{- 1}1 - \tan^{- 1}(2\log x) + \tan^{- 1}3 + \tan^{- 1}(2\log x)$ ⇒ $y = \tan^{- 1}1 + \tan^{- 1}3 \Rightarrow \frac{dy}{dx} = 0 \Rightarrow \frac{d^{n}y}{dx^{n}} = 0$.