Question

The different coefficient w.r.t.x of the function

(log­_cosx sinx)(log_sinxcosx)^–1 + arc sin$\frac{2x}{1 + x^{2}}$ at x = $\frac{\pi}{4}$ is–

• A. $\frac{8}{\log 2} + \frac{32}{\pi^{2} + 16}$
• B. –$\frac{8}{\log 2} + \frac{32}{\pi^{2} + 16}$
• C. –$\frac{8}{\log 2} - \frac{32}{\pi^{2} + 16}$
• D. None of these

Answer 1

Answer: (B)

–$\frac{8}{\log 2} + \frac{32}{\pi^{2} + 16}$

Answer 2

y = $\left( \frac{{logsin}x}{{logcos}x} \right)$ × $\left( \frac{{logcos}x}{{logsin}x} \right)^{- 1}$ + 2 tan^–1 x

y = $\left( \frac{{logsin}x}{{logcos}x} \right)^{- 2}$ + 2 tan^–1x

Now

$\left( \frac{dy}{dx} \right)_{x = \frac{\pi}{4}}$= $\left\lbrack \frac{{logcos}x \times (\cot x) + \tan x.{logsin}x}{({logcos}x)^{2}} \right\rbrack$+$\frac{2}{1 + x^{2}}$

= $\frac{- 8}{\log 2}$ + $\frac{32}{\pi^{2} + 16}$