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Mathematics Question on Statistics

If f(x)=cos⁡−1(1−(log⁡ex)21+(log⁡ex)2)f\left(x\right)= \cos^{-1} \left(\frac{1-\left(\log _{e} x\right)^{2}}{1+\left(\log _{e} x\right)^{2}}\right) , then f′(1e)= f'\left(\frac{1}{e}\right) =

A

2

B

0

C

-1

D

e

Answer

e

Explanation

Solution

f(x)=cos⁡−1[1−(log⁡ex)21+(log⁡ex)2]f\left(x\right)= \cos^{-1} \left[\frac{1-\left(\log _{e} x\right)^{2}}{1+\left(\log _{e} x\right)^{2}}\right]
Putting log⁡x=tan⁡θ⇒θ=tan⁡−1(log⁡x) i.e.,x=tan⁡θ \log x = \tan\theta \Rightarrow \theta =\tan^{-1} \left(\log x\right) \: i.e., x = ^{\tan \theta}
f(x)=cos⁡−1(1−tan⁡2θ1+tan⁡2θ)f\left(x\right)= \cos ^{-1} \left(\frac{1-\tan ^{2} \theta }{1+\tan ^{2} \theta }\right)
f(x)=cos⁡−1cos⁡2θ=2θf\left(x\right) = \cos ^{-1} \cos 2\theta = 2\theta
=2tan⁡−1(log⁡x)= 2 \tan^{-1} \left(\log x\right)
Differentiating w.r.t. x, we get
f′(x)=211+(log⁡x)2.1x=2x[1+(log⁡x)2]f'\left(x\right) = 2 \frac{1}{1+ \left(\log x\right)^{2}} . \frac{1}{x} = \frac{2}{x\left[1+ \left(\log x\right)^{2}\right]}
f′(1e)=2(1e)[1+(log⁡(1e))2]=2e(1+(−1)2)f'\left(\frac{1}{e}\right) = \frac{2}{\left(1 e\right) \left[1+\left(\log \left(1 e\right)\right)^{2}\right]} = \frac{2e}{\left(1+\left(-1\right)^{2}\right)}
=2e2=e= \frac{2e}{2} =e