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Question: Sum of maximum and minimum values of (sin<sup>–1</sup> x)<sup>4</sup> + (cos<sup>–1</sup> x)<sup>4<...

Sum of maximum and minimum values of

(sin–1 x)4 + (cos–1 x)4 is–

A

137π2128\frac{137\pi^{2}}{128}

B

π417\frac{\pi^{4}}{17}

C

17π416\frac{17\pi^{4}}{16}

D

137π4128\frac{137\pi^{4}}{128}

Answer

137π4128\frac{137\pi^{4}}{128}

Explanation

Solution

S=tan⁡−113+tan⁡−129+tan⁡−1433+....∞S = \tan^{- 1}\frac{1}{3} + \tan^{- 1}\frac{2}{9} + \tan^{- 1}\frac{4}{33} + ....\infty

tan⁡−12n−11+22n−1=tan⁡−12n−2n−11+2n.2n−1\tan^{- 1}\frac{2^{n - 1}}{1 + 2^{2n - 1}} = \tan^{- 1}\frac{2^{n} - 2^{n - 1}}{1 + 2^{n}.2^{n - 1}}

= tan⁡−12n−tan⁡−12n−1\tan^{- 1}2^{n} - \tan^{- 1}2^{n - 1}

Sn = Sum of n term of the series

(tan⁡−12−tan⁡−11)+(tan⁡−122−tan⁡−12)+(tan⁡−123−tan⁡−122)\left( \tan^{- 1}2 - \tan^{- 1}1 \right) + \left( \tan^{- 1}2^{2} - \tan^{- 1}2 \right) + \left( \tan^{- 1}2^{3} - \tan^{- 1}2^{2} \right)

…………. (tan⁡−12n−tan⁡−12n−1)\left( \tan^{- 1}2^{n} - \tan^{- 1}2^{n - 1} \right)

= tan–1 2n – tan–1 1

S=lim⁡x→∞Sn=lim⁡n→∞tan⁡–12n−π/4S = \lim_{x \rightarrow \infty}S_{n} = \lim_{n \rightarrow \infty}\tan^{–1}2^{n} - \pi/4

= π2−π4=π4\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}