BITSAT 2021
Exam BITSAT
Year 2021
Subject Mathematics
Chapter Differentiation BITSAT 2021 · Mathematics · Differentiation
(d)/(d x) ( ^ -1 (1+ (x)/(2))/(1- (x)/(2)) ) is equal to
\frac{d}{d x}
( tan − 1 1 + cos x 2 1 − cos x 2 ) \left(\tan ^{-1} \sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right) ( tan − 1 1 − c o s 2 x 1 + c o s 2 x ) is equal to
Options A − 1 4 -\frac{1}{4} − 4 1 B 1 4 \frac{1}{4} 4 1 C − 1 2 -\frac{1}{2} − 2 1 D 1 2 \frac{1}{2} 2 1 Correct answer A. − 1 4 -\frac{1}{4} − 4 1
Step-by-step solution
We have
d d x ( tan − 1 1 + cos x 2 1 − cos x 2 ) = d d x [ tan − 1 1 + ( 2 cos 2 x 4 − 1 ) 1 − ( 1 − 2 sin 2 x 4 ) ] = d d x ( tan − 1 2 cos 2 x 4 2 sin 2 x 4 ) = d d x ( tan − 1 cot 2 x 4 ) = d d x ( tan − 1 cot x 4 ) = d d x ( tan − 1 tan ( π 2 − x 4 ) ) = d d x ( π 2 − x 4 ) = 0 − 1 4 = − 1 4 \frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right)=\frac{d}{d x}\left[\tan ^{-1} \sqrt{\frac{1+\left(2 \cos ^{2} \frac{x}{4}-1\right)}{1-\left(1-2 \sin ^{2} \frac{x}{4}\right)}}\right]=\frac{d}{d x}\left(\tan ^{-1} \sqrt{\left.\frac{2 \cos ^{2} \frac{x}{4}}{2 \sin ^{2} \frac{x}{4}}\right)}\right.=\frac{d}{d x}\left(\tan ^{-1} \sqrt{\cot ^{2} \frac{x}{4}}\right)=\frac{d}{d x}\left(\tan ^{-1} \cot \frac{x}{4}\right)=\frac{d}{d x}\left(\tan ^{-1} \tan \left(\frac{\pi}{2}-\frac{x}{4}\right)\right)=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x}{4}\right)=0-\frac{1}{4}=-\frac{1}{4} d x d ( tan − 1 1 − c o s 2 x 1 + c o s 2 x ) = d x d [ tan − 1 1 − ( 1 − 2 s i n 2 4 x ) 1 + ( 2 c o s 2 4 x − 1 ) ] = d x d ( tan − 1 2 s i n 2 4 x 2 c o s 2 4 x ) = d x d ( tan − 1 cot 2 4 x ) = d x d ( tan − 1 cot 4 x ) = d x d ( tan − 1 tan ( 2 π − 4 x ) ) = d x d ( 2 π − 4 x ) = 0 − 4 1 = − 4 1
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