Question

Let $f(x)=\cos \left(2 \tan ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1-x}{x}}\right)\right)$, $0 < x < 1 $. Then:

• A. $(1- x )^{2} f '( x )-2( f ( x ))^{2}=0$
• B. $(1+x)^{2} f'(x)+2(f(x))^{2}=0$
• C. $(1- x )^{2} f '( x )+2( f ( x ))^{2}=0$
• D. $(1+x)^{2} f'(x)-2(f(x))^{2}=0$

Answer 1

Answer: (C)

$ f ( x )=\cos \left(2 \tan ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1- x }{ x }}\right)\right) $
$\cot ^{-1} \sqrt{\frac{1- x }{ x }}=\sin ^{-1} \sqrt{ x } $
or $ f ( x )=\cos \left(2 \tan ^{-1} \sqrt{ x }\right)$
$=\cos \tan ^{-1}\left(\frac{2 \sqrt{ x }}{1- x }\right) $
$f ( x )=\frac{1- x }{1+ x }$
Now $f'(x)=\frac{-2}{(1+x)^{2}}$
or $f'(x)(1-x)^{2}=-2\left(\frac{1-x}{1+x}\right)^{2}$
or $(1-x)^{2} f'(x)+2(f(x))^{2}=0$