Simplify the inner trigonometric expression
Given
f ( x ) = cos ( 2 tan − 1 ( sin ( cot − 1 1 − x x ) ) ) , 0 < x < 1. f(x)=\cos\left(2\tan^{-1}\left(\sin\left(\cot^{-1}\sqrt{\frac{1-x}{x}}\right)\right)\right), \qquad 0<x<1. f ( x ) = cos ( 2 tan − 1 ( sin ( cot − 1 x 1 − x ) ) ) , 0 < x < 1. Let
θ = cot − 1 1 − x x . \theta=\cot^{-1}\sqrt{\frac{1-x}{x}}. θ = cot − 1 x 1 − x . Then
cot θ = 1 − x x . \cot\theta=\sqrt{\frac{1-x}{x}}. cot θ = x 1 − x . Using
sin θ = 1 1 + cot 2 θ , \sin\theta=\frac{1}{\sqrt{1+\cot^2\theta}}, sin θ = 1 + cot 2 θ 1 , we get
sin θ = 1 1 + 1 − x x = 1 1 x = x . \sin\theta=\frac{1}{\sqrt{1+\frac{1-x}{x}}}
=\frac{1}{\sqrt{\frac{1}{x}}}
=\sqrt{x}. sin θ = 1 + x 1 − x 1 = x 1 1 = x . So the function becomes
f ( x ) = cos ( 2 tan − 1 ( x ) ) . f(x)=\cos\left(2\tan^{-1}(\sqrt{x})\right). f ( x ) = cos ( 2 tan − 1 ( x ) ) .
Use the identity for cos ( 2 tan − 1 t ) \cos(2\tan^{-1} t) cos ( 2 tan − 1 t )
Recall,
cos ( 2 tan − 1 t ) = 1 − t 2 1 + t 2 . \cos(2\tan^{-1} t)=\frac{1-t^2}{1+t^2}. cos ( 2 tan − 1 t ) = 1 + t 2 1 − t 2 . Here t = x t=\sqrt{x} t = x
f ( x ) = 1 − x 1 + x . f(x)=\frac{1-x}{1+x}. f ( x ) = 1 + x 1 − x .
Differentiate f ( x ) f(x) f ( x )
f ( x ) = 1 − x 1 + x . f(x)=\frac{1-x}{1+x}. f ( x ) = 1 + x 1 − x . Using the quotient rule,
f ′ ( x ) = ( 1 + x ) ( − 1 ) − ( 1 − x ) ( 1 ) ( 1 + x ) 2 = − 1 − x − 1 + x ( 1 + x ) 2 = − 2 ( 1 + x ) 2 . f'(x)=\frac{(1+x)(-1)-(1-x)(1)}{(1+x)^2}
=\frac{-1-x-1+x}{(1+x)^2}
=\frac{-2}{(1+x)^2}. f ′ ( x ) = ( 1 + x ) 2 ( 1 + x ) ( − 1 ) − ( 1 − x ) ( 1 ) = ( 1 + x ) 2 − 1 − x − 1 + x = ( 1 + x ) 2 − 2 .
Compute f ( x ) 2 f(x)^2 f ( x ) 2
( f ( x ) ) 2 = ( 1 − x 1 + x ) 2 = ( 1 − x ) 2 ( 1 + x ) 2 . (f(x))^2=\left(\frac{1-x}{1+x}\right)^2=\frac{(1-x)^2}{(1+x)^2}. ( f ( x ) ) 2 = ( 1 + x 1 − x ) 2 = ( 1 + x ) 2 ( 1 − x ) 2 .
Check each option
Option A
( 1 − x ) 2 f ′ ( x ) − 2 ( f ( x ) ) 2 = ( 1 − x ) 2 ( − 2 ( 1 + x ) 2 ) − 2 ⋅ ( 1 − x ) 2 ( 1 + x ) 2 = − 4 ( 1 − x ) 2 ( 1 + x ) 2 ≠ 0. (1-x)^2f'(x)-2(f(x))^2
= (1-x)^2\left(\frac{-2}{(1+x)^2}\right)-2\cdot\frac{(1-x)^2}{(1+x)^2}
=\frac{-4(1-x)^2}{(1+x)^2}\neq 0. ( 1 − x ) 2 f ′ ( x ) − 2 ( f ( x ) ) 2 = ( 1 − x ) 2 ( ( 1 + x ) 2 − 2 ) − 2 ⋅ ( 1 + x ) 2 ( 1 − x ) 2 = ( 1 + x ) 2 − 4 ( 1 − x ) 2 = 0. So, A is false .
Option B
( 1 + x ) 2 f ′ ( x ) + 2 ( f ( x ) ) 2 = ( 1 + x ) 2 ( − 2 ( 1 + x ) 2 ) + 2 ⋅ ( 1 − x ) 2 ( 1 + x ) 2 = − 2 + 2 ( 1 − x ) 2 ( 1 + x ) 2 . (1+x)^2f'(x)+2(f(x))^2
= (1+x)^2\left(\frac{-2}{(1+x)^2}\right)+2\cdot\frac{(1-x)^2}{(1+x)^2}
=-2+2\frac{(1-x)^2}{(1+x)^2}. ( 1 + x ) 2 f ′ ( x ) + 2 ( f ( x ) ) 2 = ( 1 + x ) 2 ( ( 1 + x ) 2 − 2 ) + 2 ⋅ ( 1 + x ) 2 ( 1 − x ) 2 = − 2 + 2 ( 1 + x ) 2 ( 1 − x ) 2 . This is not identically zero. So, B is false .
Option C
( 1 − x ) 2 f ′ ( x ) + 2 ( f ( x ) ) 2 = ( 1 − x ) 2 ( − 2 ( 1 + x ) 2 ) + 2 ⋅ ( 1 − x ) 2 ( 1 + x ) 2 = 0. (1-x)^2f'(x)+2(f(x))^2
= (1-x)^2\left(\frac{-2}{(1+x)^2}\right)+2\cdot\frac{(1-x)^2}{(1+x)^2}=0. ( 1 − x ) 2 f ′ ( x ) + 2 ( f ( x ) ) 2 = ( 1 − x ) 2 ( ( 1 + x ) 2 − 2 ) + 2 ⋅ ( 1 + x ) 2 ( 1 − x ) 2 = 0. So, C is true .
Option D
( 1 + x ) 2 f ′ ( x ) − 2 ( f ( x ) ) 2 = − 2 − 2 ( 1 − x ) 2 ( 1 + x ) 2 ≠ 0. (1+x)^2f'(x)-2(f(x))^2
= -2-2\frac{(1-x)^2}{(1+x)^2}\neq 0. ( 1 + x ) 2 f ′ ( x ) − 2 ( f ( x ) ) 2 = − 2 − 2 ( 1 + x ) 2 ( 1 − x ) 2 = 0. So, D is false .
Final Answer
The correct option is
C \boxed{\text{C}} C