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

(A) $(1-x)^{2} f^{\prime}(x)+2(f(x))^{2}=0$

(B) $(1-x)^{2} f^{\prime}(x)-2(f(x))^{2}=0$

(C) $(1+x)^{2} f^{\prime}(x)+2(f(x))^{2}=0$

(D) $(1+x)^{2} f^{\prime}(x)-2(f(x))^{2}=0$

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Inverse Trigonometric Functions

Let

f (x) = cos (2 tan^{- 1} sin (cot^{- 1} \frac{1 - x}{x})), 0 < x < 1

. Then:

Answer: (A)

Put

θ = cot^{- 1} \frac{1 - x}{x}

, so

cot^{2} θ = \frac{1 - x}{x}

and

sin^{2} θ = \frac{1}{1 + cot ^{2} θ} = x .

For

0 < x < 1

the angle

θ

is acute, so

sin θ = x

and, with

cos (2 tan^{- 1} t) = \frac{1 - t ^{2}}{1 + t ^{2}}

f (x) = cos (2 tan^{- 1} x) = \frac{1 - x}{1 + x} .

Differentiating,

f^{'} (x) = - \frac{2}{( 1 + x ) ^{2}}

, and therefore

(1 - x)^{2} f^{'} (x) + 2 (f (x))^{2} = - \frac{2 ( 1 - x ) ^{2}}{( 1 + x ) ^{2}} + \frac{2 ( 1 - x ) ^{2}}{( 1 + x ) ^{2}} = 0,

f

satisfies

(1 - x)^{2} f^{'} (x) + 2 (f (x))^{2} = 0

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option