If $f(x)=\sin \left(\cos ^{-1}\left(\frac{1-2^{2 x}}{1+2^{2 x}}\right)\right)$ and its first derivative with respect to $x$ is $-\frac{b}{a} \log _{e} 2$ when $x=1$, where $a$ and $b$ are integers, then the minimum value of $\left|a^{2}-b^{2}\right|$ is. ____

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Differentiation

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

and its first derivative with respect to

x

- \frac{b}{a} lo g_{e} 2

when

x = 1

, where

a

and

b

are integers, then the minimum value of

a^{2} - b^{2}

is. ____

Answer: 481

Put

t = 2^{x}

and

θ = cos^{- 1} \frac{1 - t ^{2}}{1 + t ^{2}}

. Since

θ \in [0, π]

sin θ

is non-negative, and

sin θ = 1 - (\frac{1 - t ^{2}}{1 + t ^{2}})^{2} = \frac{2 t}{1 + t ^{2}}, so f (x) = \frac{2 \cdot 2 ^{x}}{1 + 2 ^{2 x}} .

Differentiating with

\frac{d t}{d x} = t ln 2

f^{'} (x) = \frac{2 ( 1 - t ^{2} )}{( 1 + t ^{2} ) ^{2}} t ln 2.

x = 1

t = 2

, so

f^{'} (1) = \frac{2 ( 1 - 4 )}{25} \cdot 2 ln 2 = - \frac{12}{25} ln 2

. Hence

\frac{b}{a} = \frac{12}{25}

, so

a = 25 k

and

b = 12 k

for a non-zero integer

k

, and

a^{2} - b^{2} = ∣ 625 - 144 ∣ k^{2} = 481 k^{2}

. The minimum value is

481

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Integer answer