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Step-by-Step Solution
Step 1: Rewrite the given function using a known identity
We have
f(x) = \sin\!\Big(\cos^{-1}\!\big(\tfrac{1 - 2^{2x}}{1 + 2^{2x}}\big)\Big) .
Recall the identity
\cos\!\big(2 \tan^{-1}(t)\big) = \tfrac{1 - t^2}{1 + t^2} .
By comparison, if
\tfrac{1 - 2^{2x}}{1 + 2^{2x}} = \cos\!\big(2 \theta\big) ,
then we can write
2 \theta = 2\,\tan^{-1}\!\big(2^x\big) ,
hence
\theta = \tan^{-1}\!\big(2^x\big) .
Therefore,
\cos^{-1}\!\bigg(\tfrac{1 - 2^{2x}}{1 + 2^{2x}}\bigg) = 2\,\tan^{-1}\!\big(2^x\big).
Thus
f(x) = \sin\!\big(2\,\tan^{-1}(2^x)\big).
Step 2: Differentiate the function with respect to x
Use the chain rule. First, let
g(x) = 2\,\tan^{-1}\!\big(2^x\big).
Then
f(x) = \sin\!\big(g(x)\big).
We have
f'(x) = \cos\!\big(g(x)\big)\,g'(x).
Next, differentiate
g(x) = 2\,\tan^{-1}\!\big(2^x\big).
Use the derivative of \tan^{-1}(u) which is
\frac{1}{1 + u^2}\,\frac{du}{dx}.
Here,
u = 2^x.
Then
\frac{du}{dx} = 2^x \ln(2).
So,
g'(x) = 2 \cdot \frac{1}{1 + (2^x)^2} \cdot \big(2^x \ln(2)\big)
= \frac{2 \cdot 2^x \ln(2)}{1 + 2^{2x}}.
Hence,
f'(x) = \cos\!\Big(2\,\tan^{-1}\!\big(2^x\big)\Big)\,\frac{2 \cdot 2^x \ln(2)}{1 + 2^{2x}}.
Step 3: Evaluate the derivative at x = 1
Substitute
x = 1
into the expression for
f'(x).
We get
f'(1) = \cos\!\Big(2\,\tan^{-1}(2)\Big)
\times \frac{2 \cdot 2^1 \ln(2)}{1 + 2^{2}}.
Note that
2^1 = 2
and
2^{2} = 4.
So
f'(1) = \cos\!\Big(2\,\tan^{-1}(2)\Big)
\times \frac{2 \times 2 \ln(2)}{1 + 4}
= \cos\!\Big(2\,\tan^{-1}(2)\Big)
\times \frac{4 \ln(2)}{5}.
Step 4: Relate the expression to the given form and simplify
It is known that
\cos\!\big(2\,\tan^{-1}(2)\big)
= \tfrac{1 - (2)^2}{1 + (2)^2}
= \tfrac{1 - 4}{1 + 4}
= -\tfrac{3}{5}.
However, be careful to find the exact value:
\frac{1 - 4}{1 + 4} = \frac{-3}{5},
so indeed
\cos\!\Big(2\,\tan^{-1}(2)\Big) = -\tfrac{3}{5}.
Thus
f'(1)
= \Big(-\tfrac{3}{5}\Big)\times \frac{4 \ln(2)}{5}
= -\tfrac{12}{25}\ln(2).
We compare this with the given form
-\tfrac{b}{a}\ln(2).
Clearly,
\tfrac{b}{a} = \tfrac{12}{25},
so
a = 25
and
b = 12.
Step 5: Compute the desired absolute difference
We require
\big|a^2 - b^2\big|
= \big|25^2 - 12^2\big|
= \big|625 - 144\big|
= 481.
Thus, the minimum value of
\big|a^2 - b^2\big|
is
481.
