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(B) Given $e^{-x} f(x) = 2 + \int_0^x \sqrt{t^4 + 1} \, dt \dots (i)$At $x = 0$,$e^0 f(0) = 2 + \int_0^0 \sqrt{t^4 + 1} \, dt \implies f(0) = 2$.Since

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$1/3$

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(B) Given $e^{-x} f(x) = 2 + \int_0^x \sqrt{t^4 + 1} \, dt \dots (i)$At $x = 0$,$e^0 f(0) = 2 + \int_0^0 \sqrt{t^4 + 1} \, dt \implies f(0) = 2$.Since $f(0) = 2$,it follows that $f^{-1}(2) = 0$.We know that $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$.For $y = 2$,$(f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(0)}$.Differentiating equation $(i)$ with respect to $x$ using the product rule and the Fundamental Theorem of Calculus:$-e^{-x} f(x) + e^{-x} f'(x) = \sqrt{x^4 + 1}$.At $x = 0$:$-e^0 f(0) + e^0 f'(0) = \sqrt{0^4 + 1}$$-1(2) + 1(f'(0)) = 1$$-2 + f'(0) = 1 \implies f'(0) = 3$.Therefore,$(f^{-1})'(2) = \frac{1}{f'(0)} = \frac{1}{3}$.

Let $f$ be a real-valued function defined on the interval $(-1, 1)$ such that $e^{-x} f(x) = 2 + \int_0^x \sqrt{t^4 + 1} \, dt$,for all $x \in (-1, 1)$ and let $f^{-1}$ be the inverse function of $f$. Then $(f^{-1})'(2)$ is equal to

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