Let f be a twice differentiable function on R. If $\mathrm{f}^{\prime}(0)=4$ and $f(x)+\int_{0}^{x}(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x$, then $(2 \mathrm{a}+1)^{5} \mathrm{a}^{2}$ is equal to

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Differential Equations

Let f be a twice differentiable function on R. If

f^{'} (0) = 4

and

f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x

, then

(2 a + 1)^{5} a^{2}

is equal to

Answer: 8

Because

\frac{d}{d x} \int_{0}^{x} (x - t) f^{'} (t) d t = \int_{0}^{x} f^{'} (t) d t = f (x) - f (0)

, differentiating the given identity once yields

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

, where

R (x) = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x

. Evaluated at

x = 0

, the derivative of

(e^{2 x} + e^{- 2 x}) cos 2 x

(2 - 2) \cdot 1 + 2 \cdot 0 = 0

, so

R^{'} (0) = \frac{2}{a}

, while the left side collapses to

f^{'} (0) + f (0) - f (0) = f^{'} (0) = 4

. Thus

\frac{2}{a} = 4

, giving

a = \frac{1}{2}

, and

(2 a + 1)^{5} a^{2} = 2^{5} \cdot \frac{1}{4} = 8

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Integer answer

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