Let f be a twice differentiable function on m | vedclass

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(B) Given the equation: $f(x)+\int_{0}^{x}(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x$ $(i)$.At $x=0$,$f(0) + 0 = (1+

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$8$

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(B) Given the equation: $f(x)+\int_{0}^{x}(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x$ $(i)$.At $x=0$,$f(0) + 0 = (1+1)\cos(0) + 0$,so $f(0)=2$.Using the Leibniz rule for differentiation under the integral sign on the integral $\int_{0}^{x}(x-t) f^{\prime}(t) d t$,we get $\frac{d}{dx} \int_{0}^{x}(x-t) f^{\prime}(t) d t = \int_{0}^{x} f^{\prime}(t) d t + (x-x)f^{\prime}(x) = f(x) - f(0)$.Differentiating equation $(i)$ with respect to $x$:$f^{\prime}(x) + f(x) - f(0) = \frac{d}{dx} [2 \cosh(2x) \cos(2x)] + \frac{2}{a}$.$f^{\prime}(x) + f(x) - 2 = 4 \sinh(2x) \cos(2x) - 4 \cosh(2x) \sin(2x) + \frac{2}{a}$.At $x=0$,$f^{\prime}(0) + f(0) - 2 = 0 - 0 + \frac{2}{a}$.Given $f^{\prime}(0)=4$ and $f(0)=2$,we have $4 + 2 - 2 = \frac{2}{a}$,which implies $4 = \frac{2}{a}$,so $a = \frac{1}{2}$.Finally,$(2a+1)^5 a^2 = (2(\frac{1}{2})+1)^5 (\frac{1}{2})^2 = (2)^5 \cdot \frac{1}{4} = 32 \cdot \frac{1}{4} = 8$.

Let $f$ be a twice differentiable function on $\mathbb{R}$. If $f^{\prime}(0)=4$ 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 $(2 a+1)^{5} a^{2}$ is equal to $\dots\dots$

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