Let 2f(x) + f(-x) = frac{1}{x} sin left( x - | vedclass

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(A) Given $2f(x) + f(-x) = \frac{1}{x} \sin \left( x - \frac{1}{x} \right)$  .....$(1)$Replacing $x$ with $-x$,we get $2f(-x) + f(x) = \frac{1}{-x} \s

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(A) Given $2f(x) + f(-x) = \frac{1}{x} \sin \left( x - \frac{1}{x} ight)$  .....$(1)$Replacing $x$ with $-x$,we get $2f(-x) + f(x) = \frac{1}{-x} \sin \left( -x - \frac{1}{-x} ight) = \frac{1}{-x} \sin \left( - \left( x - \frac{1}{x} ight) ight) = \frac{1}{x} \sin \left( x - \frac{1}{x} ight)$  .....$(2)$Subtracting $(2)$ from $2 	imes (1)$,we get $4f(x) + 2f(-x) - 2f(-x) - f(x) = \frac{2}{x} \sin \left( x - \frac{1}{x} ight) - \frac{1}{x} \sin \left( x - \frac{1}{x} ight)$$3f(x) = \frac{1}{x} \sin \left( x - \frac{1}{x} ight) \implies f(x) = \frac{1}{3x} \sin \left( x - \frac{1}{x} ight)$Now,$I = \int_{1/e}^{e} f(x) dx = \frac{1}{3} \int_{1/e}^{e} \frac{1}{x} \sin \left( x - \frac{1}{x} ight) dx$Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,here $a+b = 1/e + e$. However,a simpler substitution $x = 1/t$ works better.Let $x = 1/t$,then $dx = -1/t^2 dt$. When $x = 1/e, t = e$ and when $x = e, t = 1/e$.$I = \frac{1}{3} \int_{e}^{1/e} \frac{1}{1/t} \sin \left( \frac{1}{t} - t ight) \left( -\frac{1}{t^2} ight) dt = \frac{1}{3} \int_{e}^{1/e} t \sin \left( - \left( t - \frac{1}{t} ight) ight) \left( -\frac{1}{t^2} ight) dt$$I = \frac{1}{3} \int_{e}^{1/e} \frac{1}{t} \sin \left( t - \frac{1}{t} ight) dt = - \frac{1}{3} \int_{1/e}^{e} \frac{1}{t} \sin \left( t - \frac{1}{t} ight) dt = -I$$2I = 0 \implies I = 0$.

Let $2f(x) + f(-x) = \frac{1}{x} \sin \left( x - \frac{1}{x} \right)$. Then the value of $\int_{1/e}^{e} f(x) dx$ is -

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