If 2 f(x)-3 fleft(frac{1}{x}right)=x,then int | vedclass

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(A) Given equation is $2 f(x)-3 f\left(\frac{1}{x}\right)=x$  ---$(1)$Replacing $x$ with $\frac{1}{x}$ in equation $(1)$,we get:$2 f\left(\frac{1}{x}\

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$-\left(\frac{2+e^2}{5}\right)$

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(A) Given equation is $2 f(x)-3 f\left(\frac{1}{x}\right)=x$  ---$(1)$Replacing $x$ with $\frac{1}{x}$ in equation $(1)$,we get:$2 f\left(\frac{1}{x}\right)-3 f(x)=\frac{1}{x}$  ---$(2)$To eliminate $f\left(\frac{1}{x}\right)$,multiply equation $(1)$ by $2$ and equation $(2)$ by $3$,then add them:$4 f(x)-6 f\left(\frac{1}{x}\right)=2 x$$6 f\left(\frac{1}{x}\right)-9 f(x)=\frac{3}{x}$Adding these two equations:$-5 f(x)=2 x+\frac{3}{x} \implies f(x)=-\frac{2}{5} x-\frac{3}{5 x}$Now,calculate the integral:$\int_1^e f(x) d x = \int_1^e \left(-\frac{2}{5} x-\frac{3}{5 x}\right) d x$$= -\frac{2}{5} \int_1^e x d x - \frac{3}{5} \int_1^e \frac{1}{x} d x$$= -\frac{2}{5} \left[\frac{x^2}{2}\right]_1^e - \frac{3}{5} [\ln x]_1^e$$= -\frac{1}{5} (e^2-1) - \frac{3}{5} (\ln e - \ln 1)$$= -\frac{e^2}{5} + \frac{1}{5} - \frac{3}{5} (1 - 0)$$= -\frac{e^2}{5} - \frac{2}{5} = -\left(\frac{2+e^2}{5}\right)$

If $2 f(x)-3 f\left(\frac{1}{x}\right)=x$,then $\int_1^e f(x) d x=$

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