If 5 f(x)+3 fleft(frac{1}{x}right)=2-frac{1}{ | vedclass

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(A) Given equation: $5 f(x)+3 f\left(\frac{1}{x}\right)=2-\frac{1}{x}$ ...$(i)$Replace $x$ with $\frac{1}{x}$ in equation $(i)$:$5 f\left(\frac{1}{x}\

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$\frac{6 \log 2-7}{32}$

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(A) Given equation: $5 f(x)+3 f\left(\frac{1}{x}\right)=2-\frac{1}{x}$ ...$(i)$Replace $x$ with $\frac{1}{x}$ in equation $(i)$:$5 f\left(\frac{1}{x}\right)+3 f(x)=2-x$ ...(ii)To eliminate $f(x)$,multiply equation (ii) by $5$ and equation $(i)$ by $3$:$25 f\left(\frac{1}{x}\right)+15 f(x)=10-5 x$ ...(iii)$15 f(x)+9 f\left(\frac{1}{x}\right)=6-\frac{3}{x}$ ...(iv)Subtract equation (iv) from equation (iii):$(25-9) f\left(\frac{1}{x}\right) = (10-5 x) - (6-\frac{3}{x})$$16 f\left(\frac{1}{x}\right) = 4-5 x+\frac{3}{x}$$f\left(\frac{1}{x}\right) = \frac{1}{16} \left(4-5 x+\frac{3}{x}\right)$Now,calculate the integral:$\int_1^2 f\left(\frac{1}{x}\right) d x = \int_1^2 \frac{1}{16} \left(4-5 x+\frac{3}{x}\right) d x$$= \frac{1}{16} \left[4 x-\frac{5 x^2}{2}+3 \ln |x|\right]_1^2$$= \frac{1}{16} \left[ \left(4(2)-\frac{5(2)^2}{2}+3 \ln 2\right) - \left(4(1)-\frac{5(1)^2}{2}+3 \ln 1\right) \right]$$= \frac{1}{16} \left[ (8-10+3 \ln 2) - (4-2.5+0) \right]$$= \frac{1}{16} \left[ (-2+3 \ln 2) - 1.5 \right]$$= \frac{1}{16} \left[ 3 \ln 2 - 3.5 \right] = \frac{1}{16} \left[ 3 \ln 2 - \frac{7}{2} \right]$$= \frac{6 \ln 2-7}{32}$

If $5 f(x)+3 f\left(\frac{1}{x}\right)=2-\frac{1}{x}, x \neq 0$,then $\int_1^2 f\left(\frac{1}{x}\right) d x=$

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