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

1. $\frac{6 \log 2-7}{32}$
2. $\frac{6 \log 2-17}{32}$
3. $\frac{6 \log 2-1}{32}$
4. $\frac{6 \log 2-7}{16}$

Solution

$5 f(x)+3 f\left(\frac{1}{x}\right)=2-\frac{1}{x}$ ...(i) Put $x=\frac{1}{x}$ in above equation: $5 f\left(\frac{1}{x}\right)+3 f(x)=2-x$ ...(ii) Equation (ii) $\times 5-$ (i) $\times 3$ : $\begin{aligned} & \Rightarrow 16 f\left(\frac{1}{x}\right)=10-5 x-6+\frac{3}{x} \\ & \Rightarrow f\left(\frac{1}{x}\right)=\frac{1}{16}\left[4-5 x+\frac{3}{x}\right]\end{aligned}$
$=\frac{1}{16}\left[4 x-\frac{5 x^2}{2}+3 \ln x\right]_1^2=\frac{6 \ln 2-7}{32}$

Asked in: AP EAMCET 2023 (17 May Shift 2)

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