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

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(B) Given the equation: $2f(x) - 3f\left( \frac{1}{x} \right) = x$ ... $(i)$Replacing $x$ with $\frac{1}{x}$ in $(i)$,we get:$2f\left( \frac{1}{x} \ri

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$-\frac{3}{5}(1 + \ln 2)$

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(B) Given the equation: $2f(x) - 3f\left( \frac{1}{x} \right) = x$ ... $(i)$Replacing $x$ with $\frac{1}{x}$ in $(i)$,we get:$2f\left( \frac{1}{x} \right) - 3f(x) = \frac{1}{x}$ ... $(ii)$To eliminate $f\left( \frac{1}{x} \right)$,multiply $(i)$ by $2$ and $(ii)$ by $3$:$4f(x) - 6f\left( \frac{1}{x} \right) = 2x$$6f\left( \frac{1}{x} \right) - 9f(x) = \frac{3}{x}$Adding these two equations:$-5f(x) = 2x + \frac{3}{x}$$f(x) = -\frac{1}{5} \left( 2x + \frac{3}{x} \right)$Now,integrate $f(x)$ from $1$ to $2$:$\int_1^2 f(x) \, dx = -\frac{1}{5} \int_1^2 \left( 2x + \frac{3}{x} \right) \, dx$$= -\frac{1}{5} \left[ x^2 + 3\ln|x| \right]_1^2$$= -\frac{1}{5} \left( (2^2 + 3\ln 2) - (1^2 + 3\ln 1) \right)$$= -\frac{1}{5} (4 + 3\ln 2 - 1 - 0)$$= -\frac{1}{5} (3 + 3\ln 2) = -\frac{3}{5}(1 + \ln 2)$.

If $2f(x) - 3f\left( \frac{1}{x} \right) = x$,then $\int_1^2 f(x) \, dx$ is equal to

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