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

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(B) Given the equation: $3 f(x)-2 f\left(\frac{1}{x}\right)=x$ Differentiating both sides with respect to $x$,we get: $3 f^{\prime}(x)-2 f^{\prime}\le

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$\frac{1}{2}$

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(B) Given the equation: $3 f(x)-2 f\left(\frac{1}{x}\right)=x$ Differentiating both sides with respect to $x$,we get: $3 f^{\prime}(x)-2 f^{\prime}\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)=1$ $3 f^{\prime}(x)+\frac{2}{x^2} f^{\prime}\left(\frac{1}{x}\right)=1$ For $x=2$: $3 f^{\prime}(2)+\frac{2}{4} f^{\prime}\left(\frac{1}{2}\right)=1 \Rightarrow 3 f^{\prime}(2)+\frac{1}{2} f^{\prime}\left(\frac{1}{2}\right)=1$ ...$(i)$ For $x=\frac{1}{2}$: $3 f^{\prime}\left(\frac{1}{2}\right)+\frac{2}{(1/4)} f^{\prime}(2)=1 \Rightarrow 3 f^{\prime}\left(\frac{1}{2}\right)+8 f^{\prime}(2)=1$ ...(ii) From $(i)$,$f^{\prime}\left(\frac{1}{2}\right)=2-6 f^{\prime}(2)$. Substituting this into (ii): $3(2-6 f^{\prime}(2))+8 f^{\prime}(2)=1$ $6-18 f^{\prime}(2)+8 f^{\prime}(2)=1$ $-10 f^{\prime}(2)=-5$ $f^{\prime}(2)=\frac{1}{2}$

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

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