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

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

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$\frac{-1}{5}$

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(A) Given equation is $2 f(x)-3 f\left(\frac{1}{x}ight)=x+1 \quad ...(i)$ Replace $x$ with $\frac{1}{x}$ in equation $(i)$: $2 f\left(\frac{1}{x}ight)-3 f(x)=\frac{1}{x}+1 \quad ...(ii)$ Multiply equation $(i)$ by $2$ and equation $(ii)$ by $3$: $4 f(x)-6 f\left(\frac{1}{x}ight)=2 x+2 \quad ...(iii)$ $6 f\left(\frac{1}{x}ight)-9 f(x)=\frac{3}{x}+3 \quad ...(iv)$ Adding equation $(iii)$ and $(iv)$: $(4 f(x)-9 f(x)) = 2 x + \frac{3}{x} + 5$ $-5 f(x) = 2 x + \frac{3}{x} + 5$ $f(x) = -\frac{2}{5} x - \frac{3}{5 x} - 1$ Differentiating with respect to $x$: $f^{\prime}(x) = -\frac{2}{5} + \frac{3}{5 x^2}$ Now,substitute $x = \sqrt{3}$: $f^{\prime}(\sqrt{3}) = -\frac{2}{5} + \frac{3}{5(\sqrt{3})^2}$ $f^{\prime}(\sqrt{3}) = -\frac{2}{5} + \frac{3}{5 	imes 3}$ $f^{\prime}(\sqrt{3}) = -\frac{2}{5} + \frac{1}{5} = -\frac{1}{5}$

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

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