If 8 f(x)+6 fleft(frac{1}{x}right)=x+5 and y= | vedclass

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(D) Given $8 f(x)+6 f\left(\frac{1}{x}\right)=x+5 \dots (i)$ Replacing $x$ with $\frac{1}{x}$,we get $8 f\left(\frac{1}{x}\right)+6 f(x)=\frac{1}{x}+5

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$-\frac{1}{14}$

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(D) Given $8 f(x)+6 f\left(\frac{1}{x}\right)=x+5 \dots (i)$ Replacing $x$ with $\frac{1}{x}$,we get $8 f\left(\frac{1}{x}\right)+6 f(x)=\frac{1}{x}+5 \dots (ii)$ Multiply $(i)$ by $4$ and $(ii)$ by $3$: $32 f(x)+24 f\left(\frac{1}{x}\right)=4x+20 \dots (iii)$ $18 f(x)+24 f\left(\frac{1}{x}\right)=\frac{3}{x}+15 \dots (iv)$ Subtracting $(iv)$ from $(iii)$: $14 f(x)=4x-\frac{3}{x}+5 \implies f(x)=\frac{1}{14}\left(4x-\frac{3}{x}+5\right)$ Then $f'(x)=\frac{1}{14}\left(4+\frac{3}{x^2}\right)$ At $x=-1$: $f(-1)=\frac{1}{14}(-4+3+5)=\frac{4}{14}=\frac{2}{7}$ $f'(-1)=\frac{1}{14}(4+3)=\frac{7}{14}=\frac{1}{2}$ Given $y=x^2 f(x)$,differentiating w.r.t $x$: $\frac{dy}{dx}=2x f(x)+x^2 f'(x)$ At $x=-1$: $\left.\frac{dy}{dx}\right|_{x=-1}=2(-1)f(-1)+(-1)^2 f'(-1)$ $= -2\left(\frac{2}{7}\right) + 1\left(\frac{1}{2}\right) = -\frac{4}{7} + \frac{1}{2} = \frac{-8+7}{14} = -\frac{1}{14}$

If $8 f(x)+6 f\left(\frac{1}{x}\right)=x+5$ and $y=x^2 f(x)$,then $\frac{d y}{d x}$ at $x=-1$ is

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