Let f be a differentiable function such that 8f(x)+6f(1/x)−x=5(x≠0) and y=x2f(x) then dydx at x=1 is

# Let $f$ be differentiable function such that $8f\left(x\right)+6f\left(1/x\right)-x=5\left(x\ne 0\right)$ and $y={x}^{2}f\left(x\right)$ then $\frac{dy}{dx}$ at $x=1$ is

1. A
2. B
3. C
4. D

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### Solution:

Differentiating the given expression, we get

Also $\frac{dy}{dx}=2xf\left(x\right)+{x}^{2}{f}^{\mathrm{\prime }}\left(x\right)$

so,  ${\frac{dy}{dx}|}_{x=1}=2f\left(1\right)+{f}^{\mathrm{\prime }}\left(1\right)$

Putting  in the given equation, we obtain

Hence      ${\frac{dy}{dx}|}_{x=1}=2,\frac{3}{7}+\frac{1}{2}=\frac{19}{14}$

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