Question

If $5f\left(x\right)+3f\left(\frac{1}{x}\right)=x+2$ and $y=xf\left(x\right)$

then $\frac{dy}{dx}$ at $x=1$

• A. $14$
• B. $\frac{7}{8}$
• C. $1$
• D. $7$

Answer 1

The correct option is C $1$

$5f\left(x\right)+3f\left(\frac{1}{x}\right)=x+2----\left(1\right)$

so $5f\left(\frac{1}{x}\right)+3\left(x\right)=\frac{1}{x}+2----\left(2\right)$

Multiply $5\left(1\right)$ & $3\left(2\right)$ & subtract thems

$\frac{25f\left(x\right)+15f\left(\frac{1}{x}\right)=5(x+2)-9f\left(x\right)+15f\left(\frac{1}{x}\right)=3(\frac{1}{x}+2)}{16f\left(x\right)=5x+10-\frac{3}{x}-6f\left(x\right)=\frac{1}{16}(5x+4-\frac{3}{x})}$

$y=xf\left(x\right)$

$=x\left(\frac{1}{16}(5x+4-\frac{3}{x})\right)$

$\frac{1}{16}(5x^2+4x-3)$

$\frac{dy}{dx}{|}_{x=1}\Rightarrow \frac{1}{16}(10x+4)$

$=\frac{1}{16}\left(10\right(1)+4)$

$=\frac{1}{16}\times 16=1$

Option $c$ is write