Question

If $5f\left(x\right)+3f\left(\frac{1}{x}\right)=x+2$ and $y=xf\left(x\right)$ then ${\left(\frac{dy}{dx}\right)}_{x=1}$ is equal to :-

• A. $\frac{14}{16}$
• B. $14$
• C. $1$
• D. None of these

Answer 1

The correct option is A $\frac{14}{16}$

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

Let $x=\frac{1}{x}$

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

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

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

$25f\left(x\right)+15f\left(\frac{1}{x}\right)=5x+10$

$9f\left(x\right)+15f\left(\frac{1}{x}\right)=\frac{3}{x}+6$

$\underset{–}{---}$

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

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

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

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

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

${\left(\frac{dy}{dx}\right)}_{x=1}=\frac{1}{16}(10+4)=\frac{14}{16}$