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. $14$
• B. $\frac{7}{8}$
• C. $1$
• D. none of these

Answer 1

Answer: (B)

$\frac{7}{8}$

$5f\left(x\right)+3f\left(\frac{1}{x}\right)=x+2\Rightarrow \left(1\right)$
Replace $x$ by $\frac{1}{x}$
$\Rightarrow 5f\left(\frac{1}{x}\right)+3f\left(x\right)=\frac{1}{x}+2\Rightarrow \left(2\right)$
$\Rightarrow 5\times \left(1\right)-3\times \left(2\right)\Rightarrow 16f\left(x\right)=5x-\frac{3}{x}+4=\frac{5x^2+4x-3}{x}$
$\therefore y=xf\left(x\right)=\frac{5x^2+4x-3}{16}$
$\Rightarrow \frac{dy}{dx}=\frac{10x+4}{16}$
$\therefore {\left(\frac{dy}{dx}\right)}_{x=1}=\frac{10+4}{16}=\frac{7}{8}$