Question

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

The correct option is B $\frac{7}{8}$

$5f\left(x\right)+3f\left(\frac{1}{x}\right)=x+2........\left(i\right)$
$Replacingxby\frac{1}{x}$
$\therefore 5f\left(\frac{1}{x}\right)+3f\left(x\right)=\frac{1}{x}+2...\left(ii\right)$
$FromEq.\left(i\right).$
$25f\left(x\right)+15f\left(\frac{1}{x}\right)=5x+10....\left(iii\right)$
$andfromEq.\left(ii\right)$
$9f\left(x\right)+15f\left(\frac{1}{x}\right)=\frac{3}{x}+6.....\left(iv\right)$
$SubtractingEq.\left(iv\right)from\left(iii\right),weget$
$\therefore 16f\left(x\right)=5x-\frac{3}{x}+4$
$\therefore xf\left(x\right)=\frac{5x^2-3+4x}{16}=y$
$\therefore \frac{dy}{dx}=\frac{10x+4}{16}$
$\therefore \frac{dy}{dx}{|}_{x=1}=\frac{10+4}{16}$
$=\frac{7}{8}$