Question

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

• A. 14
• B. 7/8
• C. 1
• D. 2

Answer 1

The correct option is B

7/8

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

Replacing x by $\frac{1}{x}$

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

From Eq. (i),

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

and from Eq. (ii),

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

Substracting Eq.(iv) from (iii), we get

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

$\therefore xf\left(x\right)$ = $\frac{5x^2-3+4x}{16}$ = 4

$\therefore \frac{dy}{dx}$ = $\frac{10x+4}{16}$

$\frac{dy}{dx}{|}_{x=1}$ = $\frac{10+4}{16}$ = $\frac{7}{8}$