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

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

Replacing x by $\frac{1}{x}$ in (i),

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

On solving equation (i) and (ii), we get,

16 f(x) = 5x – $\frac{3}{x}$+ 4,

\ 16 f ¢(x) = 5 + $\frac{3}{x^{2}}$

Q y = xf(x)

̃ $\frac{dy}{dx}$ = f(x) + xf ¢(x)

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

at x = 1, $\frac{dy}{dx}$ = $\frac{1}{16}$ (5 – 3 + 4) + $\frac{1}{16}$ (5 + 3) = $\frac{7}{8}$