If for non-zero x , $ 3 f(x) + 4 f(\frac{1}{x}) = \frac{1}{x} – 10 $ then $\int_{2}^{3}f(x) dx $ is equal to

Q: If for non-zero x , $\large 3 f(x) + 4 f(\frac{1}{x}) = \frac{1}{x} – 10 $ then $\large \int_{2}^{3}f(x) dx $ is equal to

(A) $\frac{4}{7} ln\frac{2}{3}$

(B) $\frac{3}{7} ln\frac{3}{2}$

(C) $\frac{3}{7} ln\frac{2}{3}$

(D) None of these

Sol: $\large 3 f(x) + 4 f(\frac{1}{x}) = \frac{1}{x} – 10 $

replace x by 1/x

$\large 3 f(\frac{1}{x}) + 4 f(x) = x – 10 $

solving (1) and (2)

9f(x) – 16f (x) = 3/x – 30 – 4x + 40

– 7f (x) = 10 + 3/x – 4x

$\large f(x) = -\frac{10}{7}-\frac{3}{7x} + \frac{4x}{7} $

$\large \int_{2}^{3}f(x) dx = \int_{2}^{3} (-\frac{10}{7}-\frac{3}{7x} + \frac{4x}{7}) dx $

$\large [-\frac{10}{7}x -\frac{3}{7}ln|x| + \frac{4x^2}{14}]_{2}^{3}$

$\large = \frac{3}{7}ln\frac{2}{3}$

Hence (C) is the correct answer.