If $ f(x) = \int_{x^2}^{x^3} \frac{dt}{ln t} , x > 0 , \ne 1$ then

Q: If $\large f(x) = \int_{x^2}^{x^3} \frac{dt}{ln t} , x > 0 , \ne 1$ then

(A) f(x) is an increasing function

(B) f(x) has a minima at x = 1

(C) f(x) is a decreasing function

(D) f(x) has a maxima at x = 1

Ans: (A)

Sol: $\large f(x) = \int_{x^2}^{x^3} \frac{dt}{ln t}$

For increasing or decreasing function,

$\large f'(x) = \frac{1}{ln x^3} . 3x^2 – \frac{1}{ln x^2} . 2x $ (using Leibnitz formula)

$\large f'(x) = \frac{1}{3 ln x} . 3x^2 – \frac{1}{2 ln x} . 2x $

$\large = \frac{1}{ln x}(x^2 – x) $

Since f'(x) > 0 for x > 0 , x ≠ 1 hence f(x) is increasing function.

Hence (A) is correct.