If $f(x)=\int\limits_x^{x^2} \frac{1}{(\log t)^2} d t, x \neq 0, x \neq 1$ then f(x) is monotomically
Answer & explanation
Correct answer: option 1
We have,
$f(x) =\int\limits_x^{x^2} \frac{1}{(\log t)^2} d t$
$\Rightarrow f'(x) =2 x \frac{1}{\left(\log x^2\right)^2}-\frac{1}{(\log x)^2}=\left(\frac{x-2}{2}\right) \times \frac{1}{(\log x)^2}$
Clearly, f'(x) > 0 for all x > 2.
Hence, f(x) is increasing on (2, ∞).