The function $f(x) = \frac{2x^2 - 1}{x^4}$, (where, $x > 0$) decreases in which of the following intervals?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $(1, \infty)$ ##
$ f'(x) = \frac{x^4 \cdot 4x - (2x^2 - 1) \cdot 4x^3}{x^8} = \frac{4x^5 - 8x^5 + 4x^3}{x^8}$
$= \frac{-4x^5 + 4x^3}{x^8} = \frac{4x^3(-x^2 + 1)}{x^8}$
Also, $f'(x) < 0$
$\Rightarrow \frac{4x^3(1 - x^2)}{x^8} < 0 \Rightarrow x^2 > 1$
$\Rightarrow x > \pm 1$
$∴x \in (1, \infty) \cup (-1, 0)$