If 1< x < 2, then which of the following is minimum ?

The correct answer is Option (4) → $-\frac{1}{x^2}$

$x^4>x^3>x^2$

$∴\frac{1}{x^4}<\frac{1}{x^3}<\frac{1}{x^2}$ (Since x is positive $1<x<2$)

$⇒\frac{-1}{x^4}>\frac{-1}{x^3}>\frac{-1}{x^2}$

$∴\frac{-1}{x^2}$ is minimum