CUET Preparation Today
If $1< x < 2, $ then which of the following is minimum ? |
$-\frac{1}{x^3}$ $\frac{1}{x}$ $-\frac{1}{x^4}$ $-\frac{1}{x^2}$ |
$-\frac{1}{x^2}$ |
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 |
