If $0<2 x<1$, which of the following is smallest? |
$x$ $x^2$ $\frac{1}{x}$ $\frac{1}{x^2}$ |
$x^2$ |
The correct answer is Option (2) → $x^2$ $0 < 2x < 1 \Rightarrow 0 < x < \frac{1}{2}$ $\text{For } 0 < x < 1,\ x^2 < x$ $\text{Also, } \frac{1}{x} > 1 \text{ and } \frac{1}{x^2} > \frac{1}{x}$ $\text{ therefore }x^2 < x < \frac{1}{x} < \frac{1}{x^2}$ $\text{Smallest value is } x^2$ |