Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Numbers, Quantification and Numerical Applications

Question:

If $0<2 x<1$, which of the following is smallest?

Options:

$x$

$x^2$

$\frac{1}{x}$

$\frac{1}{x^2}$

Correct Answer:

$x^2$

Explanation:

The correct answer is Option (2) → $x^2$

$0 < 2x < 1 \Rightarrow 0 < x < \frac{1}{2}$

$\text{For } 0 < x < 1,\ x^2 < x$

$\text{Also, } \frac{1}{x} > 1 \text{ and } \frac{1}{x^2} > \frac{1}{x}$

$\text{ therefore }x^2 < x < \frac{1}{x} < \frac{1}{x^2}$

$\text{Smallest value is } x^2$