Question

If $0<y<x$, which statement must be true?

• A. $x\sqrt{y}=y\sqrt{x}$
• B. $\sqrt{x}+\sqrt{x}=\sqrt{2x}$
• C. $\sqrt{x}-\sqrt{y}=\sqrt{x-y}$
• D. $\sqrt{xy}=\sqrt{x}\sqrt{y}$

Answer 1

The correct option is D $\sqrt{xy}=\sqrt{x}\sqrt{y}$

Since, $\sqrt{x}-\sqrt{y}\ne \sqrt{x-y}$
$\Rightarrow$ Option (a) is incorrect.

$\sqrt{x}+\sqrt{x}=2\sqrt{x}\ne \sqrt{2x}$
$\Rightarrow$ Option (b) is incorrect.

$x\sqrt{y}\ne y\sqrt{x}$
Two numbers $x\sqrt{y}$ and $y\sqrt{x}$ will be equal if $x=y$
But given $x>y$
$\Rightarrow$ Option (c) is incorrect.

$\sqrt{xy}=(xy{)}^{\frac{1}{2}}=(x{)}^{\frac{1}{2}}(y{)}^{\frac{1}{2}}=\sqrt{x}\sqrt{y}$ is true.

Hence, option (d) is correct.