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

MCQ

If $0<\text{y}<\text{x},$ which statement must be true?

A
$\sqrt{\text{x}}-\sqrt{\text{y}}=\sqrt{\text{x}-\text{y}}$
B
$$$\sqrt{\text{x}}+\sqrt{\text{x}}=\sqrt{2\text{x}}$
C
$\text{x}\sqrt{\text{y}}=\text{y}\sqrt{\text{x}}$
✓
$\sqrt{\text{xy}}=\sqrt{\text{x}}\sqrt{\text{y}}$

✓

Answer

Correct option: D.

$\sqrt{\text{xy}}=\sqrt{\text{x}}\sqrt{\text{y}}$

We have to find which statement must be true?
Given $0<\text{y}<\text{x},$
Option (a):
Left hand side:
$\sqrt{\text{x}}-\sqrt{\text{y}}=\sqrt{\text{x}\text{}}=\sqrt{\text{y}}$
Right Hand side:
$\sqrt{\text{x}-\text{y}}=\sqrt{\text{x}-\text{y}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (b):
$\sqrt{\text{x}}+\sqrt{\text{x}}=\sqrt{2\text{x}}$
Left hand side:
$\sqrt{\text{x}}+\sqrt{\text{x}}=2\sqrt{\text{x}}$
Right Hand side:
$\sqrt{2\text{x}}=\sqrt{2\text{x}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (c):
$\text{x}\sqrt{\text{y}}=\text{y}\sqrt{\text{x}}$
Left hand side:
$\text{x}\sqrt{\text{y}}=\text{x}\sqrt{\text{y}}$
Right Hand side:
$\text{y}\sqrt{\text{x}}=\text{y}\sqrt{\text{x}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (d):
$\sqrt{\text{xy}}=\sqrt{\text{x}}\sqrt{\text{y}}$
Left hand side:
$\sqrt{\text{xy}}=\sqrt{\text{xy}}$
Right Hand side:
$\sqrt{\text{x}}\sqrt{\text{y}}=\sqrt{\text{x}}\times\sqrt{\text{y}}$
$=\sqrt{\text{xy}}$
Left hand side is equal to right hand side
The statement is true.
Hence the correct choice is $d$.