Which of the following inequalities holds true?
(A) $\sqrt{5}+\sqrt{3}>\sqrt{6}+\sqrt{2}$.
(B) If $a > b$ and $c <0$, then $\frac{a}{c}<\frac{b}{c}$
(C) $\frac{1}{x^2}>\frac{1}{x}>1$, if $0<x<1$.
(D) If $a$ and $b$ are positive integers and $\frac{a-b}{6.25}=\frac{4}{2.5}$, then $b> a$.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
(A) $\sqrt5+\sqrt3$ and $\sqrt6+\sqrt2$
Square both sides
$(\sqrt5+\sqrt3)^2=8+2\sqrt{15}$
$(\sqrt6+\sqrt2)^2=8+2\sqrt{12}$
Since $\sqrt{15}>\sqrt{12}$, left side is greater
True
(B) If $a>b$ and $c<0$ then $\frac{a}{c}<\frac{b}{c}$
Division by a negative number reverses inequality
True
(C) If $0<\text{ x }<1$ then $\frac{1}{x^2}>1$ and $\frac{1}{x}>1$ but $\frac{1}{x^2}>\frac{1}{x}$
Hence $\frac{1}{x^2}>\frac{1}{x}>1$
True
(D) Given $\frac{a-b}{6.25}=\frac{4}{2.5}$
$a-b=6.25\times\frac{4}{2.5}=10$
$a-b=10$
$a>b$
So $b>a$ is false
False
The correct inequalities are (A), (B) and (C).