Question

If $a=\sqrt{11}+\sqrt{3},b=\sqrt{12}+\sqrt{2},c=\sqrt{6}+\sqrt{4}$, then which of the following holds true ?

• A. $c>a>b$
• B. $a>b>c$
  
• C. $a>c>b$
• D. $b>a>c$
  

Answer 1

Answer: (B)

$a>b>c$

$a=\sqrt{11}+\sqrt{3}$

$a^2=11+3+2\sqrt{33}=14+2\sqrt{33}$

$b=\sqrt{12}+\sqrt{2}$

$b^2=14+2\sqrt{24}$

As $\sqrt{33}>\sqrt{24},a^2>b^2,a>b$

$c=\sqrt{6}+\sqrt{4}$

$c^2=10+2\sqrt{24}$

As $14>10,b^2>c^2,b>c$

Hence, $a>b>c$.

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