Question

If ${\cot }^{-1}\frac{n^2-10n+21.6}{\pi }>\frac{\pi }{6},n\in N$, then $n$ can be

• A. $3$
• B. $2$
• C. $4$
• D. $8$

Answer 1

Answer: (A), (C)

The correct options are
A $3$
C $4$

We have
${\cot }^{-1}\left(\frac{n^2-10n+21.6}{\pi }\right)>\frac{\pi }{6}$
$\Rightarrow \frac{n^2-10n+21.6}{\pi }<\cot \frac{\pi }{6}$
$(\text{as}\cot x\text{is decreasing for}0<x<\pi )$
or $n^2-10n+21.6<\pi \sqrt{3}$
or $n^2-10n+25+21.6-25<\pi \sqrt{3}$
or $(n-5{)}^2<\pi \sqrt{3}+3.4$
or $-\sqrt{\sqrt{3}\pi +3.4}<n-5<\sqrt{\sqrt{3}\pi +3.4}$
or $5-\sqrt{\sqrt{3}\pi +3.4}<n<5+\sqrt{\sqrt{3}\pi +3.4}$....(i)
Since $\sqrt{3}\pi =5.5,$ nearly, $\sqrt{\sqrt{3}\pi +3.4}\sim \sqrt{8.9}\sim 2.9$
$\Rightarrow 2.1<n<7.9$
$\therefore n=3,4,5,6,7$ as $n\in N$