Question

If $\underset{x\to 0}{\lim }\frac{1-\sqrt{\cos 2x}\cdot \sqrt[3]{\cos 3x}\cdot \sqrt[4]{\cos 4x}\dots \cdot \sqrt[n]{\cos nx}}{x^2}$ has the value equal to $10$ , then the value of $n$ equals.

Answer 1

Answer: 6

$L=\underset{x\to 0}{\lim }=-\underset{x\to 0}{\lim }\frac{D\prod _{r=2}^n(\cos rx{)}^{1/r}}{2x}$
(using L’Hospital’s rule)
let $y=\prod _{r=2}^n(\cos rx{)}^{1/r}$
$\Rightarrow \ln y=\sum _{r=2}^n\left(\frac{1}{r}\ln \cos rx\right)$
$\Rightarrow \frac{1}{y}\frac{dy}{dx}=-\sum _{r=2}^n\tan (rx)$
$\Rightarrow -Dy=y\sum _{r=2}^n\tan (rx)$
$D\prod _{r=2}^n(\cos rx{)}^{1/r}$
$=-y\sum _{r=2}^n\tan (rx)$
$L=\underset{x\to 0}{\lim }\frac{y\cdot \sum _{r=2}^n\tan (rx)}{2x}$
$=\frac{1}{2}[2+3+4+\dots .+n]$
$=\frac{1}{2}\left[\frac{n(n+1)}{2}-1\right]$
$=\frac{n^2+n-2}{4}$
$\Rightarrow \frac{n^2+n-2}{4}=10$
$\Rightarrow n^2+n-42=0$
$\Rightarrow (n+7)(n-6)=0$
$\Rightarrow n=6$