Question

$\underset{x\to 0}{lim}\frac{1-\sqrt{\cos 2x}\cdot \sqrt[3]{3\cos 3x}\cdot \sqrt[4]{4\cos 4x}\cdot \cdot \cdot \sqrt[n]{n\cos nx}}{x^2}=10,$ then $n=$

• A. $6$
• B. $7$
• C. $8$
• D. $5$

Answer 1

Answer: (A)

$6$

$\underset{x\to 0}{lim}\frac{1-\sqrt{\cos 2x}\cdot \sqrt[3]{3\cos 3x}\cdot \sqrt[4]{4\cos 4x}\cdot \cdot \cdot \sqrt[n]{n\cos nx}}{x^2}$
$L=-\underset{x\to 0}{lim}\frac{D{\prod }_{r=2}^n(\cos rx{)}^{1/r}}{2x}$ (Using L' Hospital's rule, $D$ is the derivative of the term)
Let $y=\prod _{r=2}^n(\cos rx{)}^{1/r}$
$\Rightarrow \ln y=\sum _{r=2}^n\left(\frac{1}{r}\ln \right(\cos rx\left)\right)$
$\Rightarrow \frac{1}{y}\frac{dy}{dx}=-\sum _{r=2}^n\tan \left(rx\right)$
$\Rightarrow -D=y\sum _{r=2}^n\tan \left(rx\right)$
$\Rightarrow D\prod _{r=2}^n(\cos rx{)}^{1/x}=-y\sum _{r=2}^n\tan (rx)$
$\Rightarrow L=\underset{x\to 0}{lim}\frac{y\cdot {\sum }_{r=2}^n\tan \left(rx\right)}{2x}$
Using limit , $\underset{x\to 0}{lim}\frac{\tan x}{x}=1$, we get
$=\frac{1}{2}[2+3+4+.....+n]$
$=\frac{1}{2}[\frac{n(n+1)}{2}-1]$
$=\frac{n^2+n-2}{4}$
$\Rightarrow \frac{n^2+n-2}{4}=10$
$\Rightarrow n^2+n-42=0$
$\Rightarrow (n+7)(n-6)=0$
$\Rightarrow n=6$