Question

lf $e^{({\cos }^{2}x+{\cos }^{4}x+{\cos }^{6}x+\dots .)\log 3}$ satisfies $y^2-10y+9=0$ and $0\le x\le \frac{\pi }{2}$, then ${\cot }^{2}x=$

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $9$

Answer 1

The correct option is A $0$

$e^{({\cos }^{2}x+{\cos }^{4}x+\cdots )\log 3}=e^{\left(\frac{{\cos }^{2}x}{1-{\cos }^{2}x}\right)\log 3.}$
$\because ({\cos }^{2}x+{\cos }^{4}x+\cdots )$ is forming an infinite G.P. $=e^{{\cot }^{2}x\log 3.}$
$y^2-10y+9=0\Rightarrow y=9,1$
$e^{{\cot }^{2}x\log 3}=9,1$
$3^{{\cot }^{2}x}=9,1[\because e^{\log x}=x]$
$\Rightarrow {\cot }^{2}x=2,0$ but as $x\in [0,\frac{\pi }{2}]$
$\Rightarrow {\cot }^{2}x=0$

Hence, option 'A' is correct.