Question

Let $\theta$ be an acute angle such that the equation $x^3 + 4x^2 \cos \theta +x \cot \theta = 0 $ has multiple roots. Then the value of $\theta$ (in radians) is

• A. $\frac{\pi}{3}$
• B. $\frac{\pi}{8}$
• C. $\frac{\pi}{12}$ or $\frac{ 5 \pi}{12}$
• D. $\frac{\pi}{6}$ or $\frac{ 5 \pi}{12}$

Answer 1

Answer: (C)

Given, $x^{3}+4 x^{2} \cos \theta+x \cot \theta=0$
$\Rightarrow x\left(x^{2}+4 x \cos \theta+\cot \theta\right)=0$
$\Rightarrow x=0$ is root of equation.
$\Rightarrow x^{2}+4 x \cos \theta+\cot \theta=0$ has multiple roots
So, $ \Delta=0$
$\Rightarrow (4 \cos \theta)^{2}-4(1)(\cot \theta)=0 $
$ \Rightarrow 16 \cos ^{2} \theta=4 \cot \theta $
$ \Rightarrow 4 \cos ^{2} \theta=\cot \theta$
$ \Rightarrow 4 \cos ^{2} \theta=\frac{\cos \theta}{\sin \theta} $
$ \Rightarrow \cos \theta=0 $ as $ 4 \cos \theta=\frac{1}{\sin \theta} $
$ \Rightarrow 2 \sin \theta \cos \theta=\frac{1}{2}$
$ \Rightarrow \sin 2 \theta=\frac{1}{2} $
$ \Rightarrow 2 \theta=\frac{\pi}{6} $ or $ \frac{5 \pi}{6} $
$\Rightarrow \theta=\frac{\pi}{12}$ or $ \frac{5 \pi}{12}$