Question

Let $\alpha$, $\beta$ be any two positive values of x for which $2\cos x$, $\left|\cos x\right|$ and $1-3{\cos }^{2}x$ are in GP. The minimum value of $|\alpha -\beta |$

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. none of these

Answer 1

Answer: (D)

none of these

As $2\cos x,|\cos x|and(1-3{\cos }^{2}x)$ are in G.P. so
by using the property of geometric mean we get,

$2\cos x(1-3{\cos }^{2}x)=|\cos x{|}^2$

$\Rightarrow 2\cos x-6{\cos }^{3}x={\cos }^{2}x$

$\Rightarrow 6{\cos }^{3}x+{\cos }^{2}x-2\cos x=0$

$\Rightarrow \cos x(6{\cos }^{2}x+\cos x-2)=0$

$\Rightarrow \cos x(6{\cos }^{2}x+4\cos x-3\cos x-2)=0$

$\Rightarrow \cos x\left(2\cos x\right(3\cos x+2)-1(3\cos x+2\left)\right)=0$

$\Rightarrow \cos x(2\cos x-1)(3\cos x+2)=0$

$\cos x=0,\frac{1}{2},\frac{-2}{3}$

$\alpha$ and $\beta$ are positive values , so we discard the last root,

so, $\alpha ,\beta \equiv \frac{\pi }{2},\frac{\pi }{3}$

$|\alpha -\beta |=\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}$