Let α,β be any two positive values of x for which 2cos⁡x,|cos⁡x| and 1−3cos2⁡x are in G. P. The minimum value of ∣α−β∣, is

A
$π / 3$
B
$π / 4$
C
$π / 2$
D
none of these

Solution:

It is given that

$2 \cos x, | \cos x | and 1 - 3 \cos^{2} x$ are in $G.P.$

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

$\begin{array}{l} \Rightarrow \cos^{2} x = 2 \cos x - 6 \cos^{3} x \\ \Rightarrow \cos x (6 \cos^{2} x + \cos x - 2) = 0 \end{array}$

$\begin{array}{l} \Rightarrow \cos x (3 \cos x + 2) (2 \cos x - 1) = 0 \\ \Rightarrow \cos x = 0, \frac{1}{2}, - \frac{2}{3} \Rightarrow x = \frac{π}{2}, \frac{π}{3}, π + \cos^{- 1} (\frac{2}{3}) \end{array}$

Clearly, $∣ α - β ∣$ is minimum when $α = \frac{π}{2}$ and $β = \frac{π}{3}$

The minimum value of $| α - β |$ is $|\frac{π}{2} - \frac{π}{3}| = \frac{π}{6}$

Let $α, β$ be any two positive values of $x$ for which $2 \cos x, | \cos x |$ and $1 - 3 \cos^{2} x$ are in $G . P$ . The minimum value of $∣ α - β ∣$ , is

Solution:

Related content