Question

Let α,β be any two positive values of x for which 2cosx,|cosx| and (1−3 cos2x) are in G.P. The minimum value of |α−β| is

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\boxed{min|\alpha - \beta| = \pi}$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

Since $2cos(x), |cos(x)|, (1- 3cos(2x))$ are in GP,

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

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

$|cos(x)|^2 = 2cos(x)$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 6cos(x)cos(2x)$

$cos^2(x)= 2cos(x)$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 6cos(x)(2cos^2(x)-1)$

$cos^2(x)= 2cos(x)- 12cos^3(x)$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-6cos(x)$

$cos^2(x)= - 4cos(x)- 12cos^3(x)$

$12cos^3(x) + cos^2(x) + 4cos(x) =0$

$cos(x)[12cos^2(x) + cos(x) + 4]=0$

$\to cos(x) = 0$

or $\:\:[12cos^2(x) + cos(x) + 4]=0$

Discriminant of the quadractic equation is

$\:\:\:\:\:D = b^2 - 4ac$

$\to D = 1^2 - 4\times 12\times 4$

$\to D = 1 - 192$

$\to D = - 191$

$\to D ≤ 0$

Thus the quadratic equation has no real roots and can be neglected.

$\to cos(x) = 0$

$\to x = \frac{\pi} {2}, \frac{3\pi}{2}, \frac{5\pi} {2},...$ .Therefore minimum value

$\:\:\:min|\alpha - \beta|$

$\to |\frac{3\pi} {2}- \frac{\pi}{2}|$

$\to |\frac{(3-1)\pi}{2}|$

$\to \pi$