Question

Minimum value of $4x^2-4x|\sin \theta |-{\cos }^2\theta$ is:

• A. -2
• B. -1
• C. -1/2
• D. 0

Answer 1

The correct option is B -1

Consider the quadratic equation

$4x^2-4x|\sin \theta |-{\cos }^2\theta =0$
$x=\frac{4|\sin \theta |\pm \sqrt{16si{n}^2\theta +16{\cos }^2\theta }}{8}$

$=\frac{|\sin \theta |\pm 1}{2}$

Thus, $4x^2-4x|\sin \theta |-{\cos }^2\theta$

$=(2x+1-|\sin \theta |)(2x-1-|\sin \theta |)$

Thus the minimum value will be at $\sin \theta =0$ and $x=0$.

Hence minimum is $-1$.