Q. If $p = \sin^{2}{x + \cos^{4}x}$, then

A) $\frac{3}{4} \leq p \leq 1$
B) $\frac{3}{16} \leq p \leq \frac{1}{4}$
C) $\frac{1}{4} \leq p \leq 1$
D) None of these
Explanation

(a)
$p = \sin^{2}{x + \cos^{2}x(1 - \sin^{2}x)}$
$\Rightarrow p = (\sin^{2}{x + \cos^{2}{x) - \sin^{2}x\cos^{2}x}}$
$= 1 - \sin^{2}{x\cos^{2}x}$ …(i)
Which shows $p \leq 1$
Again, $p = 1 - \cos^{2}{x + \cos^{4}x}$
$p = \left( \cos^{2}{x - \frac{1}{2}} \right)^{2} + \frac{3}{4}$
Which shows
$p \geq \frac{3}{4}$ …(ii)
From Eqs. (i) and (ii), we get
$\frac{3}{4} \leq p \leq 1$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.