Question

Given $A=si{n}^2\theta +co{s}^4\theta$ then for all real values of $\theta$

• A. $1\le A\le 2$
• B. $\frac{3}{4}\le A\le 1$
• C. $\frac{13}{16}\le A\le 1$
• D. $\frac{3}{4}\le A\le \frac{13}{16}$

Answer 1

The correct option is B

$\frac{3}{4}\le A\le 1$

Explanation for the correct option:

Find the domain of $A$:

$A={\sin }^2\theta +{\cos }^4\theta$

$=1–{\cos }^2\theta +{\cos }^4\theta$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$=1–{\cos }^2\theta (1–{\cos }^2\theta )$

$=1–{\cos }^2\theta \left({\sin }^2\theta \right)$

$={\sin }^2\theta +{\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta$

$=1-\frac{1}{4}{(2\sin \theta \cos \theta )}^2$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{A}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{A}\right]$

$=1–\frac{1}{4}{\sin }^2\left(2\theta \right)$

As we know, $0\le {\sin }^2\left(2\theta \right)\le 1$

$\Rightarrow$ $0\le \frac{1}{4}{\sin }^2\left(2\theta \right)\le \frac{1}{4}$

$\Rightarrow$ $\frac{3}{4}\le 1–\frac{1}{4}{\sin }^2\left(2\theta \right)\le 1$

$\therefore \frac{\mathbf{3}}{\mathbf{4}}\mathbf{}\mathbf{\le }\mathbf{}\mathit{A}\mathbf{}\mathbf{\le }\mathbf{}\mathbf{1}$

Hence, Option ‘B’ is Correct.