Question

If $A={\sin }^2\theta +{\cos }^4\theta$, then for all real value 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

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

Step 1: Simplify the given expression using trigonometric identities.

We have, $A={\sin }^2\theta +{\cos }^4\theta$.

$\begin{array}{rcl}A& =& {\sin }^2\theta +{\cos }^4\theta \\ A& =& {\sin }^2\theta +{\cos }^2\theta {\cos }^2\theta \\ A& =& {\sin }^2\theta +{\cos }^2\theta \left(1-{\sin }^2\theta \right)\left\{\because {\sin }^2\theta +{\cos }^2\theta =1\right\}\\ A& =& {\sin }^2\theta +{\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta \\ A& =& 1-\frac{1}{4}{\left(2\sin \theta \cos \theta \right)}^2\left\{\because 2\sin \theta \cos \theta =\sin 2\theta \right\}\\ A& =& 1-\frac{1}{4}{\sin }^{2}2\theta \end{array}$

Step 2: Find the minimum value of A.

It is known that:

$\begin{array}{rcl}min\left({\sin }^{2}2\theta \right)& =& 0\\ max\left({\sin }^{2}2\theta \right)& =& 1\end{array}$

Therefore, the minimum value of A is:

$\begin{array}{rcl}min\left(A\right)& =& 1-\frac{1}{4}max\left({\sin }^{2}2\theta \right)\\ min\left(A\right)& =& 1-\frac{1}{4}\times 1\\ min\left(A\right)& =& \frac{3}{4}\end{array}$

Step 3: Find the maximum value of A.

Therefore, the maximum value of A is:

$\begin{array}{rcl}max\left(A\right)& =& 1-\frac{1}{4}min\left({\sin }^{2}2\theta \right)\\ max\left(A\right)& =& 1-\frac{1}{4}\times 0\\ max\left(A\right)& =& 1\end{array}$

Hence, the required answer is $\frac{3}{4}\le A\le 1$.

Thus, the correct option is B.