If $A=\cos^2\theta+\sin^4\theta$ then for all value of $\theta$

$\begin{array}{1 1}(a)\;1\leq A\leq 2&(b)\;\large\frac{3}{4}\normalsize\leq A\leq 1\\(c)\;\large\frac{13}{16}\normalsize\leq A\leq 1&(d)\;None\;of\;these\end{array}$

$A=\cos^2\theta+\sin^4\theta$
$\;\;=\cos^2\theta+\sin^2\theta(\sin^2\theta)$
$\sin^2\theta+\cos^2\theta=1$
$1-\cos^2\theta=\sin^2\theta$
$\;\;=\cos^2\theta+\sin^2\theta(1-\cos^2\theta)$
$\;\;=\cos^2\theta+\sin^2\theta-\sin^2\theta\cos^2\theta$
$\;\;=1-\sin^2\theta\cos^2\theta$
$\sin 2\theta=2\sin\theta\cos\theta$
$\;\;=1-\large\frac{1}{4}$$\sin^22\theta (0\leq \sin^22\theta\leq 1) \therefore \large\frac{3}{4}$$\leq A\leq 1$
Hence (b) is the right option.