Question
Given $$A = {\sin ^2}\theta + {\cos ^4}\theta $$ then for all real values of $$\theta $$
A.
$$1 \leqslant A \leqslant 2$$
B.
$$\frac{3}{4} \leqslant A \leqslant 1$$
C.
$$\frac{13}{16} \leqslant A \leqslant 1$$
D.
$$\frac{3}{4} \leqslant A \leqslant \frac{{13}}{{16}}$$
Answer :
$$\frac{3}{4} \leqslant A \leqslant 1$$
Solution :
$$\eqalign{
& A = {\sin ^2}\theta + {\cos ^4}\theta \cr
& = {\sin ^2}\theta + {\left( {1 - {{\sin }^2}\theta } \right)^2} \cr
& = {\sin ^4}\theta - {\sin ^2}\theta + 1 \cr
& \Rightarrow \,\,A = {\left( {{{\sin }^2}\theta - \frac{1}{2}} \right)^2} + \frac{3}{4} \cr
& {\text{But }}0 \leqslant {\left( {{{\sin }^2}\theta - \frac{1}{2}} \right)^2} \leqslant \frac{1}{4} \cr
& \therefore \,\,\frac{3}{4} \leqslant {\left( {{{\sin }^2}\theta - \frac{1}{2}} \right)^2} + \frac{3}{4} \leqslant 1\,\,{\text{or }}\frac{3}{4} \leqslant A \leqslant 1 \cr} $$