If A = sin2θ + cos4θ for any value of θ, then the value of A is?
A. $${\text{1}} \leqslant {\text{A}} \leqslant {\text{1}}$$
B. $$\frac{3}{4} \leqslant {\text{A}} \leqslant {\text{1}}$$
C. $$\frac{{13}}{{16}} \leqslant {\text{A}} \leqslant {\text{1}}$$
D. $$\frac{3}{4} \leqslant {\text{A}} \leqslant \frac{{13}}{{16}}$$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {\text{According to the question, }} \cr & {\text{A}} = {\sin ^2}\theta + {\text{co}}{{\text{s}}^4}\theta \cr & {\text{Put }}\theta = {90^ \circ }{\text{ for maximum value of A}} \cr & {\text{A}} = {\sin ^2}{90^ \circ } + {\text{co}}{{\text{s}}^4}{90^ \circ } \cr & {\text{A}} = 1 + 0 \cr & {\text{A}} = 1 \cr & {\text{Put }}\theta = {45^ \circ }{\text{ for minimum value of A}} \cr & {\text{A}} = {\sin ^2}{45^ \circ } + {\text{co}}{{\text{s}}^4}{45^ \circ } \cr & {\text{A}} = \frac{1}{2} + \frac{1}{4} \cr & {\text{A}} = \frac{3}{4} \cr & \therefore {\text{A lies in }}\frac{3}{4} \leqslant {\text{A}} \leqslant {\text{1}} \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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