If A=sin2x+cos4x, then ∀x∈R

A=sin2⁡x+cos2⁡x2=1−cos2x2+1+cos2x22=12−12cos2x+14+12cos2x+14cos22x=34+14cos22x 0≤cos22x≤134+0≤A≤34+1434≤A≤1

If A=sin2x+cos4x, then ∀x∈R

A=sin2⁡x+cos2⁡x2=1−cos2x2+1+cos2x22=12−12cos2x+14+12cos2x+14cos22x=34+14cos22x 0≤cos22x≤134+0≤A≤34+1434≤A≤1

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If $A = \sin^{2} x + \cos^{4} x$ , then $\forall x \in R$

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

$\frac{13}{11} \leq A \leq 1$

$\frac{3}{4} \leq A \leq \frac{13}{16}$

$1 \leq A \leq 2$

answer is D.

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Detailed Solution

$\begin{array}{l} A = \sin^{2} x + {(\cos^{2} x)}^{2} \\ = \frac{1 - \cos 2 x}{2} + {(\frac{1 + \cos 2 x}{2})}^{2} \\ = \frac{1}{2} - \frac{1}{2} \cos 2 x + \frac{1}{4} + \frac{1}{2} \cos 2 x + \frac{1}{4} \cos^{2} 2 x \\ = \frac{3}{4} + \frac{1}{4} \cos^{2} 2 x 0 \leq \cos^{2} 2 x \leq 1 \\ \frac{3}{4} + 0 \leq A \leq \frac{3}{4} + \frac{1}{4} \\ \frac{3}{4} \leq A \leq 1 \end{array}$