Q200 / 270 · NDA 2018 (Phase 2) · Mathematics

If $A=\sin ^2 \theta+\cos ^4 \theta$, the for all real $\theta$, which one of the following is correct?

  • A
    $1 \leq \mathrm{A} \leq 2$
  • Correct answer
    B
    $\frac{3}{4} \leq \mathrm{A} \leq 1$
  • C
    $\frac{13}{16} \leq \mathrm{A} \leq 1$
  • D
    $\frac{3}{4} \leq$ A $\leq \frac{13}{16}$

Solution

$\begin{aligned} A & =\sin ^2 \theta+\cos ^4 \theta \\ & =\sin ^2 \theta+\left(1-\sin ^2 \theta\right)^2 \\ & =1+\sin ^4 \theta-\sin ^2 \theta \\ & =1-\sin ^2 \theta\left(1-\sin ^2 \theta\right) \\ & =1-\sin ^2 \theta \cdot \cos ^2 \theta \\ & =\frac{4-4 \sin ^2 \theta \cdot \cos ^2 \theta}{4}=\frac{4-\sin ^2(2 \theta)}{4} \end{aligned}$ As, we know, $0 \leq \sin ^2 2 \theta \leq 1$ $\therefore \quad \mathrm{A}=\frac{4-0}{4} \text { or } \frac{4-1}{4} \Rightarrow \frac{3}{4} \leq \mathrm{A} \leq 1$

Hence, the correct answer is option: 'B'.

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