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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2018 · Mathematics · Trigonometric Ratios & Identities

If A= ^2 θ+ ^4 θ , the for all real θ , which one of the following...

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

Options

  1. A1≤A≤21 \leq \mathrm{A} \leq 2
  2. B34≤A≤1\frac{3}{4} \leq \mathrm{A} \leq 1
  3. C1316≤A≤1\frac{13}{16} \leq \mathrm{A} \leq 1
  4. D34≤\frac{3}{4} \leq A ≤1316\leq \frac{13}{16}

Correct answer

B. 34≤A≤1\frac{3}{4} \leq \mathrm{A} \leq 1

Step-by-step solution

A=sin⁡2θ+cos⁡4θ=sin⁡2θ+(1−sin⁡2θ)2=1+sin⁡4θ−sin⁡2θ=1−sin⁡2θ(1−sin⁡2θ)=1−sin⁡2θ⋅cos⁡2θ=4−4sin⁡2θ⋅cos⁡2θ4=4−sin⁡2(2θ)4\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≤sin⁡22θ≤1∴A=4−04 or  4−14⇒34≤A≤10 \leq \sin ^2 2 \theta \leq 1\therefore \quad \mathrm{A}=\frac{4-0}{4} \text{\text{ or } } \frac{4-1}{4} \Rightarrow \frac{3}{4} \leq \mathrm{A} \leq 1
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