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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2016 · Mathematics · Trigonometric Ratios & Identities

Consider the function f(θ)=4 ( ^2 θ+ ^4 θ) What is the minimum...

Consider the function f(θ)=4\mathrm{f}(\theta)=4\left(\sin ^2 \theta+\cos ^4 \theta\right) What is the minimum value of the function f(θ)f(\theta)?

Options

  1. A0
  2. B1
  3. C2
  4. D3

Correct answer

D. 3

Step-by-step solution

f(θ)=4(sin⁡2θ+cos⁡4θ)=4(sin⁡2θ+cos⁡2θ(1−sin⁡2θ))=4(sin⁡2θ+cos⁡2θ−sin⁡2θcos⁡2θ)=4(1−14sin⁡22θ)[∵sin⁡2θ=2sin⁡θcos⁡θ]\begin{aligned} f(\theta)= & 4\left(\sin ^2 \theta+\cos ^4 \theta\right) \\ = & 4\left(\sin ^2 \theta+\cos ^2 \theta\left(1-\sin ^2 \theta\right)\right) \\ = & 4\left(\sin ^2 \theta+\cos ^2 \theta-\sin ^2 \theta \cos ^2 \theta\right) \\ = & 4\left(1-\frac{1}{4} \sin ^2 2 \theta\right) \\ & {[\because \sin 2 \theta=2 \sin \theta \cos \theta] }\end{aligned} For minimum value of f(θ),sin⁡22θf(\theta), \sin ^2 2 \theta should be maximum i.e. sin⁡22θ=1\sin ^2 2 \theta=1. \left.f(\theta)\right|_{\min }=4\left(1-\frac{1}{4}(1)\right)=4 \times \frac{3}{4}=3$
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