Q160 / 270 · NDA 2016 (Phase 1) · Mathematics

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

  • A
    0
  • B
    1
  • C
    2
  • Correct answer
    D
    3

Solution

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

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

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