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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) Consider the following...

Consider the function f(θ)=4(sin⁡2θ+cos⁡4θ)f(\theta)=4\left(\sin ^2 \theta+\cos ^4 \theta\right) Consider the following statements: 1. f(θ)=2f(\theta)=2 has no solution. 2. f(θ)=72\mathrm{f}(\theta)=\frac{7}{2} has a solution. Which of the above statements is/are correct?

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

C. Both 1 and 2

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} f(θ)=24(1−14sin⁡22θ)=2⇒1−14sin⁡22θ=24⇒−14sin⁡22θ=+12−1⇒−14sin⁡22θ=−12⇒sin⁡22θ=2⇒sin⁡2θ=±2\begin{aligned} & f(\theta)=2 \\ & 4\left(1-\frac{1}{4} \sin ^2 2 \theta\right)=2 \\ & \Rightarrow 1-\frac{1}{4} \sin ^2 2 \theta=\frac{2}{4} \\ & \Rightarrow-\frac{1}{4} \sin ^2 2 \theta=+\frac{1}{2}-1 \\ & \Rightarrow-\frac{1}{4} \sin ^2 2 \theta=-\frac{1}{2} \\ & \Rightarrow \sin ^2 2 \theta=2 \\ & \Rightarrow \sin 2 \theta= \pm \sqrt{2} \end{aligned} Since sin⁡θ\sin \theta cannot have vlaue greater than 11 &\& less than -1. Hence f(θ)=2f(\theta)=2 has no solution. f(θ)=724(1−14sin⁡22θ)=72⇒(1−14sin⁡22θ)=78\begin{aligned} & f(\theta)=\frac{7}{2} \\ & 4\left(1-\frac{1}{4} \sin ^2 2 \theta\right)=\frac{7}{2} \\ & \Rightarrow\left(1-\frac{1}{4} \sin ^2 2 \theta\right)=\frac{7}{8}\end{aligned} ⇒−14sin⁡22θ=78−1⇒−14sin⁡22θ=−18⇒sin⁡22θ=12⇒sin⁡2θ=±12⇒sin⁡2θ=±sin⁡π4⇒sin⁡2θ=sin⁡(±π/4)⇒2θ=±π/4θ=(±)π8\begin{aligned} & \Rightarrow-\frac{1}{4} \sin ^2 2 \theta=\frac{7}{8}-1 \\ & \Rightarrow-\frac{1}{4} \sin ^2 2 \theta=-\frac{1}{8} \\ & \Rightarrow \sin ^2 2 \theta=\frac{1}{2} \\ & \Rightarrow \sin 2 \theta= \pm \frac{1}{\sqrt{2}} \\ & \Rightarrow \sin 2 \theta= \pm \sin \frac{\pi}{4} \\ & \Rightarrow \sin 2 \theta=\sin ( \pm \pi / 4) \\ & \Rightarrow 2 \theta= \pm \pi / 4 \\ & \theta=( \pm) \frac{\pi}{8} \end{aligned} Hence f(θ)=72f(\theta)=\frac{7}{2} has a solution.
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