Q161 / 270 · NDA 2016 (Phase 1) · Mathematics

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

  • A
    1 only
  • B
    2 only
  • Correct answer
    C
    Both 1 and 2
  • D
    Neither 1 nor 2

Solution

$\begin{aligned} f(\theta)= & 4\left(\sin ^2 \theta+\cos ^4 \theta\right) \\ = & 4\left(\sin ^2 \theta+\cos ^2 \theta\left(1-\sin ^2 \theta\right)\right) \\ = & 4\left(\sin ^2 \theta+\cos ^2 \theta-\sin ^2 \theta \cos ^2 \theta\right) \\ = & 4\left(1-\frac{1}{4} \sin ^2 2 \theta\right) \\ & {[\because \sin 2 \theta=2 \sin \theta \cos \theta] }\end{aligned}$ $\begin{aligned} & f(\theta)=2 \\ & 4\left(1-\frac{1}{4} \sin ^2 2 \theta\right)=2 \\ & \Rightarrow 1-\frac{1}{4} \sin ^2 2 \theta=\frac{2}{4} \\ & \Rightarrow-\frac{1}{4} \sin ^2 2 \theta=+\frac{1}{2}-1 \\ & \Rightarrow-\frac{1}{4} \sin ^2 2 \theta=-\frac{1}{2} \\ & \Rightarrow \sin ^2 2 \theta=2 \\ & \Rightarrow \sin 2 \theta= \pm \sqrt{2} \end{aligned}$ Since $\sin \theta$ cannot have vlaue greater than $1$ $\&$ less than -1. Hence $f(\theta)=2$ has no solution. $\begin{aligned} & f(\theta)=\frac{7}{2} \\ & 4\left(1-\frac{1}{4} \sin ^2 2 \theta\right)=\frac{7}{2} \\ & \Rightarrow\left(1-\frac{1}{4} \sin ^2 2 \theta\right)=\frac{7}{8}\end{aligned}$ $\begin{aligned} & \Rightarrow-\frac{1}{4} \sin ^2 2 \theta=\frac{7}{8}-1 \\ & \Rightarrow-\frac{1}{4} \sin ^2 2 \theta=-\frac{1}{8} \\ & \Rightarrow \sin ^2 2 \theta=\frac{1}{2} \\ & \Rightarrow \sin 2 \theta= \pm \frac{1}{\sqrt{2}} \\ & \Rightarrow \sin 2 \theta= \pm \sin \frac{\pi}{4} \\ & \Rightarrow \sin 2 \theta=\sin ( \pm \pi / 4) \\ & \Rightarrow 2 \theta= \pm \pi / 4 \\ & \theta=( \pm) \frac{\pi}{8} \end{aligned}$ Hence $f(\theta)=\frac{7}{2}$ has a solution.

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

Relevant Questions