Consider the following statements : 1. f(θ)=2 has no solution 2. f(θ)=7/2 has a solution Which of the above statements is/are correct [NDA I 2016]

Correct Answer is : Both 1 and 2

Correct Answer is : Both 1 and 2

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UPSC NDA I 2016 Mathematics Solved Paper

For the next three(3) items that follow :

Consider the function

f(\theta)=4(\sin^2\theta+\cos^4\theta)

[NDA I 2016]

Question : 11 of 120

Marks: +1, -0

f(\theta)=2

f(\theta)=\frac{7}{2}

Solution:

f(\theta)=4(\sin^2\theta+\cos^4\theta)

=4(\sin^{2}\theta+\cos^{2}\theta(1-\sin^{2}\theta))

=4(\sin^{2}\theta+\cos^{2}\theta-\sin^{2}\theta\cos^{2}\theta)

=4\left(1-\frac{1}{4}\sin^{2}2\theta\right)

[\because \sin2\theta=2\sin\theta\cos\theta]

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 \sin2\theta=\pm\sqrt{2}

Since sin θ cannot have value greater than 1 & less than

-1

Hence

f(\theta)=2

has no solution

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}

\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 \sin2\theta=\pm\frac{1}{\sqrt{2}}

\Rightarrow \sin 2\theta=\pm \sin \frac{\pi}{4}

\Rightarrow \sin2\theta=\sin\left(\pm\frac{\pi}{4}\right)

\Rightarrow 2\theta=\pm\frac{\pi}{4}

\theta=(\pm)\frac{\pi}{8}

Hence

f(\theta)=\frac{7}{2}

has a solution

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