UPSC NDA I 2016 Mathematics Solved Paper

Concept:Simplify the given trigonometric expression into a form involving $\sin^2 2\theta$ and then use the range of $\sin^2 2\theta$ to find the minimum value of $f(\theta)$.Explanation:Step 1: Start with $f(\theta) = 4(\sin^2 \theta + \cos^4 \theta)$.Step 2: Rewrite $\sin^2 \theta = 1 - \cos^2 \theta$, so $f(\theta) = 4(1 - \cos^2 \theta + \cos^4 \theta)$.Step 3: Alternatively, factor: $f(\theta) = 4[1 - \cos^2 \theta (1 - \cos^2 \theta)] = 4(1 - \sin^2 \theta \cos^2 \theta)$.Step 4: Since $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$, we get $\sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2 2\theta$.Step 5: Thus $f(\theta) = 4\left(1 - \frac{1}{4}\sin^2 2\theta\right) = 4 - \sin^2 2\theta$.Step 6: Known range: $-1 \le \sin 2\theta \le 1$, so squaring gives $0 \le \sin^2 2\theta \le 1$.Step 7: Therefore $4 - 1 \le 4 - \sin^2 2\theta \le 4 - 0$, i.e. $3 \le f(\theta) \le 4$.Step 8: The smallest value of $f(\theta)$ in this interval is $3$.Answer:$3$