🛠️ JEE➗ Maths

Let \(\alpha\) and \(\beta\) respectively be the maximum and the minimum values of the function \( f(\theta)=4\left(\sin…

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Let \(\alpha\) and \(\beta\) respectively be the maximum and the minimum values of the function

\( f(\theta)=4\left(\sin ^4\left(\frac{7 \pi}{2}-\theta\right)+\sin ^4(11 \pi+\theta)\right) -2\left(\sin ^6\left(\frac{3 \pi}{2}-\theta\right)+\sin ^6(9 \pi-\theta)\right)\). Then \(\alpha +2\beta\) is equal to:

[JEE Main 2026, 23 Jan (Shift 2)]

a

\(6\)

b

\(3\)

c

\(4\)

d

\(5\)

✓ Correct answer: d)

\(5\)

Explanation

Given:

\( f(\theta)=4\left(\sin ^4\left(\frac{7 \pi}{2}-\theta\right)+\sin ^4(11 \pi+\theta)\right) -2\left(\sin ^6\left(\frac{3 \pi}{2}-\theta\right)+\sin ^6(9 \pi-\theta)\right)\)

\(\Rightarrow f(\theta)=4\left(\cos ^4(\theta)+\sin ^4(\theta)\right)-2\left(\cos ^6 \theta+\sin ^6 \theta\right)\)

\(\Rightarrow f(\theta)=4\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)-2\left(1-3 \sin ^2 \theta \cos ^2 \theta\right)\)

\(\Rightarrow f(\theta)=2-2 \sin ^2 \theta \cos ^2 \theta\)

\(\Rightarrow f(\theta)=2-\frac{\sin ^2(2 \theta)}{2}\)

So, \(\alpha=f(\theta)_{\max }=2\),

\(\beta=f(\theta)_{\min }=\frac{3}{2}\)

\(\Rightarrow \alpha+2 \beta=5\)

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