max 0 <= x <= pi (16 sin(x/2) cos^3(x/2)) is equal to:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q12 – JEE-MAIN 2026

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Home Maths Continuity and DifferentiabilityQuestion #151

max_{0 \leq x \leq π} (16 \sin (\frac{x}{2}) \cos^{3} (\frac{x}{2})) is equal to:

Clue 1 of 3: Maxima on a Closed Interval

The problem asks for the maximum value of a function on a closed interval. This can be found by checking the function's values at its critical points and at the endpoints of the interval.

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Ninja Strategy•Bounding and Test Points

Quickly establish an upper bound for the function and evaluate it at a convenient point like $x = π / 2$ to eliminate options that are clearly too large or too small.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Define and Simplify the Function✦ Active

Let the given function be $f (x) = 16 \sin (\frac{x}{2}) \cos^{3} (\frac{x}{2})$ for $x \in [0, π]$ . To simplify, substitute $u = \frac{x}{2}$ . As $x$ varies from $0$ to $π$ , $u$ varies from $0$ to $\frac{π}{2}$ .

g (u) = 16 \sin (u) \cos^{3} (u) for u \in [0, \frac{π}{2}]

Step 2: Find Critical Points using Differentiation○ Expand

Differentiate $g (u)$ with respect to $u$ using the product rule and set the derivative to zero.

g^{'} (u) = 16 [\cos (u) \cdot \cos^{3} (u) + \sin (u) \cdot 3 \cos^{2} (u) (- \sin (u))] = 16 \cos^{2} (u) (\cos^{2} (u) - 3 \sin^{2} (u))

Setting $g^{'} (u) = 0$ gives two conditions: $\cos^{2} (u) = 0$ or $\cos^{2} (u) - 3 \sin^{2} (u) = 0$ . The first condition gives $u = \frac{π}{2}$ . The second gives $\tan^{2} (u) = \frac{1}{3}$ , which for $u \in [0, \frac{π}{2}]$ implies $\tan (u) = \frac{1}{\sqrt{3}}$ , so $u = \frac{π}{6}$ .

Step 3: Evaluate the Function at Critical Points and Endpoints○ Expand

The points to check are the endpoints $u = 0, u = \frac{π}{2}$ and the critical point $u = \frac{π}{6}$ .

g (0) = 16 \sin (0) \cos^{3} (0) = 0

g (\frac{π}{2}) = 16 \sin (\frac{π}{2}) \cos^{3} (\frac{π}{2}) = 0

g (\frac{π}{6}) = 16 \sin (\frac{π}{6}) \cos^{3} (\frac{π}{6}) = 16 (\frac{1}{2}) {(\frac{\sqrt{3}}{2})}^{3} = 16 \cdot \frac{1}{2} \cdot \frac{3 \sqrt{3}}{8} = 3 \sqrt{3}

Comparing the values ${0, 0, 3 \sqrt{3}}$ , the maximum value of the function is $3 \sqrt{3}$ .

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Maths Question 12 – JEE-MAIN 2026

Clue 1 of 3: Maxima on a Closed Interval

🧠 Full Solution Path