max(0≤ x≤ π)(16 sin(x/2)cos³(x/2)) is equal to:

Solution

Maths sir · 4 steps

Put t = x/2, differentiate sin t cos³ t, and the maximum comes at tan t = 1/√3.

1. Substitute. Let t = x/2. As x runs over [0, π], t runs over [0, π/2], and we must maximise f(t) = 16 sin t cos³ t.

   f'(t) = 16(cos⁴ t - 3 sin² t cos² t) = 16 cos² t(cos² t - 3 sin² t)

2. Critical point. Inside the interval cos t ≠ 0, so f'(t) = 0 gives tan² t = 1/3, that is t = π/6. For t < π/6, f' > 0, and for t > π/6, f' < 0, so this is a maximum.
3. Evaluate. f(π/6) = 16 × 1/2 × (√3/2)³ = 16 × 1/2 × 3√3/8 = 3√3. At the endpoints t = 0 and t = π/2, f = 0.
4. So. The maximum value is 3√3, reached at x = π/3. The answer is (b).

The idea to keep. For sinᵐ t cosⁿ t, differentiate and factor. The maximum sits where tan² t = m/n, and on a closed interval you check the endpoints too.

Answer from the NTA final answer key. Solution by PrepWiser.