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JEE Main 20264 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2026 · 4 April · Evening Shift · Mathematics · Application of Derivatives

_ 0 ≤ x ≤ π (16 ((x)/(2)) ^3((x)/(2))) is equal to:

max⁡0≤x≤π\max_{0 \leq x \leq \pi}\left(16\sin\left(\dfrac{x}{2}\right)\cos^3\left(\dfrac{x}{2}\right)\right) is equal to:

Options

  1. A332\dfrac{3\sqrt{3}}{2}
  2. B333\sqrt{3}
  3. C434\sqrt{3}
  4. D636\sqrt{3}

Correct answer

B. 333\sqrt{3}

Step-by-step solution

Let y=x2y = \dfrac{x}{2}. Since 0≤x≤π0 \leq x \leq \pi, we have 0≤y≤π20 \leq y \leq \dfrac{\pi}{2}.

The given expression becomes f(y)=16sin⁡ycos⁡3yf(y) = 16\sin y \cos^3 y.

Differentiating with respect to yy:

f′(y)=16(cos⁡y⋅cos⁡3y+sin⁡y⋅3cos⁡2y(−sin⁡y))f'(y) = 16(\cos y \cdot \cos^3 y + \sin y \cdot 3\cos^2 y(-\sin y))

f′(y)=16cos⁡2y(cos⁡2y−3sin⁡2y)f'(y) = 16\cos^2 y(\cos^2 y - 3\sin^2 y)

For maximum value, f′(y)=0f'(y) = 0:

cos⁡2y−3sin⁡2y=0⇒tan⁡2y=13\cos^2 y - 3\sin^2 y = 0 \Rightarrow \tan^2 y = \dfrac{1}{3}

Since y∈[0,π2]y \in \left[0, \dfrac{\pi}{2}\right], tan⁡y=13⇒y=π6\tan y = \dfrac{1}{\sqrt{3}} \Rightarrow y = \dfrac{\pi}{6}.

Substituting y=π6y = \dfrac{\pi}{6} in f(y)f(y):

f(π6)=16sin⁡(π6)cos⁡3(π6)f\left(\dfrac{\pi}{6}\right) = 16\sin\left(\dfrac{\pi}{6}\right)\cos^3\left(\dfrac{\pi}{6}\right)

f(π6)=16×12×(32)3f\left(\dfrac{\pi}{6}\right) = 16 \times \dfrac{1}{2} \times \left(\dfrac{\sqrt{3}}{2}\right)^3

f(π6)=8×338=33f\left(\dfrac{\pi}{6}\right) = 8 \times \dfrac{3\sqrt{3}}{8} = 3\sqrt{3}

The maximum value is 333\sqrt{3}.

Answer: 333\sqrt{3}
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