JEE Advanced · Mathematics · 25. AOD

Consider all rectangles lying in the region $\{(x, y) \in R \times R : 0 \leq x \leq \frac{π}{2} a n d 0 \leq y \leq 2 \sin (2 x)\}$ and having one side on the $x$ -axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is

A $\frac{3 π}{2}$
B $π$
C $\frac{π}{2 \sqrt{3}}$
D $\frac{π \sqrt{3}}{2}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{π}{2 \sqrt{3}}$

Step-by-step Solution

Detailed explanation

2 \sin 2 θ_{1} = 2 \sin 2 θ_{2}

\Rightarrow 2 θ_{1} = π - 2 θ_{2}

\Rightarrow θ_{2} = \frac{π}{2} - θ_{1} \dots . (1)

Now perimeter

p (θ_{1}, θ_{2}) = 2 \{(θ_{2} - θ_{1}) + 2 \sin 2 θ_{1}\}

\Rightarrow p (θ_{1}) = 2 [\frac{π}{2} - 2 θ_{1} + 2 \sin 2 θ_{1}]

\Rightarrow p^{'} (θ_{1}) = 2 (- 2 + 4 \cos 2 θ_{1})

Also,

p^{"} (θ_{1}) = 2 (- 8 \sin 2 θ_{1}) < 0

For maximum perimeter,

p^{'} (θ_{1}) = 0 & p^{"} (θ_{1}) < 0

\Rightarrow \cos 2 θ_{1} = \frac{1}{2} \Rightarrow 2 θ_{1} = \frac{π}{3} \Rightarrow θ_{1} = \frac{π}{6}

Now area at

θ_{1} = \frac{π}{6}

will be

(θ_{2} - θ_{1}) \times 2 \sin 2 θ_{1}

= (\frac{π}{2} - 2 θ_{1}) \cdot 2 \sin 2 θ_{1}

= (\frac{π}{2} - \frac{π}{3}) \times 2 \sin \frac{π}{3}

= \frac{π}{6} \cdot \sqrt{3} = \frac{π}{2 \sqrt{3}}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Consider all rectangles lying in the region x, y∈R×R:0≤x≤π2 and 0≤y≤2sin2x and having one side on the x-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is

Step-by-step Solution

Consider all rectangles lying in the region $\{(x, y) \in R \times R : 0 \leq x \leq \frac{π}{2} a n d 0 \leq y \leq 2 \sin (2 x)\}$ and having one side on the $x$ -axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is