JEE Advanced 2020 · Paper 1 · Mathematics Q6

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MathematicsSingle correct+3 / -1Question 42 of 54

Consider all rectangles lying in the region

$\left\{(x,y)\in\mathbb R\times\mathbb R : 0 \le x \le \frac\pi2 \text{ and } 0 \le y \le 2\sin(2x)\right\}$

and having one side on the $x$ -axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is

\dfrac{3\pi}{2}

\pi

\dfrac{\pi}{2\sqrt3}

\dfrac{\pi\sqrt3}{2}

Answer and worked solution

Official answer: C

Setting up. For the largest perimeter the rectangle's top touches the curve at both ends, so it is symmetric about $x=\pi/4$ . It spans $[a,\frac\pi2-a]$ with height $2\sin2a$ , $0<a\le\frac\pi4$ .

Perimeter.
$P(a)=2\left(\frac\pi2-2a\right)+4\sin2a,\qquad P'(a)=-4+8\cos2a=0\implies a=\frac\pi6.$
$P''<0$ there, so this is the maximum.

Area. The width is $\frac\pi2-\frac\pi3=\frac\pi6$ and the height is $2\sin\frac\pi3=\sqrt3$ :
$\text{Area}=\frac{\pi\sqrt3}6=\frac\pi{2\sqrt3}.$
(C)

Topics:Maxima and minima

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JEE Advanced 2020 · Paper 1, Mathematics question 6