JEE Main 2024 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 5 April 2024, Shift 1, Maths Q9

Question 9 of 90 in this shift, Maths question 9 of 30, Section A.

Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let

a

and

b

be the sides of the rectangle PQRS when its area is maximum. Then

(a+b)^2

is equal to :

(1) $72$ Official answer
(2) $64$
(3) $80$
(4) $60$

Official answer

Option 1

NTA final key.

Chapter: Application of Derivatives
Topic: Maxima and Minima
Idea tested: Max/Min of Inscribed Geometric Figures

Same idea in other shifts

Asked 3× in all

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30 Jan 2024, Shift 1 · Q7The maximum area of a triangle whose one vertex is at $(0,0)$ and the other two vertices lie on the curve $y=-2x^2+54$ at points $(x,y)$ …MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.