JEE Main 2023MathematicsArea Under CurvesHardNumerical

JEE Main 2023 — Area Under Curves Question with Solution

JEE Main 2023 (08 Apr Shift 2)

Question

Let the area enclosed by the lines x+y=2, y=0, x=0 and the curve f(x)=minx2+34,1+[x] where [x] denotes the greatest integer ≤x, be A. Then the value of 12A is

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Show full solutionCorrect answer: 17
Correct answer
17

Step-by-step explanation

Given,

The lines x+y=2, y=0, x=0 and the curve f(x)=minx2+34,1+[x] where [x] denotes the greatest integer ≤x,

Now plotting the diagram of given function we get,

 

Now from above diagram, the area enclosed is given by,

A=∫012x2+34dx+1212+32×1=512+1

⇒ 12A=17

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About this question

This is a previous-year question from JEE Main 2023, covering the Area Under Curves chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.