Let the area enclosed by the lines x + y = 2, y = 0 , x = 0 and the curve f ( x ) = min\ x ^ 2 + (3)/(4), 1 + [ x ]\ where [ x ] denotes the greatest integer ≤ x , be A . Then the value of 12A is

Correct answer: 0 Given, The lines x + y = 2, y = 0 , x = 0 and the curve f ( x ) = min\ x ^ 2 + (3)/(4), 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 = [ ∫ _ 0 ^ (1)/(2) ( x ^ 2 + (3)/(4))dx] + (1)/(2)((1)/(2) + (3)/(2)) × 1 = (5)/(12) + 1 ⇒ 12A = 17

JEE Main 20238 AprilEvening Shift

Exam: JEE Main
Year: 2023
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2023 · 8 April · Evening Shift · Mathematics · Area Under Curves

Let the area enclosed by the lines x + y = 2, y = 0 , x = 0 and the...

Let the area enclosed by the lines

x + y = 2, y = 0

x = 0

and the curve

f ( x ) = min\left\{{x}^{2} + \dfrac{3}{4}, 1 + [ x ]\right\}

where

[ x ]

denotes the greatest integer

\le x

, be

A

. Then the value of

12A

is $Let the area enclosed by the lines x + y = 2, y = 0 , x = 0 and the curve f ( x ) = min\ x ^ 2 + (3)/(4), 1 + [ x ]\$

Correct answer

Step-by-step solution

Given,
The lines $x + y = 2, y = 0$ , $x = 0$ and the curve $f ( x ) = min\left\{{x}^{2} + \dfrac{3}{4}, 1 + [ x ]\right\}$ where $[ x ]$ denotes the greatest integer $\le x$ ,
Now plotting the diagram of given function we get,
$Let the area enclosed by the lines x + y = 2, y = 0 , x = 0 and the curve f ( x ) = min\ x ^ 2 + (3)/(4), 1 + [ x ]\$

Now from above diagram, the area enclosed is given by,
$A = \left[{∫}_{0}^{\dfrac{1}{2}}\left({x}^{2} + \dfrac{3}{4}\right)dx\right] + \dfrac{1}{2}\left(\dfrac{1}{2} + \dfrac{3}{2}\right) \times 1 = \dfrac{5}{12} + 1$
$⇒ 12A = 17$

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