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JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Area Under Curves

Let the area of the region

{(x, y) : ∣2 x - 1∣ \leq y \leq x^{2} - x, 0 \leq x \leq 1}

be A. Then

(6 A + 11)^{2}

is equal to

Answer: 125

Solution

On

[0, 1]

,

∣ x^{2} - x ∣ = x - x^{2}

, and the region exists where

∣2 x - 1∣ \leq x - x^{2} .

On

[0, \frac{1}{2}]

this is

1 - 2 x \leq x - x^{2}

, giving

x \geq \frac{3 - 5}{2}

; on

[\frac{1}{2}, 1]

it is

2 x - 1 \leq x - x^{2}

, giving

x \leq \frac{5 - 1}{2} .

With

α = \frac{3 - 5}{2}, β = \frac{5 - 1}{2}

,

A = \int_{α}^{β} (x - x^{2}) d x - \int_{α}^{β} ∣2 x - 1∣ d x = \frac{16 - 7 5}{6} - (\frac{9}{2} - 25) = \frac{5 5 - 11}{6} .

Then

6 A + 11 = 55

, so

(6 A + 11)^{2} = 125.

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Integer answer

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