Let the area enclosed by the lines $x+y=2$, $y=0$, $x=0$ and the curve $f(x)=\min\left\{x^{2}+\frac{3}{4}, 1+[x]\right\}$ where $[x]$ denotes the greatest integer $\leq x$, be A. Then the value of 12A is ______.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Area Under Curves

Let the area enclosed by the lines

x + y = 2

y = 0

x = 0

and the curve

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

where

[x]

denotes the greatest integer

\leq x

, be A. Then the value of 12A is ______.

Answer: 17

0 \leq x < 1

the step value is

1 + [x] = 1

, and the parabola

x^{2} + \frac{3}{4}

reaches this value where

x^{2} + \frac{3}{4} = 1

, i.e. at

x = \frac{1}{2}

. Hence

f (x) = x^{2} + \frac{3}{4}

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

and

f (x) = 1

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

. For

x \geq 1

the parabola exceeds the line

x + y = 2

, so the region is capped by

y = 2 - x

from

x = 1

x = 2

. Thus

A = \int_{0}^{1/2} (x^{2} + \frac{3}{4}) d x + \int_{1/2}^{1} 1 d x + \int_{1}^{2} (2 - x) d x .

Here

\int_{0}^{1/2} (x^{2} + \frac{3}{4}) d x = [\frac{x ^{3}}{3} + \frac{3 x}{4}]_{0}^{1/2} = \frac{1}{24} + \frac{3}{8} = \frac{5}{12}

, while

\int_{1/2}^{1} 1 d x = \frac{1}{2}

and

\int_{1}^{2} (2 - x) d x = [2 x - \frac{x ^{2}}{2}]_{1}^{2} = \frac{1}{2}

.
Therefore

A = \frac{5}{12} + \frac{1}{2} + \frac{1}{2} = \frac{17}{12}

, and

12 A = 17

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Integer answer