The area of the region $\{(x,y): x^2 + y^2 ≤ 1 ≤ x + y\}$ is

$\frac{1}{2}\left(\frac{\pi}{2}-1\right)$ sq.unit

The correct answer is Option (3) → $\frac{1}{2}\left(\frac{\pi}{2}-1\right)$ sq.unit

The region is defined by:

• Inside the unit circle: \( x^2 + y^2 \leq 1 \)
• Above the line: \( x + y \geq 1 \)

The area bounded by both is the region of the unit circle where \( x + y \geq 1 \).

The line \( x + y = 1 \) intersects the circle \( x^2 + y^2 = 1 \) at,

Points of intersection:

Substitute \( y = 1 - x \) into the circle:

\( x^2 + (1 - x)^2 = 1 \Rightarrow x^2 + 1 - 2x + x^2 = 1 \Rightarrow 2x^2 - 2x = 0 \Rightarrow x(x - 1) = 0 \)

So intersection points are \( (0,1) \) and \( (1,0) \).

The area under the arc from (0,1) to (1,0) lies above the line \( x + y = 1 \), which forms a sector of 90° (or \( \frac{\pi}{2} \) radians) and a triangle with vertices (0,1), (1,0), and (0,0).

Area of quarter circle: \( \frac{1}{4} \pi r^2 = \frac{\pi}{4} \)

Area of triangle: \( \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \)

Required area = Area of sector − Area of triangle = \( \frac{\pi}{4} - \frac{1}{2} \)