Question

If the area of the region \[ \{(x, y) : 1 - 2x \le y \le 4 - x^2,\ x \ge 0,\ y \ge 0\} \] is \[ \frac{\alpha}{\beta}, \] \(\alpha, \beta \in \mathbb{N}\), \(\gcd(\alpha, \beta) = 1\), then the value of \[ (\alpha + \beta) \] is : 

• A. 67
• B. 85
• C. 91
• D. 73

Hint:

Always sketch the region to identify which boundary is higher and where the integration limits should split.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The region is bounded by a parabola $y = 4 - x^2$, a line $y = 1 - 2x$, and axes in the first quadrant. We integrate to find the area between these curves.

Step 2: Detailed Explanation:
Intersection of $y = 4 - x^2$ with $x$-axis is at $x = 2$.
Intersection of $y = 1 - 2x$ with $x$-axis is at $x = 1/2$.
For $0 \le x \le 1/2$, the lower boundary is the line $y = 1 - 2x$.
For $1/2 \le x \le 2$, the lower boundary is the $x$-axis ($y = 0$).
Total Area $A = \int_0^2 (4 - x^2) dx - \int_0^{1/2} (1 - 2x) dx$.
\[ \int_0^2 (4 - x^2) dx = [4x - \frac{x^3}{3}]_0^2 = 8 - \frac{8}{3} = \frac{16}{3} \]
\[ \int_0^{1/2} (1 - 2x) dx = [x - x^2]_0^{1/2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \]
Area $A = \frac{16}{3} - \frac{1}{4} = \frac{64 - 3}{12} = \frac{61}{12}$.
Here $\alpha = 61$ and $\beta = 12$. Since $\gcd(61, 12) = 1$:
$\alpha + \beta = 61 + 12 = 73$.

Step 3: Final Answer:
The value is 73.