Question

Let the line \( x = -1 \) divide the area of the region \[ \{(x,y): 1 + x^2 \le y \le 3 - x\} \] in the ratio \( m:n \), where \( \gcd(m,n)=1 \). Then \( m+n \) is equal to

• A. 27
• B. 26
• C. 25
• D. 28

Hint:

When a vertical line divides a region bounded by curves, compute the area on each side separately using definite integrals before forming the ratio.

Answer 1

The Correct Option is B

The region is bounded by the curves \[ y = 1 + x^2 \quad \text{and} \quad y = 3 - x. \] Step 1: Find the limits of integration.
The curves intersect when \[ 1 + x^2 = 3 - x \] \[ x^2 + x - 2 = 0 \] \[ (x+2)(x-1)=0 \Rightarrow x=-2,\;1 \] Thus, the region extends from \( x=-2 \) to \( x=1 \).
Step 2: Compute the area to the left of the line \( x=-1 \).
\[ A_1 = \int_{-2}^{-1} \big[(3-x)-(1+x^2)\big]\,dx = \int_{-2}^{-1} (2 - x - x^2)\,dx \] \[ A_1 = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{-1} = \frac{13}{6} \] Step 3: Compute the area to the right of the line \( x=-1 \).
\[ A_2 = \int_{-1}^{1} (2 - x - x^2)\,dx \] \[ A_2 = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-1}^{1} = \frac{13}{6} \] Step 4: Find the ratio and required sum.
\[ A_1 : A_2 = \frac{13}{6} : \frac{13}{6} = 13 : 13 \] Reducing the ratio to coprime integers, \[ m:n = 13:13 \Rightarrow m+n = 26 \] Final Answer: \[ \boxed{26} \]