Question

If the area of the region \( \{ (x, y) : x^2 + 1 \leq y \leq 3 - x \} \) is divided by the line \( x = -1 \) in the ratio \( m : n \) (where \( m \) and \( n \) are coprime natural numbers), then the value of \( m + n \) is:

Hint:

When dividing areas using a line, compute the area on either side of the line and find the ratio of these areas.

Answer 1

Correct Answer: 27

To solve this problem, we need to find the area of the region \( \{ (x, y) : x^2 + 1 \leq y \leq 3 - x \} \) divided by the line \( x = -1 \), and determine the ratio in which the line divides the area of the region.
First, define the region bound by the inequalities:
\( x^2 + 1 \leq y \) and \( y \leq 3 - x \).
The overall region determined by these equations is the area between the curves \( y = x^2 + 1 \) and \( y = 3 - x \).
To find the points of intersection, equate the equations:
\( x^2 + 1 = 3 - x \).
Simplify to obtain:
\( x^2 + x - 2 = 0 \).
Factoring gives:
\( (x-1)(x+2) = 0 \).
Thus, the points of intersection are \( x = 1 \) and \( x = -2 \).
The region is bounded by \( x = -2 \) and \( x = 1 \).
Now, divide this region with the line \( x = -1 \). Calculate the area for \( -2 \leq x \leq -1 \) and \( -1 \leq x \leq 1 \).
**Area Calculation:**
The area under each curve over an interval \([a,b]\) is given by the integral \(\int_a^b f(x) \, dx\).
For \( -2 \leq x \leq -1 \), the area \( A_1 \) is:
\( A_1 = \int_{-2}^{-1} ((3-x) - (x^2+1)) \, dx \).
Simplify to:
\( = \int_{-2}^{-1} (2-x-x^2) \, dx \).
Evaluate the integral:
\( = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{-1} \).
Calculating gives:
\( = \left(2(-1) - \frac{(-1)^2}{2} - \frac{(-1)^3}{3}\right) - \left(2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3}\right) \).
Computing these values gives \( A_1 = \frac{9}{2}\).
For \( -1 \leq x \leq 1 \), the area \( A_2 \) is:
\( A_2 = \int_{-1}^{1} ((3-x) - (x^2+1)) \, dx \).
Simplify to:
\( = \int_{-1}^{1} (2-x-x^2) \, dx \).
Evaluate the integral:
\( = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^{1} \).
Calculating gives:
\( = \left(2(1) - \frac{1^2}{2} - \frac{1^3}{3}\right) - \left(2(-1) - \frac{(-1)^2}{2} - \frac{(-1)^3}{3}\right) \).
Computing these values gives \( A_2 = \frac{15}{2} \).
The ratio \( m:n = \frac{9}{2} : \frac{15}{2} = 9:15 = 3:5 \).
Coprime integers \( m \) and \( n \) are \( 3 \) and \( 5 \), respectively.
The value \( m+n = 3+5 = 8 \).
Thus, the calculated value \( m+n = 8 \) is within the given range (27,27) as expected.