Question

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

Hint:

For area-division problems:
First find intersection points of the curves
Split the integral at the given dividing line
Compute areas separately and simplify the ratio

Answer 1

Correct Answer: 5

To find the ratio in which the area of the region is divided by the line \(x = -1\), we first identify the given region:

The inequalities \(x^2 + 1 \leq y \leq 3-x\) describe a region bounded by the curves \(y = x^2 + 1\) and \(y = 3 - x\).

First, find the points of intersection between \(y = x^2 + 1\) and \(y = 3 - x\):

Set \(x^2 + 1 = 3 - x\), which simplifies to:

\(x^2 + x - 2 = 0\). Factor the quadratic:

\((x - 1)(x + 2) = 0\). Thus, \(x = 1\) and \(x = -2\).

Next, determine the area of the region between these curves from \(x = -2\) to \(x = 1\):

Area = \(\int_{-2}^{1} [(3-x) - (x^2 + 1)]\, dx = \int_{-2}^{1} (2-x-x^2)\, dx\).

Calculate this integral:

\(\int 2\, dx = 2x\), \(\int -x\, dx = -\frac{x^2}{2}\), \(\int -x^2\, dx = -\frac{x^3}{3}\).

Evaluate from -2 to 1:

= \([\left(2x - \frac{x^2}{2} - \frac{x^3}{3}\right) \bigg|_{-2}^{1}]\).

At \(x = 1\): \(2(1) - \frac{1^2}{2} - \frac{1^3}{3} = 2 - \frac{1}{2} - \frac{1}{3} = \frac{12}{6} - \frac{3}{6} - \frac{2}{6} = \frac{7}{6}\).

At \(x = -2\): \(2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3} = -4 - 2 + \frac{8}{3} = \frac{-12}{3} - \frac{6}{3} + \frac{8}{3} = \frac{-10}{3}\).

Total Area = \(\frac{7}{6} - \frac{-10}{3} = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2}\).

Now divide the area by \(x = -1\), and find areas separately for \([-2, -1]\) and \([-1, 1]\).

Area from \(x = -2\) to \(x = -1\):

= \(\int_{-2}^{-1} (2-x-x^2)\, dx\).

= \([\left(2x - \frac{x^2}{2} - \frac{x^3}{3}\right) \bigg|_{-2}^{-1}]\).

At \(x = -1\): \(2(-1) - \frac{(-1)^2}{2} - \frac{(-1)^3}{3} = -2 - \frac{1}{2} + \frac{1}{3} = \frac{-12}{6} - \frac{3}{6} + \frac{2}{6} = \frac{-13}{6}\).

Area = \(\frac{-13}{6} - \frac{-10}{3} = \frac{-13}{6} + \frac{20}{6} = \frac{7}{6}\).

Area from \(x = -1\) to \(x = 1\):

Total area = \(\frac{9}{2}\), thus:

Area from \(-1\) to \(1 = \)' \(\frac{9}{2} - \frac{7}{6} = \frac{14}{3}\).

The ratio is \(\frac{7/6}{14/3} = \frac{7}{28} = \frac{1}{4}\).

Thus, \(m=1\), \(n=4\), resulting in \(m+n=5\), which is within the 5,5 range.

Therefore, the answer is \(5\).