Consider the functions f(x) and g(x), both defined from `R rarrR` and are defined as `f(x)=2x-x^(2) and g(x)=x^(n)` where `n in N`. If the area between f(x) and g(x) is 1/2, then the value of n is

Options

A. 12
B. 15
C. 20
D. 30

Answer

To solve the problem, we need to find the value of \( n \) such that the area between the curves \( f(x) = 2x - x^2 \) and \( g(x) = x^n \) is equal to \( \frac{1}{2} \).

### Step-by-Step Solution:

1. **Equate the Functions**:
To find the limits of integration, we first set the two functions equal to each other:
\[ 2x - x^2 = x^n \]
Rearranging gives:
\[ x^2 + x^n - 2x = 0 \]
This is a polynomial equation in \( x \).

2. **Find the Roots**:
We can factor or use the Rational Root Theorem to find the roots. Testing \( x = 0 \) and \( x = 1 \):
- For \( x = 0 \): \( 2(0) - (0)^2 = 0 \) and \( (0)^n = 0 \) (True)
- For \( x = 1 \): \( 2(1) - (1)^2 = 1 \) and \( (1)^n = 1 \) (True)

Thus, the roots are \( x = 0 \) and \( x = 1 \).

3. **Set Up the Area Integral**:
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) is given by:
\[ A = \int_{0}^{1} (f(x) - g(x)) \, dx = \int_{0}^{1} \left( (2x - x^2) - x^n \right) \, dx \]

4. **Integrate**:
We compute the integral:
\[ A = \int_{0}^{1} (2x - x^2 - x^n) \, dx \]
This can be split into three separate integrals:
\[ A = \int_{0}^{1} 2x \, dx - \int_{0}^{1} x^2 \, dx - \int_{0}^{1} x^n \, dx \]

Calculating each integral:
- \( \int_{0}^{1} 2x \, dx = [x^2]_{0}^{1} = 1 \)
- \( \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} \)
- \( \int_{0}^{1} x^n \, dx = \left[ \frac{x^{n+1}}{n+1} \right]_{0}^{1} = \frac{1}{n+1} \)

Thus, we have:
\[ A = 1 - \frac{1}{3} - \frac{1}{n+1} \]
Simplifying this gives:
\[ A = \frac{2}{3} - \frac{1}{n+1} \]

5. **Set the Area Equal to \( \frac{1}{2} \)**:
We set the area equal to \( \frac{1}{2} \):
\[ \frac{2}{3} - \frac{1}{n+1} = \frac{1}{2} \]

6. **Solve for \( n \)**:
Rearranging gives:
\[ \frac{1}{n+1} = \frac{2}{3} - \frac{1}{2} \]
Finding a common denominator (6):
\[ \frac{2}{3} = \frac{4}{6}, \quad \frac{1}{2} = \frac{3}{6} \]
Thus:
\[ \frac{1}{n+1} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \]
Therefore:
\[ n + 1 = 6 \implies n = 5 \]

### Final Answer:
The value of \( n \) is \( 5 \).