If the p.d.f of a continuous random variable X is
`f(x)=kx^(2)(1-x), 0 lt x lt 1`
= 0 otherwise
Then the value of k is

Options

A. 8
B. 10
C. 12
D. 16

Answer

To find the value of \( k \) in the probability density function (p.d.f.) given by

\[ f(x) = kx^2(1-x), \quad 0 < x < 1 \]
and \( f(x) = 0 \) otherwise, we need to use the property that the total area under the p.d.f. must equal 1. This can be expressed mathematically as:

\[ \int_0^1 f(x) \, dx = 1 \]

### Step-by-Step Solution:

1. **Set Up the Integral**:
We start by substituting the p.d.f. into the integral:

\[ \int_0^1 kx^2(1-x) \, dx = 1 \]

2. **Factor Out \( k \)**:
Since \( k \) is a constant, we can factor it out of the integral:

\[ k \int_0^1 x^2(1-x) \, dx = 1 \]

3. **Simplify the Integral**:
We need to evaluate the integral \( \int_0^1 x^2(1-x) \, dx \). We can expand the integrand:

\[ x^2(1-x) = x^2 - x^3 \]

Thus, the integral becomes:

\[ \int_0^1 (x^2 - x^3) \, dx \]

4. **Evaluate the Integral**:
We can now evaluate the integral term by term:

\[ \int_0^1 x^2 \, dx - \int_0^1 x^3 \, dx \]

Using the formula \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \):

- For \( x^2 \):

\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - 0 = \frac{1}{3} \]

- For \( x^3 \):

\[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1^4}{4} - 0 = \frac{1}{4} \]

Therefore, we have:

\[ \int_0^1 (x^2 - x^3) \, dx = \frac{1}{3} - \frac{1}{4} \]

5. **Combine the Results**:
To combine \( \frac{1}{3} \) and \( \frac{1}{4} \), we find a common denominator:

\[ \frac{1}{3} = \frac{4}{12}, \quad \frac{1}{4} = \frac{3}{12} \]

Thus,

\[ \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12} \]

6. **Substitute Back**:
Now substituting back into our equation:

\[ k \cdot \frac{1}{12} = 1 \]

7. **Solve for \( k \)**:
To find \( k \), we multiply both sides by 12:

\[ k = 12 \]

### Final Answer:
The value of \( k \) is \( \boxed{12} \).