P.d.f. of a c.r.v X is
`f(x)={(6x(1-x)","0lexle1,),(0",",elsewhere):}`
If P (`Xlt a )= P(Xgta) ` then : a =..

Options

A. 1
B. `1/2`
C. `1/3`
D. `1/4`

Answer

To solve the problem, we need to find the value of \( a \) such that \( P(X < a) = P(X > a) \) for the given probability density function (pdf):

\[ f(x) = \begin{cases} 6x(1-x) & \text{for } 0 \leq x \leq 1 \\ 0 & \text{elsewhere} \end{cases} \]

### Step 1: Understand the Condition
We start with the condition \( P(X < a) = P(X > a) \). This implies that the total probability must be split evenly at the point \( a \). Since the total probability must equal 1, we can express this as:

\[ P(X < a) = P(X > a) = \frac{1}{2} \]

### Step 2: Calculate \( P(X < a) \)
The probability \( P(X < a) \) can be calculated using the integral of the pdf from 0 to \( a \):

\[ P(X < a) = \int_0^a f(x) \, dx = \int_0^a 6x(1-x) \, dx \]

### Step 3: Evaluate the Integral
Now, we evaluate the integral:

\[ \int_0^a 6x(1-x) \, dx = \int_0^a (6x - 6x^2) \, dx \]

Calculating the integral:

\[ = \left[ 3x^2 - 2x^3 \right]_0^a = 3a^2 - 2a^3 \]

### Step 4: Set the Equation
We set the equation for \( P(X < a) \) equal to \( \frac{1}{2} \):

\[ 3a^2 - 2a^3 = \frac{1}{2} \]

### Step 5: Rearrange the Equation
Rearranging gives us:

\[ 2a^3 - 3a^2 + \frac{1}{2} = 0 \]

Multiplying through by 2 to eliminate the fraction:

\[ 4a^3 - 6a^2 + 1 = 0 \]

### Step 6: Solve the Cubic Equation
Now, we need to solve the cubic equation \( 4a^3 - 6a^2 + 1 = 0 \). We can use the Rational Root Theorem or trial and error to find possible rational roots. Testing \( a = \frac{1}{2} \):

\[ 4\left(\frac{1}{2}\right)^3 - 6\left(\frac{1}{2}\right)^2 + 1 = 4 \cdot \frac{1}{8} - 6 \cdot \frac{1}{4} + 1 = \frac{1}{2} - \frac{3}{2} + 1 = 0 \]

Thus, \( a = \frac{1}{2} \) is a root.

### Step 7: Factor the Cubic Polynomial
We can factor \( 4a^3 - 6a^2 + 1 \) as:

\[ 4a^3 - 6a^2 + 1 = (a - \frac{1}{2})(4a^2 - 4a - 2) \]

### Step 8: Solve the Quadratic
Now, we can solve the quadratic \( 4a^2 - 4a - 2 = 0 \) using the quadratic formula:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4} \]

Calculating the discriminant:

\[ = \frac{4 \pm \sqrt{16 + 32}}{8} = \frac{4 \pm \sqrt{48}}{8} = \frac{4 \pm 4\sqrt{3}}{8} = \frac{1 \pm \sqrt{3}}{2} \]

### Step 9: Conclusion
Thus, the possible values for \( a \) are:

1. \( a = \frac{1}{2} \)
2. \( a = \frac{1 + \sqrt{3}}{2} \)
3. \( a = \frac{1 - \sqrt{3}}{2} \)

However, since \( a \) must be between 0 and 1, we discard \( \frac{1 - \sqrt{3}}{2} \) (which is negative). Therefore, the valid values for \( a \) are:

\[ a = \frac{1}{2} \quad \text{(valid)} \quad \text{and} \quad a = \frac{1 + \sqrt{3}}{2} \quad \text{(invalid, as it exceeds 1)} \]

Thus, the only valid solution is:

\[ \boxed{\frac{1}{2}} \]