The function `'f'` is defined by `f(x)=x^p(1-x)^q` for all `x\ in R ,` where `p ,\ q` are positive integers, has a maximum value, for `x` equal to : `(p q)/(p+q)` (b) 1 (c) 0 (d) `p/(p+q)`

Options

A. `(pq)/(p+q)`
B. 1
C. 0
D. `(p)/(p+q)`

Answer

To solve the problem, we need to find the maximum value of the function \( f(x) = x^p (1-x)^q \) where \( p \) and \( q \) are positive integers. We will follow these steps:

### Step 1: Differentiate the function
We start by differentiating \( f(x) \) with respect to \( x \).

\[ f'(x) = \frac{d}{dx}(x^p (1-x)^q) \]

Using the product rule, we have:

\[ f'(x) = x^p \cdot \frac{d}{dx}((1-x)^q) + (1-x)^q \cdot \frac{d}{dx}(x^p) \]

Calculating the derivatives:

\[ \frac{d}{dx}((1-x)^q) = -q(1-x)^{q-1} \]
\[ \frac{d}{dx}(x^p) = p x^{p-1} \]

Substituting back, we get:

\[ f'(x) = x^p \cdot (-q(1-x)^{q-1}) + (1-x)^q \cdot (p x^{p-1}) \]

### Step 2: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):

\[ -x^p q(1-x)^{q-1} + p x^{p-1} (1-x)^q = 0 \]

Rearranging gives:

\[ p x^{p-1} (1-x)^q = q x^p (1-x)^{q-1} \]

### Step 3: Factor out common terms
Factoring out \( x^{p-1} (1-x)^{q-1} \) from both sides:

\[ p (1-x) = q x \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:

\[ p - p x = q x \]
\[ p = (p + q)x \]
\[ x = \frac{p}{p+q} \]

### Step 5: Verify if it is a maximum
To confirm that this critical point is indeed a maximum, we can check the second derivative \( f''(x) \) or use the first derivative test. However, since we are only asked for the value of \( x \) at which the maximum occurs, we can conclude here.

### Final Answer
The maximum value of the function occurs at:

\[ x = \frac{p}{p+q} \]