If `phi`(x) is continuous at x= `a` such that
`f(x)=(ax-a^2-x^2)phi(x)`for all x, then f(x) is

Options

A. increasing in the neighbourhood of `x = alpha `
B. decresing in the neighbourhood of `x = apha `
C. constant in the neighbourhood of `x=alpha`
D. minimum at `x=alpha`

Answer

To solve the problem, we need to analyze the function \( f(x) = (ax - a^2 - x^2) \phi(x) \) and determine its behavior around the point \( x = a \).

### Step 1: Evaluate \( f(a) \)
First, we substitute \( x = a \) into the function \( f(x) \):

\[ f(a) = (a \cdot a - a^2 - a^2) \phi(a) = (a^2 - a^2 - a^2) \phi(a) = -a^2 \phi(a) \]

### Step 2: Find the left-hand limit \( f(a^-) \)
Next, we compute the left-hand limit as \( x \) approaches \( a \) from the left:

\[ f(a^-) = (a - h) a - a^2 - (a - h)^2) \phi(a - h) \]
where \( h \) is a small positive number. Expanding this gives:

\[ f(a^-) = ((a - h)a - a^2 - (a^2 - 2ah + h^2)) \phi(a - h) \]
\[ = (a^2 - ah - a^2 - a^2 + 2ah - h^2) \phi(a - h) \]
\[ = (-a^2 + ah + 2ah - h^2) \phi(a - h) \]
\[ = (-a^2 + 3ah - h^2) \phi(a - h) \]

### Step 3: Find the right-hand limit \( f(a^+) \)
Now, we compute the right-hand limit as \( x \) approaches \( a \) from the right:

\[ f(a^+) = (a + h)a - a^2 - (a + h)^2) \phi(a + h) \]
Expanding this gives:

\[ f(a^+) = ((a + h)a - a^2 - (a^2 + 2ah + h^2)) \phi(a + h) \]
\[ = (a^2 + ah - a^2 - a^2 - 2ah - h^2) \phi(a + h) \]
\[ = (-a^2 - ah - h^2) \phi(a + h) \]

### Step 4: Compare the left-hand and right-hand limits
Now, we compare \( f(a^-) \) and \( f(a^+) \):

- As \( h \to 0 \), \( \phi(a - h) \) approaches \( \phi(a) \) and \( \phi(a + h) \) approaches \( \phi(a) \) since \( \phi(x) \) is continuous at \( x = a \).
- The left-hand limit \( f(a^-) \) approaches \( -a^2 \phi(a) \).
- The right-hand limit \( f(a^+) \) also approaches \( -a^2 \phi(a) \).

### Step 5: Determine if \( f(x) \) is increasing or decreasing
To determine whether \( f(x) \) is increasing or decreasing around \( x = a \), we can analyze the derivative \( f'(x) \). However, based on the behavior of the left-hand and right-hand limits, we can conclude that:

- If \( \phi(a) > 0 \), \( f(x) \) is increasing around \( x = a \).
- If \( \phi(a) < 0 \), \( f(x) \) is decreasing around \( x = a \).
- If \( \phi(a) = 0 \), \( f(x) \) is constant around \( x = a \).

### Conclusion
Thus, the function \( f(x) \) is increasing in the neighborhood of \( x = a \) if \( \phi(a) > 0 \), decreasing if \( \phi(a) < 0 \), and constant if \( \phi(a) = 0 \).