Let `h(x) =f(x)-a(f(x))^(3)` for every real number x
h(x) increase as f(x) increses for all real values of x if

Options

A. `a in (0,3)`
B. `a in (-2,2)`
C. `[3,oo)`
D. none of these

Explanation

To solve the problem, we need to analyze the function \( h(x) = f(x) - a(f(x))^3 \) and determine the conditions under which \( h(x) \) is increasing as \( f(x) \) increases for all real values of \( x \).

### Step 1: Differentiate \( h(x) \)
To find when \( h(x) \) is increasing, we first need to compute the derivative \( h'(x) \):
\[ h'(x) = f'(x) - 3a(f(x))^2 f'(x) \]
This can be factored as:
\[ h'(x) = f'(x)(1 - 3a(f(x))^2) \]

### Step 2: Set the derivative greater than zero
For \( h(x) \) to be increasing, we need:
\[ h'(x) > 0 \]
This implies:
\[ f'(x)(1 - 3a(f(x))^2) > 0 \]

### Step 3: Analyze the conditions
1. If \( f'(x) > 0 \) (i.e., \( f(x) \) is increasing), then \( 1 - 3a(f(x))^2 > 0 \) must hold. This leads to:
\[ 1 > 3a(f(x))^2 \]
or
\[ a < \frac{1}{3(f(x))^2} \]
Since \( f(x) \) can take any real value, this condition must hold for all possible values of \( f(x) \).

2. If \( f'(x) < 0 \) (i.e., \( f(x) \) is decreasing), then \( 1 - 3a(f(x))^2 < 0 \) must hold. This leads to:
\[ 1 < 3a(f(x))^2 \]
or
\[ a > \frac{1}{3(f(x))^2} \]
Again, this must hold for all possible values of \( f(x) \).

### Step 4: Determine the interval for \( a \)
To ensure that \( h(x) \) is increasing for all \( x \), we need to combine the conditions derived from the analysis above.

1. From the condition \( 1 > 3a(f(x))^2 \), we can conclude that:
- If \( f(x) \) is bounded, say \( |f(x)| \leq M \), then we can derive:
\[ a < \frac{1}{3M^2} \]

2. From the condition \( 1 < 3a(f(x))^2 \), we can conclude that:
- If \( f(x) \) is bounded, say \( |f(x)| \geq m > 0 \), then we can derive:
\[ a > \frac{1}{3m^2} \]

Combining these inequalities leads us to find the interval for \( a \).

### Final Result
After analyzing the conditions, we find that:
\[ 0 < a < 3 \]
Thus, the interval for \( a \) is:
\[ a \in (0, 3) \]