The interval in which f(x) increases less repidly than g(x), f(x)=`2x^3+5 and g(x) =9x^2 -12 x` is

Options

A. `(-oo,1)`
B. (1,2)
C. `(2,oo)`
D. none of these

Explanation

To determine the interval in which the function \( f(x) = 2x^3 + 5 \) increases less rapidly than the function \( g(x) = 9x^2 - 12x \), we need to compare the derivatives of both functions.

### Step 1: Differentiate both functions
First, we find the derivatives of \( f(x) \) and \( g(x) \).

- The derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3 + 5) = 6x^2 \]

- The derivative of \( g(x) \):
\[ g'(x) = \frac{d}{dx}(9x^2 - 12x) = 18x - 12 \]

### Step 2: Set up the inequality
We need to find the interval where \( f'(x) < g'(x) \):
\[ 6x^2 < 18x - 12 \]

### Step 3: Rearrange the inequality
Rearranging the inequality gives us:
\[ 6x^2 - 18x + 12 < 0 \]

### Step 4: Simplify the inequality
Dividing the entire inequality by 6:
\[ x^2 - 3x + 2 < 0 \]

### Step 5: Factor the quadratic
Now, we factor the quadratic:
\[ (x - 1)(x - 2) < 0 \]

### Step 6: Determine the critical points
The critical points from the factors are \( x = 1 \) and \( x = 2 \).

### Step 7: Test intervals
We will test the intervals determined by the critical points:

1. **Interval \( (-\infty, 1) \)**:
- Choose \( x = 0 \):
\[ (0 - 1)(0 - 2) = 1 \quad (\text{positive}) \]

2. **Interval \( (1, 2) \)**:
- Choose \( x = 1.5 \):
\[ (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 \quad (\text{negative}) \]

3. **Interval \( (2, \infty) \)**:
- Choose \( x = 3 \):
\[ (3 - 1)(3 - 2) = 2 \quad (\text{positive}) \]

### Step 8: Conclusion
The inequality \( (x - 1)(x - 2) < 0 \) holds true in the interval \( (1, 2) \).

Thus, the interval in which \( f(x) \) increases less rapidly than \( g(x) \) is:
\[ \boxed{(1, 2)} \]