consider the function f(X) =`3x^(4)+4x^(3)-12x^(2)`
The range of values of a for which f(x) = a has no real

Options

A. `(4,oo)`
B. `(10,oo)`
C. `(20,oo)`
D. none of these

Explanation

To solve the problem of finding the range of values of \( a \) for which the equation \( f(x) = a \) has no real solutions, we will follow these steps:

### Step 1: Define the function
We start with the function:
\[ f(x) = 3x^4 + 4x^3 - 12x^2 \]

### Step 2: Find the first derivative
To analyze the function, we first find its first derivative:
\[ f'(x) = \frac{d}{dx}(3x^4 + 4x^3 - 12x^2) = 12x^3 + 12x^2 - 24x \]
Factoring out the common terms:
\[ f'(x) = 12(x^3 + x^2 - 2x) = 12x(x^2 + x - 2) \]
Further factoring gives:
\[ f'(x) = 12x(x - 1)(x + 2) \]

### Step 3: Find critical points
Setting the first derivative to zero to find critical points:
\[ 12x(x - 1)(x + 2) = 0 \]
This gives us the critical points:
\[ x = 0, \quad x = 1, \quad x = -2 \]

### Step 4: Determine the nature of critical points
We will evaluate the function at these critical points to find the minimum value:
- For \( x = -2 \):
\[ f(-2) = 3(-2)^4 + 4(-2)^3 - 12(-2)^2 = 3(16) + 4(-8) - 12(4) = 48 - 32 - 48 = -32 \]

- For \( x = 0 \):
\[ f(0) = 3(0)^4 + 4(0)^3 - 12(0)^2 = 0 \]

- For \( x = 1 \):
\[ f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 = 3 + 4 - 12 = -5 \]

### Step 5: Identify the minimum value
From our evaluations, we find:
- \( f(-2) = -32 \)
- \( f(0) = 0 \)
- \( f(1) = -5 \)

The minimum value of \( f(x) \) is \( -32 \) at \( x = -2 \).

### Step 6: Determine the range of \( a \)
Since the function \( f(x) \) is a polynomial of degree 4 with a positive leading coefficient, it opens upwards. Therefore, the range of \( f(x) \) is:
\[ [-32, \infty) \]
Thus, for \( f(x) = a \) to have no real solutions, \( a \) must be less than the minimum value:
\[ a < -32 \]

### Final Answer
The range of values of \( a \) for which \( f(x) = a \) has no real solutions is:
\[ (-\infty, -32) \]

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