The image of [-1,3] under f is not the interval `[f(-1),f(3)]`
Statement-2: f is not an injective map.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the function \( f(x) = 12x^2 - 12 \) and determine the image of the interval \([-1, 3]\) under this function. We will also check if the statement that the image is not the interval \([f(-1), f(3)]\) implies that \( f \) is not an injective map.

### Step 1: Calculate \( f(-1) \) and \( f(3) \)

First, we will evaluate the function at the endpoints of the interval:

\[ f(-1) = 12(-1)^2 - 12 = 12(1) - 12 = 0 \]

\[ f(3) = 12(3)^2 - 12 = 12(9) - 12 = 108 - 12 = 96 \]

### Step 2: Determine the image of the interval \([-1, 3]\)

Next, we need to find the behavior of the function on the interval \([-1, 3]\). To do this, we will differentiate the function to find critical points.

### Step 3: Differentiate \( f(x) \)

The derivative of \( f(x) \) is:

\[ f'(x) = \frac{d}{dx}(12x^2 - 12) = 24x \]

### Step 4: Find critical points

Set the derivative equal to zero to find critical points:

\[ 24x = 0 \implies x = 0 \]

### Step 5: Analyze the function's behavior

Now we will evaluate \( f(x) \) at the critical point \( x = 0 \):

\[ f(0) = 12(0)^2 - 12 = -12 \]

Now we have the values of \( f \) at the critical points and endpoints:

- \( f(-1) = 0 \)
- \( f(0) = -12 \)
- \( f(3) = 96 \)

### Step 6: Determine the image of \([-1, 3]\)

The function decreases from \( f(-1) = 0 \) to \( f(0) = -12 \) and then increases from \( f(0) = -12 \) to \( f(3) = 96 \). Therefore, the image of the interval \([-1, 3]\) is:

\[ [-12, 96] \]

### Step 7: Compare with \([f(-1), f(3)]\)

The interval \([f(-1), f(3)]\) is:

\[ [0, 96] \]

Since the image of \([-1, 3]\) is \([-12, 96]\), which is not equal to \([0, 96]\), we conclude that the image of \([-1, 3]\) under \( f \) is indeed not the interval \([f(-1), f(3)]\).

### Step 8: Determine if \( f \) is injective

Since the function decreases and then increases, it takes the same value at two different points (for example, \( f(-1) = 0 \) and \( f(1) = 0 \)). Thus, \( f \) is not an injective map.

### Conclusion

The statements are correct: the image of \([-1, 3]\) under \( f \) is not the interval \([f(-1), f(3)]\), and \( f \) is not an injective map.

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