From the identity `"sin" 3x = 3 "sin"x - 4"sin"^(3)x` it follows that if x is real and `|x| lt 1`, then

Options

A. `(3x-4x^(3)) lt 1`
B. `(3x-4x^(3)) le 1`
C. `(3x-4x^(3)) lt -1`
D. none of these

Answer

To solve the problem based on the identity \( \sin 3x = 3 \sin x - 4 \sin^3 x \) and the condition \( |x| < 1 \), we need to analyze the expression \( 3x - 4x^3 \) and determine if it is less than or equal to 1 for real values of \( x \) within the specified range.

### Step 1: Define the function
Let \( f(x) = 3x - 4x^3 \).

### Step 2: Find the derivative
To find the maximum and minimum values of \( f(x) \), we first compute the derivative:
\[ f'(x) = 3 - 12x^2 \]

### Step 3: Set the derivative to zero
To find critical points, we set the derivative equal to zero:
\[ 3 - 12x^2 = 0 \]
\[ 12x^2 = 3 \]
\[ x^2 = \frac{1}{4} \]
\[ x = \pm \frac{1}{2} \]

### Step 4: Evaluate the second derivative
Next, we compute the second derivative to determine the nature of the critical points:
\[ f''(x) = -24x \]

### Step 5: Analyze critical points
- For \( x = \frac{1}{2} \):
\[ f''\left(\frac{1}{2}\right) = -24 \cdot \frac{1}{2} = -12 \quad (\text{local maximum}) \]

- For \( x = -\frac{1}{2} \):
\[ f''\left(-\frac{1}{2}\right) = -24 \cdot \left(-\frac{1}{2}\right) = 12 \quad (\text{local minimum}) \]

### Step 6: Evaluate the function at critical points
Now we evaluate \( f(x) \) at the critical points and the endpoints of the interval \( |x| < 1 \).

- At \( x = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = 3\left(\frac{1}{2}\right) - 4\left(\frac{1}{2}\right)^3 = \frac{3}{2} - 4 \cdot \frac{1}{8} = \frac{3}{2} - \frac{1}{2} = 1 \]

- At \( x = -\frac{1}{2} \):
\[ f\left(-\frac{1}{2}\right) = 3\left(-\frac{1}{2}\right) - 4\left(-\frac{1}{2}\right)^3 = -\frac{3}{2} + 4 \cdot \frac{1}{8} = -\frac{3}{2} + \frac{1}{2} = -1 \]

### Step 7: Conclusion
The maximum value of \( f(x) \) occurs at \( x = \frac{1}{2} \) and is equal to 1. The minimum value occurs at \( x = -\frac{1}{2} \) and is equal to -1. Therefore, for \( |x| < 1 \):
\[ f(x) = 3x - 4x^3 \leq 1 \]

Thus, we conclude that for all real \( x \) such that \( |x| < 1 \), the expression \( 3x - 4x^3 \) is always less than or equal to 1.