Let `f(x)=4cossqrt(x^2-pi^2/9` Then the range of `f(x)` is

Options

A. [-1,1]
B. [-4,4]
C. [0,1]
D. None of these

Answer

To find the range of the function \( f(x) = 4 \cos \left( \sqrt{x^2 - \frac{\pi^2}{9}} \right) \), we will follow these steps:

### Step 1: Analyze the expression inside the cosine function
The expression inside the cosine function is \( \sqrt{x^2 - \frac{\pi^2}{9}} \).

### Step 2: Determine the domain of the square root
For the square root to be defined, the expression inside must be non-negative:
\[ x^2 - \frac{\pi^2}{9} \geq 0 \]
This implies:
\[ x^2 \geq \frac{\pi^2}{9} \]
Taking the square root of both sides gives:
\[ |x| \geq \frac{\pi}{3} \]
Thus, \( x \) must be in the intervals:
\[ (-\infty, -\frac{\pi}{3}] \cup [\frac{\pi}{3}, \infty) \]

### Step 3: Determine the range of the angle
The angle \( \theta = \sqrt{x^2 - \frac{\pi^2}{9}} \) will take values starting from \( 0 \) (when \( |x| = \frac{\pi}{3} \)) to \( \infty \) (as \( |x| \) increases). Therefore, the range of \( \theta \) is:
\[ [0, \infty) \]

### Step 4: Find the range of the cosine function
The cosine function, \( \cos(\theta) \), has a range of:
\[ [-1, 1] \]
for \( \theta \) in the interval \( [0, \infty) \).

### Step 5: Scale the range by 4
Since \( f(x) = 4 \cos(\theta) \), we multiply the range of \( \cos(\theta) \) by 4:
\[ 4 \cdot [-1, 1] = [-4, 4] \]

### Conclusion
The range of the function \( f(x) = 4 \cos \left( \sqrt{x^2 - \frac{\pi^2}{9}} \right) \) is:
\[ [-4, 4] \]