Consider the function f(x)=`cos x^(2)` then

Options

A. f is of period `2pi`
B. f is of period `sqrt(2t)`
C. f is not periodic
D. f is of period `pi`

Answer

To determine the periodicity of the function \( f(x) = \cos(x^2) \), we need to analyze the function and see if there exists a period \( T \) such that:

\[ f(x + T) = f(x) \]

for all \( x \).

### Step 1: Define the function
We start with the function:

\[ f(x) = \cos(x^2) \]

### Step 2: Check for periodicity
To check if \( f(x) \) is periodic, we need to find \( T \) such that:

\[ f(x + T) = f(x) \]

This means:

\[ \cos((x + T)^2) = \cos(x^2) \]

### Step 3: Expand the left-hand side
Expanding \( (x + T)^2 \):

\[ (x + T)^2 = x^2 + 2xT + T^2 \]

So we have:

\[ \cos(x^2 + 2xT + T^2) = \cos(x^2) \]

### Step 4: Use the property of cosine
The cosine function is periodic with a period of \( 2\pi \). Therefore, for the equality to hold, we need:

\[ x^2 + 2xT + T^2 = x^2 + 2k\pi \]

for some integer \( k \).

### Step 5: Simplify the equation
This simplifies to:

\[ 2xT + T^2 = 2k\pi \]

### Step 6: Analyze the equation
Notice that the left-hand side \( 2xT + T^2 \) depends on \( x \). For the equation to hold for all \( x \), the coefficient of \( x \) must be zero, which implies:

\[ T = 0 \]

This means that there is no non-zero \( T \) that satisfies the equation for all \( x \).

### Step 7: Conclusion
Since we cannot find a non-zero period \( T \) such that \( f(x + T) = f(x) \) for all \( x \), we conclude that the function \( f(x) = \cos(x^2) \) is not periodic.

### Final Answer
The function \( f(x) = \cos(x^2) \) is **not periodic**.

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