Period of ` cos (x^2)` is

Options

A. ` 2pi`
B. ` 4 pi^2`
C. ` pi^2 //4`
D. None

Explanation

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

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

This means:

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

### Step 1: Expand the left-hand side
First, we expand the left-hand side:

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

Thus, we have:

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

### Step 2: Apply the cosine periodicity condition
For the equality \( \cos(A) = \cos(B) \) to hold, we can use the property of cosine:

\[ A = B + 2n\pi \quad \text{or} \quad A = -B + 2n\pi \quad \text{for some integer } n \]

Applying this to our equation:

\[ x^2 + 2xT + T^2 = x^2 + 2n\pi \quad \text{(1)} \]
or
\[ x^2 + 2xT + T^2 = -x^2 + 2n\pi \quad \text{(2)} \]

### Step 3: Simplify equation (1)
From equation (1):

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

This can be rearranged to:

\[ T^2 + 2xT - 2n\pi = 0 \]

### Step 4: Identify the nature of the equation
This is a quadratic equation in \( T \). The solutions for \( T \) can be found using the quadratic formula:

\[ T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = 2x \), and \( c = -2n\pi \):

\[ T = \frac{-2x \pm \sqrt{(2x)^2 - 4 \cdot 1 \cdot (-2n\pi)}}{2 \cdot 1} \]
\[ T = \frac{-2x \pm \sqrt{4x^2 + 8n\pi}}{2} \]
\[ T = -x \pm \sqrt{x^2 + 2n\pi} \]

### Step 5: Analyze the results
The value of \( T \) depends on \( x \) and \( n \). This means that for different values of \( x \), we can get different values of \( T \).

### Conclusion
Since the period \( T \) is not constant and depends on \( x \), we conclude that the function \( \cos(x^2) \) does not have a fixed period.

Thus, the period of \( \cos(x^2) \) does not exist.