A function is said to be even if f(x)=f(-x). Which of the following is not an even function?

Options

A. y=|x|
B. y=secx
C. y=log`x^(2)`
D. `y=x^(2)+sinx`

Explanation

To determine which of the given functions is not an even function, we need to check if each function satisfies the condition \( f(x) = f(-x) \).

### Step-by-Step Solution:

1. **Understanding Even Functions**:
A function \( f(x) \) is considered even if for every \( x \), \( f(x) = f(-x) \). This means the function's graph is symmetric about the y-axis.

2. **Analyzing Each Function**:
- **Function 1: \( f(x) = |x| \)**
To check if it's even:
\[ f(-x) = |-x| = |x| = f(x) \]
This function is even.

- **Function 2: \( f(x) = \sec x \)**
To check if it's even:
\[ f(-x) = \sec(-x) = \frac{1}{\cos(-x)} = \frac{1}{\cos x} = \sec x = f(x) \]
This function is even.

- **Function 3: \( f(x) = \log(x^2) \)**
To check if it's even:
\[ f(-x) = \log((-x)^2) = \log(x^2) = f(x) \]
This function is even.

- **Function 4: \( f(x) = x^2 + \sin x \)**
To check if it's even:
\[ f(-x) = (-x)^2 + \sin(-x) = x^2 - \sin x \]
Since \( \sin(-x) = -\sin x \), we have:
\[ f(-x) = x^2 - \sin x \neq f(x) = x^2 + \sin x \]
This function is not even.

3. **Conclusion**:
The function that is not even is \( f(x) = x^2 + \sin x \).

### Final Answer:
The function that is not an even function is \( f(x) = x^2 + \sin x \).

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