If `f(x)=|x+1|` then the true statement from the following is :

Options

A. `f(x^(2))={f(x)}^(2)`
B. `f(x+y)=f(x)+f(y)`
C. `f(|x|)=|f(x)|`
D. None of these

Answer

To solve the problem, we need to analyze the function \( f(x) = |x + 1| \) and evaluate the given statements to determine which one is true.

### Step-by-Step Solution:

1. **Evaluate \( f(x^2) \) and \( (f(x))^2 \)**:
- First, we compute \( f(x^2) \):
\[ f(x^2) = |x^2 + 1| \]
Since \( x^2 \) is always non-negative, \( x^2 + 1 \) is always greater than 0. Therefore:
\[ f(x^2) = x^2 + 1 \]

- Next, we compute \( (f(x))^2 \):
\[ f(x) = |x + 1| \implies (f(x))^2 = (|x + 1|)^2 = (x + 1)^2 \]
Expanding this gives:
\[ (f(x))^2 = x^2 + 2x + 1 \]

- Now we compare \( f(x^2) \) and \( (f(x))^2 \):
\[ f(x^2) = x^2 + 1 \quad \text{and} \quad (f(x))^2 = x^2 + 2x + 1 \]
Clearly, \( f(x^2) \neq (f(x))^2 \) unless \( 2x = 0 \) (i.e., \( x = 0 \)). Thus, this statement is **false**.

2. **Evaluate \( f(x + y) \) and \( f(x) + f(y) \)**:
- Compute \( f(x + y) \):
\[ f(x + y) = |x + y + 1| \]
- Compute \( f(x) + f(y) \):
\[ f(x) + f(y) = |x + 1| + |y + 1| \]

- To check if these are equal, let's test with \( x = 0 \) and \( y = 0 \):
\[ f(0 + 0) = |0 + 0 + 1| = 1 \]
\[ f(0) + f(0) = |0 + 1| + |0 + 1| = 1 + 1 = 2 \]
Since \( 1 \neq 2 \), this statement is also **false**.

3. **Evaluate \( f(|x|) \) and \( |f(x)| \)**:
- Compute \( f(|x|) \):
\[ f(|x|) = ||x| + 1| = |x| + 1 \quad \text{(since } |x| + 1 \text{ is non-negative)} \]
- Compute \( |f(x)| \):
\[ f(x) = |x + 1| \implies |f(x)| = ||x + 1| | = |x + 1| \]

- To check if these are equal, let's test with \( x = -2 \):
\[ f(-2) = |-2 + 1| = | -1 | = 1 \]
\[ f(|-2|) = f(2) = |2 + 1| = 3 \]
Since \( 1 \neq 3 \), this statement is also **false**.

### Conclusion:
After evaluating all statements, we find that none of the statements are true.