If f(x) is an odd function, which of the following must be even?A. f(x + 1)B.| f(x)|C.None of theseD.- f(x)E. f(|x - 1|)

To determine which of the given options must be even if $f(x)$ is an odd function, we need to recall the definition of an odd function. A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$.

Let's analyze each option:

A. $f(x + 1)$

• To check if $f(x + 1)$ is even, we need to see if $f(x + 1) = f(-(x + 1))$.
• $f(-(x + 1)) = f(-x - 1)$.
• Since $f(x)$ is odd, $f(-x - 1) = -f(x + 1)$.
• Therefore, $f(x + 1)$ is not necessarily even.

B. $|f(x)|$

• To check if $|f(x)|$ is even, we need to see if $|f(x)| = |f(-x)|$.
• Since $f(x)$ is odd, $f(-x) = -f(x)$.
• Therefore, $|f(-x)| = |-f(x)| = |f(x)|$.
• Hence, $|f(x)|$ is even.

C. None of these

• This option is incorrect because we have found that $|f(x)|$ is even.

D. $-f(x)$

• To check if $-f(x)$ is even, we need to see if $-f(x) = -f(-x)$.
• Since $f(x)$ is odd, $f(-x) = -f(x)$.
• Therefore, $-f(-x) = -(-f(x)) = f(x)$.
• Hence, $-f(x)$ is not necessarily even.

E. $f(|x - 1|)$

• To check if $f(|x - 1|)$ is even, we need to see if $f(|x - 1|) = f(|-(x - 1)|)$.
• Since $|-(x - 1)| = |x - 1|$, $f(|x - 1|) = f(|x - 1|)$.
• This does not guarantee that $f(|x - 1|)$ is even because it depends on the form of $f(x)$.

Therefore, the correct answer is:

B. $|f(x)|$