Question

If f(x) is an odd function, then |f(x)| is
[a] An odd function
[b] An even function
[c] Neither even nor odd
[d] Even and odd

Answer 1

Hint: Use the fact that the absolute value of -x is x. Recall the definitions of odd function and even function and check which of the definitions does the function follow. For an odd function, we know that f(-x) = -f(x) for all values of x in the domain of the function Hence find whether |f(x)| is an odd or even function. Verify with a known odd function, e.g. sinx.

Complete step-by-step solution -
Before solving the question, we need to understand what an odd function is and what an even function is.
Odd function: A function f(x) is said to be an odd function if $\mathrm{\forall }x\in \text{Domain}\left(f\right)$, f(-x) = -f(x). The graph of an odd function is symmetrical in about origin.
Even function: A function f(x) is said to be an even function if $\mathrm{\forall }x\in \text{Domain}\left(f\right),$ f(-x) = f(x). The graph of an even function is symmetrical about the y-axis.
Also, we have
$$|-x|=\{\begin{array}{c}-x,-x\ge 0\\ -(-x).-x\le 0\end{array}=\{\begin{array}{c}-x,x\le 0\\ x,x\ge 0\end{array}=\left|x\right|$$
Hence, we have
|-x| = |x|
Now, we have
H(x) = |f(x)|
H(-x) = |f(-x)|
Since f(x) is odd function, we have f(-x) = -f(x)
Hence, we have
H(-x) = |-f(x)|
Since |-x| = x, we have
H(-x) = |f(x)|.
Hence, we have H(-x) = H(x)
Hence, H(x) is even function (From the definition of an even function)
Hence option [b] is correct.

Note: Alternative Solution:
We use graphical transformations to determine the nature of |f(x)| when f(x) is odd.
Transformation Rule: If we know the graph of f(x), then the graph of |f(x)| can be drawn by taking simply the mirror image of the graph of f(x) below x-axis about the y-axis.
Since the graph of an odd function is symmetrical about the origin, a typical form of an odd function is as shown below

Using the above transformation rule, we draw the graph of |f(x)| as follows

Clearly, the graph is symmetric about the y-axis.
Hence |f(x)| is even.
It, however, should be noted that this not the proof of the question rather a faster way of realisation of the above result.