Question

Let $f_1\left(x\right)=\{\begin{array}{c}x\text{for}0\le x\le 1\\ 1\text{for}x>1\\ 0\text{otherwise}\end{array}$
and $f_2\left(x\right)=f_1(-x)$ for all $x$
$f_3\left(x\right)=-f_2\left(x\right)$ for all $x$
$f_4\left(x\right)=f_3(–x)$ for all $x$
Which of the following is necessarily true ?

• A. $f_4\left(x\right)=f_1\left(x\right)$ for all $x$
• B. $f_1\left(x\right)=-f_3(-x)$ for all $x$
• C. $f_2(-x)=f_4\left(x\right)$ for all $x$
• D. $f_1\left(x\right)+f_3\left(x\right)=0$ for all $x$

Answer 1

The correct option is B $f_1\left(x\right)=-f_3(-x)$ for all $x$


So, from the above graphs, we get
$f_1\left(x\right)=-f_3(-x)$ for all $x$