Question

Let A= {−1,0,1,2}, B={−4,−2,0,2} and f,g: A→B be functions defined by $f(x)=x^2-x, \,x\in A\, and \,g(x)=2\mid\frac{ x-1}{2}\mid-1,x\in A.$. Are f and g equal? Justify your answer. (Hint: One may note that two function $f:A\to B \,and \: g:A\to B$ such that $f(a)=g(a) \forall \,a \in\,A,$ are called equal functions).

Answer 1

The correct answer is: f and g are equal
It is given that A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2}.
Also, it is given that $f(x)=x^2-x, \,x\in A\, and \,g(x)=2\mid\frac{ x-1}{2}\mid-1,x\in A.$
It is observed that:
$f(-1)=(-1)^2-(-1)=1+1=2$.
$g(-1)=2\mid(-1)-\frac{1}{2}\mid-1=2(\frac{3}{2})$
$\Rightarrow(-1)=g(-1)$
$f(0)=(0)^2-0=0$
$g(0)=2\mid0=\frac{1}{2}\mid=2(\frac{1}{2})-1=0$
$f(0)=g(0)$
$f(1)=(1)^2-1=1$
$g(1)=2\mid1=\frac{1}{2}\mid=2(\frac{1}{2})-1=0$
$\Rightarrow(1)=g(1)$
$f(2)=(2)^2-2=2$
$g(2)=2\mid\frac{2-1}{2}\mid-1=2(\frac{3}{2})-1=2$
$\Rightarrow f(2)=g(2)$
$\therefore f(a)=g(a)\, \forall\, a\,\in\,A$
Hence, the functions f and g are equal.