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.