Q. 15 Let A = \{- 1, 0, 1, 2\}, B = \{- 4, - 2, 0, 2\} and f, g : A \rightarrow B be functions defined  by f(x) = x^2 -x, x\in A and g(x) = 2\left |x - \frac{1}{2} \right | - 1, x\in A. Are f and g equal?

Justify your answer. (Hint: One may note that two functions f : A \rightarrow B and
g : A \rightarrow B such that f (a) = g (a) \;\forall a \in A, are called equal functions).

Answers (1)

Given :

A = \{- 1, 0, 1, 2\},B = \{- 4, - 2, 0, 2\}

     f, g : A \rightarrow B are defined  by f(x) = x^2 -x, x\in A and g(x) = 2\left |x - \frac{1}{2} \right | - 1, x\in A.

It can be observed that 

f(-1) = (-1)^2 -(-1)=1+1=2

g(-1) = 2\left |-1 - \frac{1}{2} \right | - 1= 2\left | \frac{-3}{2} \right |-1=3-1=2

      f(-1)=g(-1)

f(0) = (0)^2 -(0)=0+0=0

g(0) = 2\left |0 - \frac{1}{2} \right | - 1= 2\left | \frac{-1}{2} \right |-1=1-1=0

       f(0)=g(0)

f(1) = (1)^2 -(1)=1-1=0

g(1) = 2\left |1 - \frac{1}{2} \right | - 1= 2\left | \frac{1}{2} \right |-1=1-1=0

      f(1)=g(1)

f(2) = (2)^2 -(2)=4-2=2

g(2) = 2\left |2 - \frac{1}{2} \right | - 1= 2\left | \frac{3}{2} \right |-1=3-1=2

        f(2)=g(2)

\therefore \, \, \, f(a)=g(a) \forall a\in A

Hence, f and g are equal functions.

 

 

 

 

 

 

 

 

 

 

 

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