Q

Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined by f x equals x square mius x,x belongs A and g x equals 2 absolute value of x minus 1 over 2 minus 1 x belongs to A are f and g equal justify your answer

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).

Views

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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