Let A$= \{$ - $1, 0, 1, 2\}$, B$= \{$ - $4,$ - $2, 0, 2\}$ and f, g : A $\rightarrow$ B be function defined by f(x)$= {x^2} - x,x \in A$and$g(x) = 2\left| {x - \cfrac{1}{2}} \right| - 1,x \in A.$ 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).
Let A$= \{$ - $1, 0, 1, 2\}$, B$= \{$ - $4,$ - $2, 0, 2\}$ and f, g : A $\rightarrow$ B be function defined by f(x)$= {x^2} - x,x \in A$and$g(x) = 2\left| {x - \cfrac{1}{2}} \right| - 1,x \in A.$ 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).
Official Solution
When x $=$ $-$1, f($-$1) $=$ 12 + 1 $=$ 2
and $g( - 1) = 2\left| { - 1 - \cfrac{1}{2}} \right| - 1 = 2$
When x $=$ 0, f(0) $=$ 0 and g(0) $=$ $2\left| { - \cfrac{1}{2}} \right| - 1 = 2 \times \cfrac{1}{2} - 1 = 0$
When x $=$1, f(1)$=$ ${1^2} - 1$ $=$0
and g(1) $= 2\left| {1 - \cfrac{1}{2}} \right| - 1 = 2 \times \cfrac{1}{2} - 1 = 0$
When x $=$ 2, f(2)$=$22$-$2$=$2 and g(2) $=$ $2\left| {2 - \cfrac{1}{2}} \right| - 1 = 3 - 1 = 2$
Thus, for each a $\in$ A, f(a) $=$ g(a). Hence, f and g are equal functions.
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