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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 $\mathrm{g}(\mathrm{x})=2\left|\mathrm{x}-\frac{1}{2}\right|-1, x \in A$. Are f and g equal? Justify your answer.
Hint : One may not 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).
Hint : One may not 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).
Solution:
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Verified Answer
At $x=-1, f(x)=1^2+1=2$ and
$$
\mathrm{g}(\mathrm{x})=2\left|-1-\frac{1}{2}\right|-1=2 \times \frac{3}{2}-1=2
$$
At $x=0, f(0)=0$ and $g(0)=2\left|-\frac{1}{2}\right|-1$
$$
=2 \times \frac{1}{2}-1=0, \text { At } x=1, f(1)=1^2-1=0
$$
$g(1)=2\left|1-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0 \quad \operatorname{Atx}=2, \mathrm{f}(2)=2^2-2=2$
$\mathrm{g}(2)=2\left|2-\frac{1}{2}\right|-1=3-1=2$
Thus, for each $a \in A, f(a)=g(x) \Rightarrow f$ and $g$ are equal function.
$$
\mathrm{g}(\mathrm{x})=2\left|-1-\frac{1}{2}\right|-1=2 \times \frac{3}{2}-1=2
$$
At $x=0, f(0)=0$ and $g(0)=2\left|-\frac{1}{2}\right|-1$
$$
=2 \times \frac{1}{2}-1=0, \text { At } x=1, f(1)=1^2-1=0
$$
$g(1)=2\left|1-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0 \quad \operatorname{Atx}=2, \mathrm{f}(2)=2^2-2=2$
$\mathrm{g}(2)=2\left|2-\frac{1}{2}\right|-1=3-1=2$
Thus, for each $a \in A, f(a)=g(x) \Rightarrow f$ and $g$ are equal function.
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