draw the graph of signum function f:R->R defined by f(x)={1,if x>0}, {0,if x=0}
{-1, if x <0}
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Step-by-step explanation:
: R → R, given by
f(x)={1,if x>0}, {0,if x=0}
{-1, if x <0}
It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴f is not one-one.
Now, as f(x) takes only 3 values (1, 0, or - 1) for the element - 2 in co-domain R, there does not exist any x in domain R such that f(x) = - 2.
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.
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