1. sol:
P | q | ~p | ~p ^ q |
| T T F F | T F T F | F F T T | F F T F |
Here, f: A—>B
f(x)= (x-1)/(x+2) ; x -2
A= {-1,0,1,2,3,4}
B= {-2,1,-1/2,0,1/2,1/4,2/5}
Range = {-2,-1/2,0,1/4,1/2,2/5}
As range is not equal to codomain so the given function is not bijective.
We can make it bijective by omitting {1} from set B
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