Solution:
A bijective function has no unpaired elements and satisfies both injective (one-to-one) and surjective (onto) mapping of a set A to a set B. Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. In mathematical terms, let f: A → B is a function; then, f will be bijective if every element ‘b’ in the co-domain B, has exactly one element ‘a’ in the domain A, such that f (a) =b.
Given
f:A→Adefined by f(x)=(2x+3)(mod7)whereA=N7={0,1,2,3,4,5,6}
We have f(0)=3,f(1)=5,f(2)=0,f(3)=2,f(4)=4,f(5)=6,f(6)=1 ...(i)
Then f is one-one and onto.
⇒ f is bijection.
Then, from (i) 0=f−1(3),1=f−1(5),2=f−1(0),3=f−1(2),4=f−1(4),5=f−1(6),6=f−1(1),
∴f−1(x)={0,1,2,3,4,5,6}
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