Questions on SETS (b) Let fand g be functions from A-(1,2,3,4) to B= {a, b,c,d)

Question

Questions on SETS

(b) Let fand g be functions from A-(1,2,3,4) to B= {a, b,c,d) and from B to A, respectively, with f 1 = d, f(2) = c, f(3) = a, and f(4)-b, and g(a) = 2, g(b) = 1, g(c) = 3, and g(d) = 2. (i) Is f one-to-one? Is g one-to-one? () Is f onto? Is g onto? (ii) Which one f and g has an inverse function, and what is this inverse? (iv) Let Sy: P(A) P(B), where S, (X) = vre X,y= f(x)), P(A) and P(B) are power sets. 4 marks] [4 marks] 4 marks] Determine the values of Sy( 1,2) and Sr(0). [5 marks]

Explanation / Answer

(i) A function is One-to-one if every element of codomain is mapped to not more than 1 element of domain.

f(1) = d, f(2) = c, f(3) = a, f(4) = b

This is one-to-one function

g(a) = 2, g(b) = 1, g(c) =3, g(d) =2

This is not one to one since "2" is mapped to both "a" and "d".

(ii) A function is Onto if all the elements of codomain are present in range set i.e. mapped to some element of domain.

f(1) = d, f(2) = c, f(3) = a, f(4) = b

codomain = {a,b,c,d}

range = {a,b,c,d}

so f is onto

g(a) = 2, g(b) = 1, g(c) =3, g(d) =2

codomain = {1,2,3,4}

range = {1,2,3}

so g is not onto

(iii)

f has inverse function because it is both One to One and Onto.

While g is not.

let's say Inverse of f is h, so

h(d) = 1

h(c) = 2

h(a) = 3

h(b) = 4

(iv)

Sf{1,2} = {f(1), f(2)} = {d,c}

Sf{phi} = {phi} = {phi}

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