Question

Suppose f : Z-> Z is a function defined by f(x) = Ax + B.
for which of the following integer values of A and B,
is the given function Bijective
a) A=0, B= {z | z € Z}
b) A = { - 1, 1}, B = { z | z € Z}
c) B = { -1, 1}, A= 0
d) B= 0, A= { z | z € Z} ​

Answer 1

Answer:

b

Step-by-step explanation:it is function of line y=mx+c it has always bijective  for A€R (except 0) and B€R

so oly optin b don"t have value 0 for A

Answer 2

In summary, A = 1 or A = -1 and B = 0 are the only values of A and B for which the function f(x) = Ax + B is bijective.

As per the question given,

For a function to be bijective, it must be both injective and surjective.

a) A=0, B= {z | z € Z}

In this case, the function is f(x) = B, which is constant. It is not injective, as f(x) = f(y) for any x and y. Therefore, the function is not bijective.

b) A = { - 1, 1}, B = { z | z € Z}

In this case, we need to consider two functions:

If A = 1 and B = 0, then f(x) = x, which is bijective.

If A = -1 and B = 0, then f(x) = -x, which is also bijective.

Therefore, there exist values of A and B for which the function is bijective.

c) B = { -1, 1}, A= 0

In this case, the function is f(x) = B, which is constant. It is not injective, as f(x) = f(y) for any x and y. Therefore, the function is not bijective.

d) B= 0, A= { z | z € Z}

In this case, the function is f(x) = Ax, which depends only on the input value x. If A is non-zero, then the function is not injective, as f(x) = f(y) for any x and y such that x = y/A. If A is zero, then the function is f(x) = 0, which is not injective either. Therefore, the function is not bijective.

In summary, the only values of A and B for which the function f(x) = Ax + B is bijective are A = 1 or A = -1 and B = 0.

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