Question -
Answer -
(i) Given f: A →A, given by f (x) = x/2
Now we have to show thatthe given function is one-one and on-to
Injection test:
Let x and y beany two elements in the domain (A), such that f(x) = f(y).
f(x) = f(y)
x/2 = y/2
x = y
So, f isone-one.
Surjection test:
Let y be anyelement in the co-domain (A), such that f(x) = y for someelement x in A (domain)
f(x) = y
x/2 = y
x = 2y, which maynot be in A.
For example,if y = 1, then
x = 2, which is not in A.
So, f is not onto.
So, f is not bijective.
(ii) Given g: A →A, given by g (x) = |x|
Now we have to showthat the given function is one-one and on-to
Injection test:
Let x and y be any two elements in the domain (A), suchthat f(x) = f(y).
g(x) = g(y)
|x| = |y|
x = ± y
So, f is notone-one.
Surjection test:
For y = -1,there is no value of x in A.
So, g is notonto.
So, g is not bijective.
(iii) Given h:A → A, given by h (x) = x2
Now we have to showthat the given function is one-one and on-to
Injection test:
Let x and y beany two elements in the domain (A), such that h(x) = h(y).
h(x) = h(y)
x2 = y2
x = ±y
So, f is notone-one.
Surjection test:
For y = – 1,there is no value of x in A.
So, h is not onto.
So, h is not bijective.