Question -
Answer -
(i) Given f: A┬атЖТA, given┬аby f (x) = x/2
Now we have to show thatthe given function is one-one and on-to
Injection test:
Let┬аx┬аand┬аy┬аbeany two elements in the domain (A), such that┬аf(x) =┬аf(y).
f(x) =┬аf(y)
x/2 = y/2
x = y
So,┬аf┬аisone-one.
Surjection test:
Let┬аy┬аbe anyelement in the co-domain (A), such that┬аf(x) =┬аy┬аfor someelement┬аx┬аin┬аA (domain)
f(x) =┬аy
x/2 = y
x┬а= 2y, which maynot be in┬аA.
For example,if┬аy┬а= 1, then
x┬а= 2, which is not in┬аA.
So,┬аf┬аis not onto.
So,┬аf┬аis not bijective.
(ii) Given g: A┬атЖТA, given┬аby g (x) = |x|
Now we have to showthat the given function is one-one and on-to
Injection test:
Let┬аx┬аand┬аy┬аbe any two elements in the domain (A), suchthat┬аf(x) =┬аf(y).
g(x) =┬аg(y)
|x| = |y|
x┬а=┬а┬▒ y
So,┬аf┬аis notone-one.
Surjection test:
For┬аy┬а= -1,there is no value of┬аx┬аin┬аA.
So,┬аg┬аis notonto.
So,┬аg┬аis not bijective.
(iii) Given h:A┬атЖТ A, given┬аby h (x) = x2
Now we have to showthat the given function is one-one and on-to
Injection test:
Let┬аx┬аand┬аy┬аbeany two elements in the domain (A), such that┬аh(x) =┬аh(y).
h(x) =┬аh(y)
x2┬а=┬аy2
x┬а= ┬▒y
So,┬аf┬аis notone-one.
Surjection test:
For┬аy┬а= тАУ 1,there is no value of┬аx┬аin┬аA.
So,┬аh┬аis not onto.
So,┬аh┬аis not bijective.