functions -which of following is false

Question

Let A={x|x∈R and x!=2} and B={x|x∈R and x!=1} define f:A->B and g:B->A by 

f(x)=x/(x-2) and g(x)=2x/(x-1) then

which of following is false?

1)f∘g=g∘f    2)f∘g=Ib     Ib is identity function on set B

3)f=g^-1      4) f∘g is bijection

 

Answer 1

Yes, 1) will be the correct answer.

Domain of fog will be same as the domain of g(x) & range of fog will be same as the range of f(x):

so fog can be defined as:

fog: B --> B

fog(x) = x, such that x ∈ B.

So If any real number except 1, will be given to fog, it will produce exactly the same output as the input number.

=> fog(x) is Identity function on set B.

f = g^-1 is also true since we getting fog = x.

fog = x implies that f is cancelling out the transformations applied on x by function g, that's why we are getting x as output.

fog is clearly a bijection, otherwise we would unable to define its inverse.

Now, domain of gof will be same as the domain of f(x) and the range of gof will be same as the co domain of g(x).

gof can be defined as:

gof: A --> A

gof(x) = x, such that x ∈ A.

So If any real number except 2, will be given to gof, it will produce exactly the same output as the input number.

=> gof(x) is Identity function on set A.

Here although expressions of  fog and gof are equivalent but since their domain & co domain differ,they are not equal.

If two functions are equal then they must generate same output on EVERY input,

here if we give x = 1 to gof it will give 1 as output, but if we give x = 1 to fog, since it is not defined for x = 1, it will produce "Not Defined" as an output.

similar case for x = 2.

Thus since fog and gof are not generating same outputs on every input so they are not equal.

Also the graph of fog will be discontinuous at x = 1, & the graph of gof will be discontinuous at x = 2,(Removable Discontinuities).

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# Examples of some functions that actually are not equal:

1) P(x) = 1, Q(x) = x/x : P(0) = 1, & Q(0) = Not Defined.

2) P(x) = x, Q(x) = √x²: P(-4) = -4 & Q(-4) = +4.

3) P(x) = x, Q(x) = (√x)²: P(-4) = -4 & Q(-4) = Not Defined.

4) P(x) = x, Q(x) = e^lnx: P(-1) = -1 & Q(-1) = Not Defined

and so on.

Answer 2

option 1 : f(g(x))=x

g(f(x))=x

therefore true.

option 3 :

g'(x)=x/(x-2) which is f(x)

therefore true.

option 4 :

for every value in f we have a unique value in g similarly for every value in g we have a unique value in f .

therfore f0g is also bijective and hence true.

option 2 must be the answer . Though i did not understand what is IB . :|

Answer 3

what is the second option fog = i b . or it's just b . all except b are correct .

Comments

answer given is a

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fog mens f(g(X)). and gof means g(fx)). i think u have solved it and got both to be x. so that option is correct. there might be a miss print.

Answer 4

Yes it will be option A because its a property composition of two function are not commutative.

Functions f and g fail to commute if for some x, g(f(x))≠f(g(x)). Take any f such that f(x)≠x for some x. Now g(f(x)) can be chosen independently of g(x), and in particular it can be some element other than f(g(x)).

for example let x=2 in fog = -2 but in gof we cant even take 2 coz A={x|x∈R and x!=2}

so fog≠gof