Relations and Functions Test - 7

f(x) = Roll number of student x
f(x₁ ) = Roll number of student x₁
f(x₂ ) = Roll number of student x₂
Putting f(x₁ ) = f(x₂ )
Roll number of student x₁ = Roll number of student x₂
Since no two student of a class have same roll number. 
Therefore, x₁ ≠x₂
So, f is one-one.

 f:A ×B →B ×A is defined as f(a,b)=(b,a).
Let (a1,b1),(a2,b2)∈A ×B such that f(a1,b1)=f(a2,b2).
⇒(b1,a1)=(b2,a2)
⇒(b1=b2) and (a1=a2)
⇒(a1,b1)=(a2,b2)
∴f is one-one.
Now, let (b,a)∈B ×A be any element.
Then, there exists (a,b)∈A ×B such that
f(a,b)=(b,a). [By definition of f]
∴f is onto.

The function is bijective (one-to-one and onto or one-to-one correspondence) if each element of the codomain is mapped to by exactly one element of the domain. (That is, the function is both injective and surjective.) A bijective function is a bijection.

Correct Answer :- b

Explanation : f: R →R is given by,

f(x) = [x+1]

It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.

∴f(1.2) = f(1.9), but 1.2 ≠1.9.

∴f is not one-one.

Now, consider 0.7 ∈R.

It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈R such that f(x) = 0.7.

∴f is not onto.

Hence, the greatest integer function is neither one-one nor onto.

Taking set {1,2,3}
Since f is onto,all elements of{1,2,3} have unique pre-image.
Total no. of one-one function = 3 ×2 ×1=6
Eg:- Since f is onto,all elements of {1,2,3} have unique pre-image.
total no. of onto functions = n ×n −1 ×n −2 ×2 ×1
= n!

For a finite set S, there is a bijection between the set of possible total orderings of the elements and the set of bijections from S to S. That is to say, the number of permutations of elements of S is the same as the number of total orderings of that set, i.e. n!
We have set A that contains  106 elements, so the number of bijective functions from set A to itself is  106!

The composition of f and g is given by f (g(x)) which is equal to 2 (3x + 4) + 1.