Sets And Functions
Question 1 
3m  
3n  
2m+1  
2n+1 
Question 1 Explanation:
No. of subsets of size 3 is = ^{m}C_{3}
n = ^{m}C_{3}
Which subsets contains element i then size is
= ^{(m1)}C_{2}
Because 1 element is already known
n = ^{m}C_{3}
Which subsets contains element i then size is
= ^{(m1)}C_{2}
Because 1 element is already known
Question 2 
Let f: B → C and g: A → B be two functions and let h = f∘g. Given that h is an onto function. Which one of the following is TRUE?
f and g should both be onto functions
 
f should be onto but g need not be onto
 
g should be onto but f need not be onto
 
both f and g need not be onto

Question 2 Explanation:
Given:
f: B→C and g: A→B are two functions.
h = f∘g = f(g(x))
→ If his onto function, that means for every value in C, there must be value in A.
→ Here, we are mapping C to A using B, that means for every value in C there is a value in B then f is onto function.
→ But g may (or) may not be the onto function i.e., so values in B which may doesn't map with A.
f: B→C and g: A→B are two functions.
h = f∘g = f(g(x))
→ If his onto function, that means for every value in C, there must be value in A.
→ Here, we are mapping C to A using B, that means for every value in C there is a value in B then f is onto function.
→ But g may (or) may not be the onto function i.e., so values in B which may doesn't map with A.
Question 3 
g(h(D)) ⊆ D
 
g(h(D)) ⊇ D
 
g(h(D)) ∩ D = ɸ
 
g(h(D)) ∩ (B—D) ≠ ɸ 
Question 3 Explanation:
f: A→B ba an injective (onetoone)
→ g: 2^{A}→2^{B} be also one to one function and g(C) = f(x)x∈C}, for all subsets C of A.
The range of this function is n(2^{A}).
→ h: 2^{B}→2^{A} it is not a one to one function and given h(D) = {xx∈A, f(x)∈D}, for all subsets D of B.
The range of this function is also n(2^{A}).
→ The function g(h(D)) also have the range n(2^{A}) that implies n(A)≤n(B), i.e., n(2^{A}) is less than n(2^{B}).
Then this result is g(h(D))⊆D.
→ g: 2^{A}→2^{B} be also one to one function and g(C) = f(x)x∈C}, for all subsets C of A.
The range of this function is n(2^{A}).
→ h: 2^{B}→2^{A} it is not a one to one function and given h(D) = {xx∈A, f(x)∈D}, for all subsets D of B.
The range of this function is also n(2^{A}).
→ The function g(h(D)) also have the range n(2^{A}) that implies n(A)≤n(B), i.e., n(2^{A}) is less than n(2^{B}).
Then this result is g(h(D))⊆D.
Question 4 
The number of functions from an m element set to an n element set is
m + n  
m^{n}  
n^{m}  
m*n 
Question 4 Explanation:
Here each m element we have n choices i.e.,
n×n×n×n×...×n (m times)
= n^{m}
n×n×n×n×...×n (m times)
= n^{m}
Question 5 
Suppose X and Y are sets and X Y and are their respective cardinalities. It is given that there are exactly 97 functions from X to Y. from this one can conclude that
X = 1, Y = 97  
X = 97, Y = 1  
X = 97, Y = 97  
None of the above 
Question 5 Explanation:
From the given information we can write,
Y^{X} = 97
→ Option A only satisfies.
Y^{X} = 97
→ Option A only satisfies.
Question 6 
Let R denotes the set of real numbers. Let f: R×R → R×R be a bijective function defined by f(x, y) = (x+y, xy). the inverse function of f is given by
Question 6 Explanation:
There are 7 questions to complete.