### Relations and Functions Miscellaneous Exercise

## Question:

Let A={1,2,3,4},B={1,5,9,11,15,16} and f={(1,5),(2,9),(3,1),(4,5),(2,11)} Are the following true? (i) f is a relation from A to B (ii) f is a function from A to B Justify your answer in each case

## Answer:

(i) True: f is a relation from A to B because it contains ordered pairs (1,5), (2,9), (3,1), (4,5), (2,11) which have elements from both A and B.

(ii) False: f is not a function from A to B because it contains more than one ordered pair with the same first element (2). This violates the definition of a function, which states that each element in the domain can only have one corresponding element in the range.

## Question:

Let f be the subset of Z×Z defined by f={(ab,a+b):a,b∈Z}. Is f a function from Z to Z : justify your answer

## Answer:

Answer: No, f is not a function from Z to Z. This is because, for any given element in Z, there may be more than one pair in f that satisfies the given condition. For example, if a = 2 and b = 3, then the pair (6,5) is an element of f. But if a = -2 and b = 3, then the pair (-6,1) is also an element of f. Since there are two pairs that satisfy the given condition, f cannot be a function from Z to Z.

## Question:

Let f={(1,1),(2,3),(0,−1),(−1,−3)} be a function from Z to Z defined by f(x)=ax+b for some integers a,b. Determine a,b.

## Answer:

- Substitute x=1 into the equation f(x)=ax+b.
- f(1)=a+b
- Since (1,1) is a point in the given set, a+b=1
- Substitute x=2 into the equation f(x)=ax+b.
- f(2)=2a+b
- Since (2,3) is a point in the given set, 2a+b=3
- Substitute x=0 into the equation f(x)=ax+b.
- f(0)=b
- Since (0,-1) is a point in the given set, b=-1
- Substitute x=-1 into the equation f(x)=ax+b.
- f(-1)=-a+(-1)
- Since (-1,-3) is a point in the given set, -a-1=-3
- Solve -a-1=-3 for a.
- a=-4
- Substitute a=-4 into a+b=1
- -4+b=1
- Solve -4+b=1 for b.
- b=5
- Therefore, a=-4 and b=5.

## Question:

Let A={9,10,11,12,13} and let f:A→N be defined by f(n)= the highest prime factor of n. Find the range of f

## Answer:

A = {9, 10, 11, 12, 13}

f(9) = 3 (the highest prime factor of 9) f(10) = 5 (the highest prime factor of 10) f(11) = 11 (the highest prime factor of 11) f(12) = 3 (the highest prime factor of 12) f(13) = 13 (the highest prime factor of 13)

Therefore, the range of f is {3, 5, 11, 13}.

## Question:

Find the domain and range of the real function f defined by f(x)=∣x−1∣

## Answer:

Domain: x ∈ (-∞,∞) Range: y ∈ [0,∞)