Prep 2 Weekly Assessment Week 4 – Solved Models
Model (A)
First: Choose the correct answer:
1) if \( R = \{(1, 3), (2, 5), (4, 3)\} \) is a function, then the domain is ...
2) The opposite figure represent a function on X, then the domain is ...
Second: Answer the following:
1) If \( X=\{0, 1, 2, 3\} \), \( Y=\{1, 2, 4, 5, 6\} \) and R is a relation from X to Y where aRb means \( (a+b=6) \) for every \( a \in X \), \( b \in Y \).
Answer: \( R = \{(0, 6), (1, 5), (2, 4)\} \). Domain = {0, 1, 2}, Range = {4, 5, 6}.2) If \( X=\{1, 2, 3\} \), and R is a relation on X where aRb means that: \( b+a = \) an odd number for all \( a \in X \), \( b \in X \).
Answer: \( R = \{(1, 2), (2, 1), (2, 3), (3, 2)\} \). Domain = {1, 2, 3}, Range = {1, 2, 3}.3) if \( f(x) = 3x + b, f(4) = 15 \) find the value of \( 5b^2 + 2b - 20 \)
Answer: \( f(4) = 3(4) + b = 15 \Rightarrow 12 + b = 15 \Rightarrow b = 3 \).Value = \( 5(3)^2 + 2(3) - 20 = 45 + 6 - 20 = 31 \).
Model (B)
First: Choose the correct answer:
1) if \( R = \{(3, 1), (4, 2), (5, 4)\} \) is a function, then the domain is ...
2) The opposite figure represent a function on X, then the domain is ...
Second: Answer the following:
1) If \( X=\{0, 1, 2, 3, 4\} \), \( Y=\{1, 2, 4, 5, 6, 7\} \) and R is a relation from X to Y where aRb means \( (a+b=7) \) for every \( a \in X \), \( b \in Y \).
Answer: \( R = \{(0, 7), (1, 6), (2, 5), (3, 4)\} \). It is a relation but NOT a function (4 has no image).2) If \( X=\{4, 3, 2\} \) and R is a relation on X where aRb means \( b+a = \) an even number, for every \( a \in X \), \( b \in X \). Answer: \( R = \{(4, 4), (4, 2), (3, 3), (2, 4), (2, 2)\} \). It is NOT a function.
3) if \( f(x) = 5x + b, f(2) = 12 \) find the value of \( b^2 - 2b - 10 \)
Answer: \( f(2) = 5(2) + b = 12 \Rightarrow b = 2 \).Value = \( 2^2 - 2(2) - 10 = 4 - 4 - 10 = -10 \).
Model (C)
First: Choose the correct answer:
1) \( R = \{(1, 3), (2, 3), (4, 3)\} \) is a function, then the domain is ...
2) Which of the following diagrams doesn't represent a function?
Answer: b) because element 1 has two images (two arrows).
Second: Answer the following:
1) If \( X=\{0, 1, 2, 3\} \), \( Y=\{1, 2, 4, 5, 6\} \) and R is a relation from X to Y where aRb means \( (a+b=5) \) for every \( a \in X \), \( b \in Y \).
Answer: \( R = \{(0, 5), (1, 4), (3, 2)\} \). Not a function because 2 has no image.2) If \( X=\{3, 5, 9\} \) and R is a relation on X where aRb means \( b+a = \) a number divisible by 3, for every \( a \in X \), \( b \in X \).
Answer: \( R = \{(3, 3), (3, 9), (9, 3), (9, 9)\} \). Not a function because 5 has no image.3) if \( f(x) = 5x + b, f(2) = 12 \) find the value of \( 3b^2 - 5b + 7 \)
Answer: \( f(2) = 5(2) + b = 12 \Rightarrow b = 2 \).Value = \( 3(2)^2 - 5(2) + 7 = 12 - 10 + 7 = 9 \).