FUNCTIONS QUESTIONS (With detailed answers)

Functions Questions

1. Function Analysis

(a) Sketch the graph of the function f(x) = (x + 3)/[(x + 2)(x - 3)]

(b) Show that x² + 6x - 10 can be expressed in the form (x - α)(x - β) + 2(x - α) + 3x in two different ways and find the values of α and β in each case.

(c) Find the composite function f∘g(x) and g∘f(x) given that:
f = {(3,6), (5,7), (9,0)} and g = {(2,3), (4,5), (6,7)}

2. Function Graphs and Composition

(a) Sketch the graph of the function y = (x³ - 2x)/[2(x² - 5)] hence determine the domain and range.

(b) The functions f and g are defined as follows:
f(x) = e⁻ˣ and g(x) = 1/(1 - x).
Find the ranges of x³ - 3x - 20 = 0 and g∘f.

3. Function Operations

(a) If x = 1,2,3,4,5, and:
f(x) = {(1,3), (2,5), (3,3), (4,1), (5,2)}
g(x) = {(1,4), (2,1), (4,2), (5,3)}
Find:
(i) fog(x)
(ii) gog(x)

(b) Sketch the graph of f(x) = (2x³ - x² - 25x - 12)/(x³ - x² - 5x + 5)

4. Function Analysis and Composition

(a) Sketch the graph of the function y = (x³ - 2x)/[2(x² - 5)] hence determine the domain and range.

(b) Sketch and state domain and range of the graph if y = (x² - 4)/(x² - 1).

(c) Find the composite function f∘g(x) and g∘f(x) given that:
f = {(3,6), (5,7), (9,0)} and g = {(2,3), (4,5), (6,7)}

(d) Find the range of the function f: x → |x - 1| whose domain is given by |x| ≤ 3.

5. Advanced Function Concepts

(a) Given that f(x + 2) = x² + 2. Evaluate the function f(x).

(b) The functions:
f(x) = (eˣ - e⁻ˣ)/(eˣ + e⁻ˣ)
g(x) = ln√[(1 + x)/(1 - x)]
are said to be inverses of each other. Show that:
f∘g(x) = g∘f(x)

(c) Sketch the graph of f(x) = 4x/(x² - 4).

Complete Sets Questions Solutions

Question A

(a) Simplify each of the following sets operation:

(i) Simplify A - (A² - B)

First, we interpret A² as A' (the complement of A), as this is a common notation issue

Rewrite the expression: A - (A' - B)

Understand that X - Y = X ∩ Y' (set difference definition)

Apply to A' - B: A' - B = A' ∩ B'

Now the expression becomes: A - (A' ∩ B') = A ∩ (A' ∩ B')'

Apply De Morgan's Law: (A' ∩ B')' = A ∪ B

Now we have: A ∩ (A ∪ B)

By the absorption law: A ∩ (A ∪ B) = A

Note: The absorption law states that for any sets A and B, A ∩ (A ∪ B) = A

Final Answer: A

(ii) Simplify [A ∩ (A² ∪ B)] ∪ [B ∩ (A² ∩ B²)]

Assume A² = A' (complement of A) and B² = B' (complement of B)

First part: A ∩ (A' ∪ B)

Distribute: (A ∩ A') ∪ (A ∩ B)

A ∩ A' = ∅ (empty set)

So first part becomes: ∅ ∪ (A ∩ B) = A ∩ B

Second part: B ∩ (A' ∩ B')

Associate: B ∩ A' ∩ B'

Since B ∩ B' = ∅, the entire expression becomes ∅

Combine both parts: (A ∩ B) ∪ ∅ = A ∩ B

Final Answer: A ∩ B

(b) Given sets:

• A = {X : X ∈ Z⁺, X ≤ 10} = {1,2,3,4,5,6,7,8,9,10}
• B = {X : X ∈ Z⁺, X is even and X ≤ 10} = {2,4,6,8,10}
• μ = {X : X ∈ Z⁺, X ≤ 20} (universal set)

i. Find A - B

A - B means elements in A that are not in B

A contains all positive integers from 1 to 10

B contains even numbers from 2 to 10

Therefore, A - B = all odd numbers in A

Final Answer: {1,3,5,7,9}

ii. Find A² ∩ B²

A² = A' = μ - A = {11,12,...,20}

B² = B' = μ - B = all numbers not in B

B' includes all odd numbers 1-20 plus even numbers 12-20

A' ∩ B' = numbers that are in both A' and B'

This is exactly {11,12,...,20} because A' is entirely contained within B'

Final Answer: {11,12,13,14,15,16,17,18,19,20}

iii. Find A² ∪ B

A² = A' = {11,12,...,20}

B = {2,4,6,8,10}

A' ∪ B = combination of both sets

Final Answer: {2,4,6,8,10,11,12,...,20}

(c) Venn Diagram Problem

Given information:

• Total people: 40
• Men: 20 → Women: 20
• Taking Mathematics: 18
• Women not taking Mathematics: 6
• Men not taking Chemistry: 4
• Taking Chemistry: 13
• Of 18 taking Mathematics, 13 are taking Chemistry

i. The number of women taking both subjects

Total women = 20

Women not taking Math = 6 → Women taking Math = 20 - 6 = 14

Total taking Math = 18 → Men taking Math = 18 - 14 = 4

Total taking Chemistry = 13

Of Math takers taking Chemistry: 13 (given)

Men taking Chemistry: Total men - men not taking Chemistry = 20 - 4 = 16

But only 4 men take Math, so maximum men taking both Math and Chemistry is 4

Thus, women taking both = Total both takers - men taking both = 13 - 4 = 9

Key insight: The number of men taking both subjects cannot exceed the number of men taking Math

Final Answer: 9 women take both Mathematics and Chemistry

ii. The number of men who take mathematics

Total taking Math = 18

Women taking Math = 14 (from part i)

Therefore, men taking Math = 18 - 14 = 4

Final Answer: 4 men take Mathematics

Question B

(a) Describe set A by roster form: A = {x²: x ∈ z⁺ and 0 ≤ x < 6}

x can be: 1, 2, 3, 4, 5 (since x is positive integer and 0 ≤ x < 6)

Calculate squares: 1²=1, 2²=4, 3²=9, 4²=16, 5²=25

Final Answer: {1, 4, 9, 16, 25}

(b) Venn diagram problem with universal set U = {a,b,c,d,e,f,g,h,i,j}

Given: A ∩ U = {d,e,h,i} → This likely means A = {d,e,h,i}

A ∩ B = {d,h} → elements common to both A and B

(A∪B)' = {f,g,j} → elements not in A or B

i. Represent in Venn diagram

Draw three regions: A only, B only, A∩B

A only: {e,i} (elements in A not in B)

A∩B: {d,h} (given)

(A∪B)' = {f,g,j} → outside both circles

Remaining elements must be in B only: {a,b,c}

Note: The universal set contains all elements not accounted for elsewhere

ii. Find number of elements in B and A'∩B

B = B only + A∩B = {a,b,c} + {d,h} = {a,b,c,d,h}

Number of elements in B: 5

A'∩B = elements in B not in A = {a,b,c}

Number of elements in A'∩B: 3

Final Answers:
Elements in B: 5
Elements in A'∩B: 3

Question C

In a class of 58 students: 30 boys and 28 girls. 29 study Mathematics, 27 study Physics. Among girls: 3 study both, 9 study Math only, 4 study neither. Among boys: 5 study neither.

a. Summarize using Venn diagram

Total students = 58

Girls: 28 (Boys: 30)

For Girls (28 total):

Both subjects: 3

Math only: 9

Neither: 4

Therefore, Physics only = 28 - (3+9+4) = 12

For Boys (30 total):

Neither: 5

Total in subjects: 30 - 5 = 25

Total Math students = 29 → Boys in Math = 29 - (9 girls math + 3 girls both) = 17

Total Physics students = 27 → Boys in Physics = 27 - (12 girls physics + 3 girls both) = 12

Boys in both = Total boys in subjects - (boys math only + boys physics only)

But we need to find overlap: Let x = boys in both

Then: (17 - x) + (12 - x) + x = 25 → 29 - x = 25 → x = 4

Venn Diagram Values:
Girls: Math only=9, Physics only=12, Both=3, Neither=4
Boys: Math only=13, Physics only=8, Both=4, Neither=5

b. How many students study both subjects?

Girls both: 3

Boys both: 4

Final Answer: 7 students study both Mathematics and Physics

c. How many boys study physics only?

From the calculation above: Boys Physics only = 12 - 4 (both) = 8

Final Answer: 8 boys study Physics only