INTRODUCTION TO MATHEMATICAL METHODS TEST ONE JAN INTAKE 2024

DATE : 4th March, 2024

DURATION: 2 HOURS

INSTRUCTIONS:

1. There are TWO questions in this Test paper.

2. Answer ALL the questions.

3. Full credit will only be given if all the working is shown. Answers alone whether correct or wrong, without any working shown, will earn a zero mark.

4. Indicate the question attempted CLEARLY

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Question One

a) Define the following terms
i. An Infinite set [2]
ii. An Identity element [2]
iii. Bijective function [2]

Answer:

1. Infinite set: A set with an unlimited number of elements (e.g., $\mathbb{N}$, $\mathbb{Z}$).
2. Identity element: For a binary operation $*$ on set $S$, an element $e \in S$ such that $a * e = e * a = a$ for all $a \in S$.
3. Bijective function: A function that is both injective (one-to-one) and surjective (onto).

OR

Answer:

1. Infinite set: A set that is not finite, meaning its elements cannot be put into a one-to-one correspondence with a finite subset of natural numbers. Example: $N=\{1,2,3,\dots \}$.
2. Identity element: For a binary operation $\ast$ on a set $S$, an element $e\in S$ such that for all $a\in S$, $\overline{)a\ast e=e\ast a=a}$ Example: For addition on $R$, $e=0$ since $a+0=a$.
3. Bijective function: A function that is both injective (one-to-one) and surjective (onto). That is, every element in the codomain is mapped to by exactly one element in the domain.

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b) For each binary operation, find an identity element.
i. On $\mathbb{R}$, define $*$ by $a * b = a + b - ab$ [2]
ii. A binary operation on $\bar{A}$ is defined by $p A q = p + q + \sqrt[3]{15}$ [2]

Answer:
i. Let $e$ be the identity element. Then:

$$a * e = a + e - ae = a \quad \forall a \in \mathbb{R}.$$

Solve:

$$a + e - ae = a \implies e - ae = 0 \implies e(1 - a) = 0.$$

For this to hold for all $a$, $e = 0$.
Verify: $a * 0 = a + 0 - a \cdot 0 = a$.
Identity element is $0$

$0$

ii. Let $e$ be the identity. Then:

$$p A e = p + e + \sqrt[3]{15} = p \quad \forall p \in \bar{A}.$$

Solve:

Subtract $pp$ from both sides:

$$p+e+\sqrt[3]{15}=p$$

$$⟹e+\sqrt[3]{15}=0$$

$$p + e + \sqrt[3]{15} = p \implies e + \sqrt[3]{15} = 0 \implies e = -\sqrt[3]{15}.$$

Identity element is $-\sqrt[3]{15}$.

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c) Show that the cardinality of the power set of the power set of $A$ with one element is 4 and list its elements. [4]

Answer:

• Let $A = \{a\}$.
• Power set of $A$, $\mathcal{P}(A) = \{ \emptyset, \{a\} \}$.
• Power set of $\mathcal{P}(A)$, $\mathcal{P}(\mathcal{P}(A)) = \mathcal{P}\left( \{ \emptyset, \{a\} \} \right)$.
• Elements:

$$\mathcal{P}(\mathcal{P}(A)) = \left\{ \emptyset, \{\emptyset\}, \{\{a\}\}, \{ \emptyset, \{a\} \} \right\}.$$

• Cardinality: $4$.

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d) Suppose that a group of 70 patients are surveyed and asked which disease they have suffered most. There are three diseases in our survey; one for Malaria, one for Cholera, and one for Asthma.40 patients suffered from malaria; 20 patients suffered from cholera; 9 atients suffered from Asthma; 11 patients suffered from Malaria and Asthma; 9 patients suffered from Cholera and Asthma; 15 patients suffered from Malaria and Cholera and 5 patients suffered from all three diseases.

• Malaria (M): 40
• Cholera (C): 20
• Asthma (A): 9
• M and A: 11
• C and A: 9
• M and C: 15
• All three: 5

(i) How many patients did not suffer from any of these diseases? [3]

(ii) How many patients suffered from Cholera AND Malaria BUT NOT Asthma? [3]

Answer:
Use the inclusion-exclusion principle.

• Total patients = 70.
• Patients with at least one disease:

$$\mathrm{\mid }M\cup C\cup A\mathrm{\mid }=\mathrm{\mid }M\mathrm{\mid }+\mathrm{\mid }C\mathrm{\mid }+\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }M\cap C\mathrm{\mid }-\mathrm{\mid }M\cap A\mathrm{\mid }-\mathrm{\mid }C\cap A\mathrm{\mid }+\mathrm{\mid }M\cap C\cap A\mathrm{\mid }$$

$$|M \cup C \cup A| = |M| + |C| + |A| - |M \cap C| - |M \cap A| - |C \cap A| + |M \cap C \cap A|$$ $$= 40 + 20 + 9 - 15 - 11 - 9 + 5 = 39.$$

• (i) No disease: $70 - 39 = \boxed{31}$.

• (ii) Cholera and Malaria but not Asthma:

$$\mathrm{\mid }M\cap C\mathrm{\mid }-\mathrm{\mid }M\cap C\cap A\mathrm{\mid }$$

$$=15-5$$

$$|M \cap C| - |M \cap C \cap A| = 15 - 5 = \boxed{10}.$$

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e) Let $f$ be the function defined as:

$$f(x) = \begin{cases} 3x + 1 & \text{for } x \leq -2 \\ x^2 - 1 & \text{for } -2 < x \leq 2 \\ 1 - 2x & \text{for } x > 2 \end{cases}$$

(i) Show that $f(-4) = f(6)$ [2]
(ii) Sketch the graph of $f(x)$. [3]

Answer:
(i)

• $f(-4)$: Since $-4 \leq -2$, use $3x + 1$:

$$f(-4)=3(-4)+1$$

$$=-12+1$$

$$f(-4) = 3(-4) + 1 = -12 + 1 = -11.$$

• $f(6)$: Since $6 > 2$, use $1 - 2x$:

$$f(6)=1-2(6)$$

$$=1-12$$

$$f(6) = 1 - 2(6) = 1 - 12 = -11.$$

Thus, $f(-4) = f(6) = -11$.

(ii) Graph of $f(x)$:

• Description:
  • For $x \leq -2$: Line $y = 3x + 1$ (passes through $(-2, -5)$, $(-4, -11)$).
  • For $-2 < x \leq 2$: Parabola $y = x^2 - 1$ (vertex at $(0, -1)$, endpoints at $(-2, 3)$, $(2, 3)$).
  • For $x > 2$: Line $y = 1 - 2x$ (passes through $(2, -3)$, $(3, -5)$, etc.).
• 

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Question Two

a) Define the following
i. Surjective function [2]
ii. Singleton set [2]
iii. Equivalence relation [2]

Answer:

1. Surjective function: A function $f: A \to B$ where every element of $B$ is mapped to by at least one element in $A$ (i.e., $\text{range}(f) = B$).
2. Singleton set: A set with exactly one element (e.g., $\{5\}$).
3. Equivalence relation: A relation that is reflexive, symmetric, and transitive.

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b) Determine the domain and range of $f(x) = \frac{\sqrt{2x+3}}{5x-3}$. Hence sketch its graph. [6]

Answer:

• Domain:

  • $\sqrt{2x+3}$ requires $2x + 3 \geq 0 \implies x \geq -\frac{3}{2}$.
  • Denominator $5x - 3 \neq 0 \implies x \neq \frac{3}{5}$.
  • Domain: $\left[-\frac{3}{2}, \frac{3}{5}\right) \cup \left(\frac{3}{5}, \infty\right)$.

• Range:

  • As $x \to \infty$, $f(x) \to 0$.
  • As $x \to \frac{3}{5}^+$, $f(x) \to -\infty$; as $x \to \frac{3}{5}^-$, $f(x) \to +\infty$.
  • Minimum value: At $x = -\frac{3}{2}$, $f(x) = 0$.
  • Range: $(-\infty, 0] \cup (0, \infty) = \mathbb{R} \setminus \{0\}$? Correction: Since numerator is always non-negative and denominator changes sign, range is $(-\infty, 0) \cup [0, \infty) = \mathbb{R}$, but verify:
    • For $x > \frac{3}{5}$, $f(x) > 0$.
    • For $-\frac{3}{2} \leq x < \frac{3}{5}$, $f(x) \leq 0$.
  • Range: $\mathbb{R}$ (all real numbers).

• Graph:
  

  • Key points:
    • Vertical asymptote at $x = \frac{3}{5}$.
    • Horizontal asymptote at $y = 0$.
    • Root at $x = -\frac{3}{2}$.
  • 

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c) Prove $f(x) = \frac{2x+1}{3x+2}$ is one-to-one. Hence find its inverse. [6]

Answer:

• One-to-one proof: Assume $f(a) = f(b)$:

$$\frac{2a+1}{3a+2} = \frac{2b+1}{3b+2}.$$

Cross-multiply:

$$(2a+1)(3b+2)=(2b+1)(3a+2)$$

$$(2a+1)(3b+2) = (2b+1)(3a+2)$$ $$6ab+4a+3b+2=6ab+4b+3a+2$$

$$6ab + 4a + 3b + 2 = 6ab + 4b + 3a + 2$$ $$4a+3b=4b+3a$$

$$4a + 3b = 4b + 3a \implies a = b.$$

Thus, injective.

• Inverse: Solve $y = \frac{2x+1}{3x+2}$ for $x$:

$$y(3x+2)=2x+1$$

$$⟹3xy+2y=2x+1$$

$$y(3x+2) = 2x+1 \implies 3xy + 2y = 2x + 1$$ $$3xy-2x=1-2y$$

$$⟹x(3y-2)=1-2y$$

$$3xy - 2x = 1 - 2y \implies x(3y - 2) = 1 - 2y$$ $$x = \frac{1 - 2y}{3y - 2}.$$

So, $f^{-1}(x) = \frac{1 - 2x}{3x - 2}$.

OR

Answer:
To prove that the function

$f(x)=\frac{2x+1}{3x+2}$

is one-to-one, and then find its inverse, we’ll follow these steps:

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✅ Part 1: Prove that $f\left(x\right)$ is one-to-one

A function is one-to-one if different inputs always produce different outputs. Mathematically, this means:

If $f\left(a\right)=f\left(b\right)$, then $a=b$

Let’s assume:

$f\left(a\right)=f\left(b\right)\Rightarrow \frac{2a+1}{3a+2}=\frac{2b+1}{3b+2}$

Cross-multiply:

$(2a+1)(3b+2)=(2b+1)(3a+2)$

Expand both sides:

Left side:

$2a\left(3b\right)+2a\left(2\right)+1\left(3b\right)+1\left(2\right)=6ab+4a+3b+2$

Right side:

$2b\left(3a\right)+2b\left(2\right)+1\left(3a\right)+1\left(2\right)=6ab+4b+3a+2$

Now equate:

$6ab+4a+3b+2=6ab+4b+3a+2$

Subtract $6ab+2$ from both sides:

$4a+3b=4b+3a$

Rearrange:

$4a-3a=4b-3b$

$\Rightarrow a=b$

✅ Therefore, $f\left(x\right)$ is one-to-one.

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🔄 Part 2: Find the inverse of $f\left(x\right)$

Let $y=f\left(x\right)=\frac{2x+1}{3x+2}$

We solve for $x$ in terms of $y$:

$y=\frac{2x+1}{3x+2}$

Multiply both sides by $3x+2$:

$y(3x+2)=2x+1$

Expand:

$3xy+2y=2x+1$

Group terms with $x$ on one side:

$3xy-2x=1-2y$

Factor out $x$:

$x(3y-2)=1-2y$

Solve for $x$:

$x=\frac{1-2y}{3y-2}$

So the inverse function is:

$f^{-1}\left(x\right)=\frac{1-2x}{3x-2}$

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✅ Final Answer:

• One-to-one proof: Shown using algebraic equality.
• Inverse function:

$f^{-1}\left(x\right)=\frac{1-2x}{3x-2}$

d) Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$, $A = \{2, 4, 7\}$, $B = \{3, 5, 7, 9, 11\}$, $C = \{7, 8, 9, 10, 11\}$.
Verify $(A \cap B)' = A' \cup B'$ [4]

Answer:

• $A \cap B = \{7\}$ (common element).
• $(A \cap B)' = U \setminus \{7\} = \{1,2,3,4,5,6,8,9,10,11\}$.
• $A' = U \setminus A = \{1,3,5,6,8,9,10,11\}$.
• $B' = U \setminus B = \{1,2,4,6,8,10\}$.
• $A' \cup B' = \{1,2,3,4,5,6,8,9,10,11\}$.
  Both sides equal, so verified.

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e) Let $A, B,$ and $C$ be sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$. Show that $B = C$. [3]

Answer:
Prove $B \subseteq C$ and $C \subseteq B$.

• Show $B \subseteq C$:
  Let $x \in B$.
  • Case 1: $x \in A$. Then $x \in A \cap B = A \cap C$, so $x \in C$.
  • Case 2: $x \notin A$. Since $x \in B$, $x \in A \cup B = A \cup C$. As $x \notin A$, $x \in C$.
• Show $C \subseteq B$:
  Let $x \in C$.
  • Case 1: $x \in A$. Then $x \in A \cap C = A \cap B$, so $x \in B$.
  • Case 2: $x \notin A$. Since $x \in C$, $x \in A \cup C = A \cup B$. As $x \notin A$, $x \in B$.
    Thus, $B = C$.

@Dr. Microbiota

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END OF SOLUTIONS