INTRODUCTION TO MATHEMATICAL METHODS TEST ONE JULY INTAKE 2024

DATE : 9TH SEP 2024

INSTRUCTIONS

DURATION: 1 hr and 30 minutes .

• Answer all questions.

• Show all your working, answers with no working will not earn marks.

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QUESTION ONE

a. Let $A = [2,3]$, $B = [2,5]$ and $C = (3, \infty)$, also let $\mathbb{R}$ be the universal set, find $(C^c \cup A)\cap B$, and show the answer on the number line. [5]

Solution:

• $C = (3, \infty)$, so $C^c = (-\infty, 3]$ (complement includes all real numbers not in $C$).
• $A = [2,3]$, so $C^c \cup A = (-\infty, 3] \cup [2,3]$. Since $[2,3] \subseteq (-\infty, 3]$, this union simplifies to $(-\infty, 3]$.
• $B = [2,5]$, so $(C^c \cup A) \cap B = (-\infty, 3] \cap [2,5]$. The intersection is $[2,3]$ because:
  • Lower bound: $\max(2, -\infty) = 2$ (inclusive, as both sets include 2).
  • Upper bound: $\min(3,5) = 3$ (inclusive, as both sets include 3).
• Thus, $(C^c \cup A) \cap B = [2,3]$.

Number Line Representation:

• Draw a horizontal number line with points at 2, 3, and 5.
• Shade the interval from 2 to 3 inclusive, indicated by solid dots at both endpoints.

Number line:  
<---|------●======●------|--->  
    1      2      3      4      5  

Answer: $[2,3]$

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b. Write the following sets in roster form and set builder form respectively.
i. $P = \{ x \mid x \in \mathbb{Z}^+ \text{ which is less than } 10 \text{ and } 2x - 1 \text{ is an odd number.} \}$ [3]

Solution:

• Condition 1: $x$ is a positive integer less than 10, so $x \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
• Condition 2: $2x - 1$ is odd. For any integer $x$, $2x$ is even, so $2x - 1$ is always odd. This condition is always true and redundant.
• Roster form: $P = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
• Set builder form is already given in the question.

Answer for i:

• Roster form: $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

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ii. $H = \{ 5, 25, 125, 625, \ldots \}$ [3]

Solution:

• The sequence is $5^1, 5^2, 5^3, 5^4, \ldots$, so it consists of powers of 5 starting from exponent 1.
• Set builder form: $H = \{ 5^n \mid n \in \mathbb{Z}^+ \}$.

Answer for ii:

• Set builder form: $\{ 5^n \mid n \in \mathbb{Z}^+ \}$

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c. Express $-12.3231$ in the form $\frac{a}{b}$ where $a$ and $b$ are integers with no common factors such that $b \neq 0$. [4]

Solution:

• $-12.3231 = -\frac{123231}{10000}$ because:
  • $12.3231 = \frac{123231}{10000}$ (since 12.3231 has 4 decimal places, so denominator is $10^4 = 10000$).
  • Thus, $-12.3231 = -\frac{123231}{10000}$.
• Simplify $\frac{123231}{10000}$ by finding gcd of 123231 and 10000:
  • Factorize denominator: $10000 = 10^4 = (2 \times 5)^4 = 2^4 \times 5^4$.
  • Numerator 123231:
    • Sum of digits: $1 + 2 + 3 + 2 + 3 + 1 = 12$, divisible by 3, so $123231 \div 3 = 41077$.
    • $41077$ has no common factors with $10000$ (since it ends with 1, not divisible by 2 or 5; and $4 + 1 + 0 + 7 + 7 = 19$, not divisible by 3).
  • gcd(123231, 10000) = 1, so the fraction is already in simplest terms.
• Thus, $-\frac{123231}{10000}$ with $a = -123231$, $b = 10000$, and no common factors.

Answer: $-\frac{123231}{10000}$

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d. Rationalize the denominator and simplify $\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{7}}$. [5]

Solution:

• Multiply numerator and denominator by the conjugate of the denominator: $3\sqrt{5} + 2\sqrt{7}$.
• Numerator: $(2\sqrt{6} - \sqrt{5}) \cdot (3\sqrt{5} + 2\sqrt{7})$: $$\begin{array}{rl}{\displaystyle }& {\displaystyle =2\sqrt{6}X3\sqrt{5}+2\sqrt{6}X2\sqrt{7}-\sqrt{5}X3\sqrt{5}-\sqrt{5}X2\sqrt{7}}\\ {\displaystyle }& {\displaystyle }\end{array}$$

$$\begin{array}{r}{\displaystyle =6\sqrt{30}+4\sqrt{42}-3\mathrm{X\; 5}-2\sqrt{35}}\\ {\displaystyle }& {\displaystyle }\end{array}$$

$$\begin{align*} &= 2\sqrt{6} \cdot 3\sqrt{5} + 2\sqrt{6} \cdot 2\sqrt{7} - \sqrt{5} \cdot 3\sqrt{5} - \sqrt{5} \cdot 2\sqrt{7} \\ &= 6\sqrt{30} + 4\sqrt{42} - 3 \cdot 5 - 2\sqrt{35} \\ &= 6\sqrt{30} + 4\sqrt{42} - 15 - 2\sqrt{35}. \end{align*}$$

• Denominator: $(3\sqrt{5}{)}^2-(2\sqrt{7}{)}^2$
• $\; \; \; \; \; \; \; \; \; \; \; \; \; \; =9X5-4X7$
• $\; \; \; \; \; \; \; \; \; \; \; \; \; \; =45-28$
• $(3\sqrt{5})^2 - (2\sqrt{7})^2 = 9 \cdot 5 - 4 \cdot 7 = 45 - 28 = 17$.
• Simplified fraction: $$\frac{6\sqrt{30} + 4\sqrt{42} - 2\sqrt{35} - 15}{17}$$
• No further simplification is possible, as the terms under the radicals have no common factors.

Answer: $\frac{6\sqrt{30} + 4\sqrt{42} - 2\sqrt{35} - 15}{17}$

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e. There are 200 individuals with a skin disorder, 120 had been exposed to chemical $C_1$, 50 to chemical $C_2$, and 30 to both the chemicals $C_1$ and $C_2$. Find the number of individuals exposed to chemical $C_1$ but not chemical $C_2$. [5]

Solution:

• Let $A$ = set exposed to $C_1$, $|A| = 120$.
• Let $B$ = set exposed to $C_2$, $|B| = 50$.
• $|A \cap B| = 30$ (exposed to both).
• Number exposed to $C_1$ but not $C_2$ is $|A - B| = |A| - |A \cap B| = 120 - 30 = 90$.
• Total individuals (200) is not needed for this calculation.

Answer: 90

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QUESTION TWO

a. Let * be an operation defined on set $A = \{1, 3, 5, 7\}$ as shown in the table below.

*	1	3	5	7
1	1	3	5	7
3	3	1	7	5
5	5	7	1	3
7	7	5	3	1

i. Determine if the operation * is binary on set $A$. [3]

Solution:

• A binary operation requires that for every pair $(a, b) \in A \times A$, $a * b$ is defined and in $A$.
• The table covers all $4 \times 4 = 16$ combinations, and all results are in $\{1, 3, 5, 7\} = A$.
• Thus, * is binary on $A$.

Answer: Yes

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ii. Determine the identity elements if exist. [3]

Solution:

• An identity element $e$ satisfies $a * e = a$ and $e * a = a$ for all $a \in A$.
• Check $e = 1$:
  • $a * 1$: For $a=1$, $1*1=1$; $a=3$, $3*1=3$; $a=5$, $5*1=5$; $a=7$, $7*1=7$. So $a * 1 = a$.
  • $1 * a$: For $a=1$, $1*1=1$; $a=3$, $1*3=3$; $a=5$, $1*5=5$; $a=7$, $1*7=7$. So $1 * a = a$.
• Check other elements (e.g., $e=3$): $1*3=3 \neq 1$, so not identity. Similarly, $e=5$ and $e=7$ fail.
• Thus, the identity element is 1.

Answer: 1

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iii. Solve $(3 * 5) * 7$ [2]

Solution:

• From table: $3 * 5 = 7$ (row 3, column 5).
• Then $7 * 7 = 1$ (row 7, column 7).
• So $(3 * 5) * 7 = 1$.

Answer: 1

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b. Let $R = \{(1,1), (1,3), (1,4), (2,2), (3,1), (3,4), (4,1), (4,3)\}$ be relation defined on set $A = \{1,2,3,4\}$. Determine whether $R$ is equivalent. Justify your answer. [6]

Solution:

• "Equivalent" means equivalence relation (reflexive, symmetric, transitive).
• Reflexive? Requires $(a,a) \in R$ for all $a \in A$.
  • $(1,1) \in R$, $(2,2) \in R$, but $(3,3) \notin R$ and $(4,4) \notin R$.
  • Not reflexive.
• Symmetric? If $(a,b) \in R$, then $(b,a) \in R$.
  • Check: $(1,3) \in R$ and $(3,1) \in R$; $(1,4) \in R$ and $(4,1) \in R$; $(3,4) \in R$ and $(4,3) \in R$; $(1,1)$ and $(2,2)$ are symmetric.
  • Symmetric holds, but reflexivity fails.
• Transitive? Not required since not reflexive.
• Conclusion: $R$ is not an equivalence relation because it is not reflexive.

Answer: No, not equivalence relation (fails reflexivity).

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c. Let $f(x) = 2x - 3$ and $g(x) = x + 1$. Find $(g \circ f)^{-1}(x)$. [3]

Solution:

• Compute $g \circ f$: $$g \circ f(x) = g(f(x)) = g(2x - 3) = (2x - 3) + 1 = 2x - 2.$$
• Let $h(x) = 2x - 2$. Find $h^{-1}(x)$: $$\begin{align*} y &= 2x - 2, \\ y + 2 &= 2x, \\ x &= \frac{y + 2}{2}. \end{align*}$$ So $h^{-1}(x) = \frac{x + 2}{2}$.
• Thus, $(g \circ f)^{-1}(x) = \frac{x + 2}{2}$.

Answer: $\frac{x + 2}{2}$

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d. State the domain and range of the function below.
i. $f(x) = x^2 + 2$ [4]

Solution:

• Domain: No restrictions; defined for all real $x$. So domain is $(-\infty, \infty)$.
• Range: $x^2 \geq 0$, so $x^2 + 2 \geq 2$. Minimum value is 2 at $x = 0$, and as $x \to \pm\infty$, $f(x) \to \infty$. So range is $[2, \infty)$.

Answer: Domain: $\mathbb{R}$ or $(-\infty, \infty)$; Range: $[2, \infty)$

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ii. $g(x) = \frac{x+4}{2x-1}$ [4]

Solution:

• Domain: Denominator $2x - 1 \neq 0$, so $x \neq \frac{1}{2}$. Domain is $\left( -\infty, \frac{1}{2} \right) \cup \left( \frac{1}{2}, \infty \right)$.
• Range: Solve for $x$ in terms of $y$: $$\begin{array}{rl}{\displaystyle y}& {\displaystyle =\frac{x+4}{2x-\; 1\mathrm{}}}\end{array}$$$$\begin{array}{r}{\displaystyle ,}\\ {\displaystyle y(2x-1)}& {\displaystyle =x+4,}\\ {\displaystyle \mathrm{}}\end{array}$$

$$\begin{array}{rl}{\displaystyle 2xy-y}& {\displaystyle =x+4,}\\ {\displaystyle \mathrm{}}\end{array}$$

$$\begin{array}{rl}{\displaystyle 2xy-x}& {\displaystyle =y+4,}\\ {\displaystyle }\end{array}$$

$$\begin{align*} y &= \frac{x+4}{2x-1}, \\ y(2x - 1) &= x + 4, \\ 2xy - y &= x + 4, \\ 2xy - x &= y + 4, \\ x(2y - 1) &= y + 4, \\ x &= \frac{y + 4}{2y - 1}. \end{align*}$$ This is defined when $2y - 1 \neq 0$, i.e., $y \neq \frac{1}{2}$. For all other $y$, there is a corresponding $x$,

so range is all reals except $\frac{1}{2}$: $\left( -\infty, \frac{1}{2} \right) \cup \left( \frac{1}{2}, \infty \right)$.

Answer: Domain: $\left( -\infty, \frac{1}{2} \right) \cup \left( \frac{1}{2}, \infty \right)$; Range: $\left( -\infty, \frac{1}{2} \right) \cup \left( \frac{1}{2}, \infty \right)$

@Dr. Microbiota

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END OF SOLUTIONS