INTRODUCTION TO MATHEMATICAL METHODS TEST ONE JULY INTAKE 2022

DATE : 9th March, 2022.

INSTRUCTIONS:

• Indicate your Name, Computer number and Tutorial group on your answer paper
• Answer all questions
• Show your working to earn full marks
  TIME ALLOWED: 2 Hours
  

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1. (a) Define the following:

(i) A function
A function $f$ from set $A$ to set $B$ is a rule that assigns to each element $x \in A$ exactly one element $y \in B$, denoted $y = f(x)$.

(ii) A one-to-one function
A function $f: A \to B$ is one-to-one (injective) if for every $x_1, x_2 \in A$, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

(iii) Binary operation
A binary operation $*$ on a set $S$ is a function $*: S \times S \to S$ that maps each ordered pair $(a, b) \in S \times S$ to an element $a * b \in S$.

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(b) Rationalize the denominator of the following:

(i) $\frac{5}{2 + \sqrt{5}}$
Solution:

$$\frac{5}{2+\sqrt{5}}X\frac{2-\sqrt{5}}{2-\sqrt{5}}$$

$$=\frac{5(2-\sqrt{5})}{(2{)}^2-(\sqrt{5}{)}^2}$$

$$=\frac{10-5\sqrt{5}}{4-5}$$

$$=\frac{10-5\sqrt{5}}{-1}$$

$$\frac{5}{2 + \sqrt{5}} \cdot \frac{2 - \sqrt{5}}{2 - \sqrt{5}} = \frac{5(2 - \sqrt{5})}{(2)^2 - (\sqrt{5})^2} = \frac{10 - 5\sqrt{5}}{4 - 5} = \frac{10 - 5\sqrt{5}}{-1} = -10 + 5\sqrt{5}$$

Final Answer: $5\sqrt{5} - 10$

(ii) $\frac{3 - \sqrt{5}}{1 + 3\sqrt{5}}$
Solution:

$$\frac{3-\sqrt{5}}{1+3\sqrt{5}}X\frac{1-3\sqrt{5}}{1-3\sqrt{5}}$$

$$=\frac{(3-\sqrt{5})(1-3\sqrt{5})}{(1{)}^2-(3\sqrt{5}{)}^2}$$

$$\frac{3 - \sqrt{5}}{1 + 3\sqrt{5}} \cdot \frac{1 - 3\sqrt{5}}{1 - 3\sqrt{5}} = \frac{(3 - \sqrt{5})(1 - 3\sqrt{5})}{(1)^2 - (3\sqrt{5})^2} = \frac{3 \cdot 1 + 3 \cdot (-3\sqrt{5}) + (-\sqrt{5}) \cdot 1 + (-\sqrt{5}) \cdot (-3\sqrt{5})}{1 - 45}$$

Numerator:
$3-9\sqrt{5}-\sqrt{5}+15$

$3 - 9\sqrt{5} - \sqrt{5} + 15 = 18 - 10\sqrt{5}$

Denominator:
$1 - 45 = -44$

$$\frac{18-10\sqrt{5}}{-44}$$

$$\frac{18 - 10\sqrt{5}}{-44} = \frac{10\sqrt{5} - 18}{44} = \frac{5\sqrt{5} - 9}{22}$$

Final Answer: $\frac{5\sqrt{5} - 9}{22}$

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(c) Express the following in the form $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$.

(i) $1.666\ldots$
Solution:
Let $x = 1.666\ldots$.
Then $10x = 16.666\ldots$.
Subtract:

$$10x-x=\mathrm{16.666\; -}1.666\dots$$

$$⟹9x=15$$

$$⟹x=\frac{15}{9}$$

$$10x - x = 16.666\ldots - 1.666\ldots \implies 9x = 15 \implies x = \frac{15}{9} = \frac{5}{3}$$

Final Answer: $\frac{5}{3}$

(ii) $0.87$
Solution:

$$0.87 = \frac{87}{100}$$

GCD of 87 and 100 is 1 (87 = 3 × 29, 100 = 2² × 5²).
Final Answer: $\frac{87}{100}$

(iii) $-5.463$
Solution:

$$-5.463 = -\frac{5463}{1000}$$

GCD of 5463 and 1000 is 1 (5463 = 3² × 607, 1000 = 2³ × 5³).
Final Answer: $-\frac{5463}{1000}$

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2. (a) Let $S = \{0, \frac{1}{3}, -2, 1.414, \frac{22}{7}, \sqrt{2}, \sqrt{3}, 0.75, \pi, \sqrt{5}, 2, \frac{2}{5}, -3\}$.

Let $A = \{\text{integers in } S\}$, $B = \{\text{rational numbers in } S\}$, $C = \{\text{irrational numbers in } S\}$.
Write the elements of:
(i) $B \cap C$

Solution:

• Rationals $B$: $0, \frac{1}{3}, -2, 1.414, \frac{22}{7}, 0.75, 2, \frac{2}{5}, -3$
• Irrationals $C$: $\sqrt{2}, \sqrt{3}, \pi, \sqrt{5}$
• $B \cap C = \emptyset$ (no number is both rational and irrational).
  Final Answer: $\emptyset$

(ii) $A \cup B'$ (where universal set is $S$, so $B' = C$)

Solution:

• Integers $A$: $0, -2, 2, -3$
• $B' = C = \{\sqrt{2}, \sqrt{3}, \pi, \sqrt{5}\}$
• $A \cup B' = \{0, -2, 2, -3, \sqrt{2}, \sqrt{3}, \pi, \sqrt{5}\}$
  Final Answer: $\{0, -2, 2, -3, \sqrt{2}, \sqrt{3}, \pi, \sqrt{5}\}$

(iii) $B \cup C'$ (where $C' = B$)

Solution:

• $C' = B$ (since $S = B \cup C$ and $B \cap C = \emptyset$)
• $B \cup C' = B \cup B = B = \{0, \frac{1}{3}, -2, 1.414, \frac{22}{7}, 0.75, 2, \frac{2}{5}, -3\}$
  

Final Answer: $\{0, \frac{1}{3}, -2, 1.414, \frac{22}{7}, 0.75, 2, \frac{2}{5}, -3\}$

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(b) Let $f(x) = \frac{5}{3x-1}$, find:

(i) $f(2)$
Solution:

$$f(2)=\frac{5}{3(2)-1}$$

$$=\frac{5}{6-1}$$

$$=\frac{5}{5}$$

$$f(2) = \frac{5}{3(2) - 1} = \frac{5}{6 - 1} = \frac{5}{5} = 1$$

Final Answer: $1$

(ii) The range of $f$
Solution:
Let $y = \frac{5}{3x - 1}$. Solve for $x$:

$$y(3x-1)=5$$

$$⟹3xy-y=5$$

$$y(3x - 1) = 5 \implies 3xy - y = 5 \implies x = \frac{y + 5}{3y}$$

Defined for all $y \neq 0$.
Final Answer: $\mathbb{R} \setminus \{0\}$

(iii) Solve $\log(x) = \frac{2}{x-1}$ where $g(x) = x^{-1}$ (assumed typo, intended $g(x) = \frac{2}{x-1}$)
Solution:
Assume equation is $g(x) = \frac{2}{x-1}$ with $g(x) = x^{-1} = \frac{1}{x}$:

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

Cross-multiply:

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

$$⟹-1=2x-x$$

$$(x - 1) = 2x \implies -1 = 2x - x \implies x = -1$$

Verify: $g(-1) = \frac{1}{-1} = -1$, 

$\frac{2}{-1-1}$

$=\frac{2}{-2}$

$\frac{2}{-1-1} = \frac{2}{-2} = -1$.
Final Answer: $x = -1$

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(c) Binary operation $*$: $a * b = a^2 + b^2 - 2$

(i) Commutative?

Solution:
$a * b = a^2 + b^2 - 2$, $b * a = b^2 + a^2 - 2 = a^2 + b^2 - 2$.
Yes, commutative.
Final Answer: Yes

(ii) Associative?

Solution:
Check $(a * b) * c \overset{?}{=} a * (b * c)$:

• $(a * b) * c = (a^2 + b^2 - 2)^2 + c^2 - 2$
• $a * (b * c) = a^2 + (b^2 + c^2 - 2)^2 - 2$
  Test $a=1, b=2, c=3$:
  $(1 * 2) * 3 = (1 + 4 - 2) * 3 = 3 * 3 = 9 + 9 - 2 = 16$
  $1 * (2 * 3) = 1 * (4 + 9 - 2) = 1 * 11 = 1 + 121 - 2 = 120 \neq 16$.
  Not associative.
  Final Answer: No

(iii) Evaluate $(3 * 2) * 5$

Solution:
$3 * 2 = 3^2 + 2^2 - 2 = 9 + 4 - 2 = 11$
$11 * 5 = 11^2 + 5^2 - 2 = 121 + 25 - 2 = 144$
Final Answer: $144$

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3. (a) Universal set $X = [-3, 10]$, subsets $A = (-2, 3]$, $B = [0, 5]$, $C = [-1, 7]$.

Write in interval form and display on number line.
(i) $C' \cap A$

Solution:

• $C' = X \setminus C = [-3, -1) \cup (7, 10]$
• $A = (-2, 3]$
• $C' \cap A = (-2, -1)$
  Number Line:

  -3    -2     -1      3      7    10
  [-----o------o)     ...    (o-----]
  Shaded: (-2, -1)

(ii) $(A \cap B) - C$

Solution:

• $A \cap B = (-2, 3] \cap [0, 5] = [0, 3]$
• $(A \cap B) - C = [0, 3] \setminus [-1, 7] = \emptyset$ (since $[0, 3] \subset [-1, 7]$)
  Final Answer: $\emptyset$

(iii) $A \cap (A \cup B)'$

Solution:

• $A \cup B = (-2, 3] \cup [0, 5] = (-2, 5]$
• $(A \cup B)' = X \setminus (-2, 5] = [-3, -2] \cup (5, 10]$
• $A \cap (A \cup B)' = (-2, 3] \cap ([-3, -2] \cup (5, 10]) = \emptyset$
  Final Answer: $\emptyset$

(iv) $X'$
Solution:
Complement of universal set is empty set.
Final Answer: $\emptyset$

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(b) Define:

(i) Into relation
A relation $R: A \to B$ is into if at least one element of $B$ is not mapped to by any element of $A$ (i.e., $\text{Range}(R) \subsetneqq B$).

(ii) Domain of a function
The set of all input values for which the function is defined.

(iii) Range of a function
The set of all output values produced by the function.

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(c) $f(x) = 3x + 4$

(i) Show $f$ is one-to-one
Solution:
Assume $f(a) = f(b)$:
$3a + 4 = 3b + 4 \implies 3a = 3b \implies a = b$.
Thus, injective.

(ii) Find $f^{-1}$ and its domain
Solution:
Let $y = 3x + 4$. Solve for $x$:
$y - 4 = 3x \implies x = \frac{y - 4}{3}$.
So $f^{-1}(x) = \frac{x - 4}{3}$.
Domain: Range of $f$ is $\mathbb{R}$, so domain of $f^{-1}$ is $\mathbb{R}$.

(iii) Solve $f^{-1}(x) = g(x)$ where $g(x) = 3x + 2$

Solution:

$$\frac{x - 4}{3} = 3x + 2$$

Multiply by 3:
$x-4=9x+6$

$⟹-4-6=9x-x$

$⟹-10=8x$

$⟹x=-\frac{10}{8}$

$x - 4 = 9x + 6 \implies -4 - 6 = 9x - x \implies -10 = 8x \implies x = -\frac{10}{8} = -\frac{5}{4}$.

Final Answer: $x = -\frac{5}{4}$

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4. (a) Find domain and range:

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

Solution:

• Domain: $x + 1 \neq 0 \implies x \neq -1$, so $(-\infty, -1) \cup (-1, \infty)$.
• Range: Let $y = \frac{1}{x+1}$. Solve for $x$: $x = \frac{1}{y} - 1$. Defined for $y \neq 0$.
  Final Answer: Domain: $\mathbb{R} \setminus \{-1\}$, Range: $\mathbb{R} \setminus \{0\}$.

(ii) $f(x) = \sqrt{1 - x^2}$

Solution:

• Domain: $1 - x^2 \geq 0 \implies x^2 \leq 1 \implies [-1, 1]$.
• Range: For $x \in [-1, 1]$, $f(x) \in [0, 1]$.
  Final Answer: Domain: $[-1, 1]$, Range: $[0, 1]$.

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

(i) Find $f^{-1}$

Solution:
Let $y = \frac{2x+1}{x-3}$. Solve for $x$:

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

$$⟹yx-3y=2x+1$$

$$⟹yx-2x=3y+1$$

$$⟹x(y-2)=3y+1$$

$$y(x - 3) = 2x + 1 \implies yx - 3y = 2x + 1 \implies yx - 2x = 3y + 1 \implies x(y - 2) = 3y + 1 \implies x = \frac{3y + 1}{y - 2}$$

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

(ii) State where $f^{-1}$ is undefined

Solution:
Denominator zero when $x - 2 = 0 \implies x = 2$.
Final Answer: $x = 2$

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(c) Determine if odd, even, or neither:

(i) $f(x) = x^3 + x$

Solution:
Check $f(-x) = (-x)^3 + (-x) = -x^3 - x = -(x^3 + x) = -f(x)$.
Odd.

(ii) $f(x) = x^9 - x$

Solution:
$f(-x) = (-x)^9 - (-x) = -x^9 + x = -(x^9 - x) = -f(x)$.
Odd.

(iii) $f(x) = -\sqrt{9 - x^2}$

Solution:
$f(-x) = -\sqrt{9 - (-x)^2} = -\sqrt{9 - x^2} = f(x)$.
Even.

@Dr. Microbiota

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END OF SOLUTIONS

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