INTRODUCTION TO MATHEMATICAL METHODS

TEST ONE JULY INTAKE 2023

DATE : 24th August, 2023

INSTRUCTIONS

1. Duration: 2 Hours.
2. Write your Computer Number on each answer paper used. Answer all questions and show all the necessary work to earn full marks.

QUESTION ONE

(a) Define:
(i) Singleton set
A set with exactly one element.
(ii) Power set
The set of all subsets of a given set.

(b) Write set $A=\{x:x^2-4=0\}$ in Roster form.
Solution:
$x^2-4=0⟹x=\pm 2$.
Final Answer: $A=\{-2,2\}$

(c) Rationalize denominator $\frac{\sqrt{7}+1}{1-3\sqrt{5}}$

Solution:

$\frac{\sqrt{7}+1}{1-3\sqrt{5}}X\frac{1+3\sqrt{5}}{1+3\sqrt{5}}$

$=\frac{(\sqrt{7}+1)(1+3\sqrt{5})}{(1{)}^2-(3\sqrt{5}{)}^2}$

$=\frac{\sqrt{7}+3\sqrt{35}+1+3\sqrt{5}}{1-45}$

$=\frac{1+\sqrt{7}+3\sqrt{5}+3\sqrt{35}}{-44}$

Final Answer: $-\frac{1+\sqrt{7}+3\sqrt{5}+3\sqrt{35}}{44}$

(d) Simplify $\left[\right(A^c\cup B{)}^c\cap (A\cap B^c){]}^c$

Solution:

$(A^c\cup B{)}^c=A\cap B^c\text{(De Morgan)}$

$\left[\right(A\cap B^c)\cap (A\cap B^c){]}^c$

$=(A\cap B^c{)}^c$

$=A^c\cup B\text{(De Morgan)}$

Final Answer: $A^c\cup B$

(e) $E=(-\infty ,+\infty )$, $A=[-15,10]$, $C=[5,30)$. Find $A-C$

Solution:
$A-C=\{x\in A\mid x\notin C\}=[-15,10]\setminus [5,30)=[-15,5)$.
Final Answer: $[-15,5)$

To find the set difference $A-C$, we subtract the elements of set $C=[5,30)$ from set $A=[-15,10]$.

🔍 Step-by-step:

• Set $A$ includes all real numbers from $-15$ to $10$, inclusive.
• Set $C$ includes all real numbers from $5$ to just below $30$, not including 30.
• The overlap between $A$ and $C$ is from $5$ to $10$, since both sets include that interval.

📘 So:

$A-C=[-15,5)$

This is the portion of $A$ that does not intersect with $C$.

✅ Final Answer:

$A-C=[-15,5)$

$$

(f) Binary operation $a\ast b=a^2+b^2$ on rationals. Find $(1\ast 3)\ast 2$ and $1\ast (3\ast 2)$. Is it associative?

Solution:

• $1\ast 3=1^2+3^2=1+9=10$, then $10\ast 2=100+4=104$.
• $3\ast 2=9+4=13$, then $1\ast 13=1+169=170$.
• $104\ne 170$, so not associative.
  Final Answer: $(1\ast 3)\ast 2=104$, $1\ast (3\ast 2)=170$, not associative.

QUESTION TWO

(a) Define:
(i) Identity relation
A relation where every element is related only to itself: $\left\{\right(a,a)\mid a\in A\}$.
(ii) Domain
The set of all first elements in the ordered pairs of a relation.

(b) Relation $R$ on $A=\{1,2,3,4\}$ given by $R=\left\{\right(x,y):x\text{ is divisible by }y\}$.
Solution:

• $R=\left\{\right(1,1),(2,1),(2,2),(3,1),(3,3),(4,1),(4,2),(4,4\left)\right\}$
• Identity? No (not only self-pairs).
• Reflexive? Yes (all self-pairs present).
• Symmetric? No (e.g., $(2,1)\in R$ but $(1,2)\notin R$).
• Transitive? Yes (e.g., $(4,2),(2,1)⟹(4,1)\in R$).
• Equivalence? No (not symmetric).

(c) Show $f\left(x\right)=\frac{2x+3}{x-3}$ is one-to-one and find inverse.

Solution:

• One-to-one: Assume $f\left(a\right)=f\left(b\right)$:

$\frac{2a+3}{a-3}=\frac{2b+3}{b-3}$

$⟹(2a+3)(b-3)$

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

Expand: $2ab-6a+3b-9=2ab-6b+3a-9$

$⟹-6a+3b=-6b+3a$

$⟹-9a=-9b$

$⟹a=b$.

• Inverse: Let $y=\frac{2x+3}{x-3}$. Solve:

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

$⟹yx-3y=2x+3$

$⟹yx-2x=3y+3$

$⟹x(y-2)=3(y+1)$

$⟹x=\frac{3(y+1)}{y-2}$

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

(d) $A=\{1,2,3,4\}$, $f=\left\{\right(1,4),(2,1),(3,3),(4,2\left)\right\}$, $g=\left\{\right(1,3),(2,1),(3,2),(4,4\left)\right\}$. Find $g\circ f$

Solution:
$g\circ f$: Apply $f$ then $g$.

• $f\left(1\right)=4$, $g\left(4\right)=4⟹(1,4)$
• $f\left(2\right)=1$, $g\left(1\right)=3⟹(2,3)$
• $f\left(3\right)=3$, $g\left(3\right)=2⟹(3,2)$
• $f\left(4\right)=2$, $g\left(2\right)=1⟹(4,1)$
  Final Answer: $\left\{\right(1,4),(2,3),(3,2),(4,1\left)\right\}$

(e) Domain and range of $f\left(x\right)=4+\sqrt{3x-1}$

Solution:

• Domain: $3x-1\ge 0⟹x\ge \frac{1}{3}$, so $[\frac{1}{3},\infty )$.
• Range: Minimum at $x=\frac{1}{3}$: $f\left(\frac{1}{3}\right)=4+0=4$, increases to $\infty$, so $[4,\infty )$.
  Final Answer: Domain: $[\frac{1}{3},\infty )$, Range: $[4,\infty )$.

@Dr. Microbiota

END OF SOLUTIONS