INTRODUCTION TO MATHEMATICAL METHODS

FINAL EXAM JAN INTAKE 2023

DATE : 18th May, 2023

DURATION: 3 HOURS

INSTRUCTIONS:

1. Write your Computer Number and TG on each answer sheet used.
2. There are five questions; answer any four.
3. Full credit only if all working is shown.
4. Write details on answer booklet.
5. Use back for rough work.

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QUESTION ONE

(a) Define:
(i) Singleton set
A set with exactly one element.
(ii) Null set
The empty set, denoted $\varnothing$, with no elements.

(b) $A=[2,6]$, $B=(4,10)$, universal set $U=[0,12]$.
(i) $A^c\cup B$
Solution:
$A^c=U\setminus A=[0,2)\cup (6,12]$, $B=(4,10)$, so $A^c\cup B=[0,2)\cup (4,12]$.

Number Line:

To solve (i) $A^c\cup B$, we need to:

🔍 Step-by-step Breakdown

1. Complement of A within universal set $U=[0,12]$

Set $A=[2,6]$, so its complement $A^c$ is everything in $U$ except $A$:

$A^c=[0,2)\cup (6,12]$

2. Set B

$B=(4,10)$

3. Union: $A^c\cup B$

We now combine:

• $A^c=[0,2)\cup (6,12]$
• $B=(4,10)$

Let’s merge these intervals:

• $[0,2)$ stays as-is.
• $(4,10)\cup (6,12]=(4,12]$ since they overlap.
• So the full union is:

$A^c\cup B=[0,2)\cup (4,12]$

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✅ Final Answer:

$A^c\cup B=[0,2)\cup (4,12]$

Number Line:

  0     2      4      6      10     12
  [-----)     (-------------] 

(ii) $B-A$
Solution:
$B-A=(4,10)\setminus [2,6]=(6,10)$.   so if you see ∖ its the same as minus(-)

Let’s solve (ii) $B-A$ using the sets:

• $A=[2,6]$
• $B=(4,10)$

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🔍 Step-by-step Breakdown

1. Understand $B-A$

This means: all elements in $B$ that are not in $A$.

• $B=(4,10)$
• $A=[2,6]$

So we subtract the overlap:

• The overlap between $A$ and $B$ is $(4,6]$
• Removing this from $B$ gives:

$B-A=(6,10)$

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✅ Final Answer:

$B-A=(6,10)$

Number Line:

  4      6      10
  (------) 

(c) $U=\{a,b,c,d,e,f,g,h,i,j\}$, $A=\{b,d,f,h,j\}$, $B=\{a,c,d,e,g\}$, $C=\{g,h,j\}$.
(i) $(A\cap B)\cup C^c$
Solution:
$A\cap B=\left\{d\right\}$, $C^c=U\mathrm{ -\; C}=\{a,b,c,d,e,f,i\}$, so $(A\cap B)\cup C^c=\{a,b,c,d,e,f,i\}$.

(ii) $(B\cap A{)}^c\cap C$
Solution:
$B\cap A=\left\{d\right\}$, $(B\cap A{)}^c=U-\{d\}=\{a,b,c,e,f,g,h,i,j\}$, then $\cap C=\{g,h,j\}$.

(d) Survey of 100 students:

• Only periodicals: 18
• Web and books: 29
• Books, web, periodicals: 15
• Books and periodicals: 40
• Web and periodicals: 20
• Used books: 60
• Used none: 7
  (i) Venn Diagram
  Solution:
  Let $B=$ books, $W=$ web, $P=$ periodicals.
• $B\cap W\cap P=15$
• $B\cap P$ only: $40-15=25$
• $W\cap P$ only: $20-15=5$
• $B\cap W$ only: $29-15=14$
• Only $P$: 18
• Only $B$: $60-(25+15+14)=6$
• Only $W$: Let $W_{\text{only}}=x$. Total: $6+x+18+25+14+5+15+7=100⟹x+90=100⟹x=10$

Venn Diagram Values:

• $B_{\text{only}}=6$, $W_{\text{only}}=10$, $P_{\text{only}}=18$
• $B\cap W_{\text{only}}=14$, $B\cap P_{\text{only}}=25$, $W\cap P_{\text{only}}=5$
• $B\cap W\cap P=15$
• None = 7

(ii) Students using web
Solution:
$n\left(W\right)=10+14+5+15=44$.

(iii) Students using books or periodicals

Solution:
$n(B\cup P)=n\left(B\right)+n\left(P\right)-n(B\cap P)$

$=60+(18+25+5+15)-40$

$=60+63-40$

$=83$.

(e) $A=\{2,3,4,7\}$, $B=\{2,3,4\}$, $C=\{1,2,3\}$
(i) $(A\cap B)\times (B\cap C)$

Solution:
$A\cap B=\{2,3,4\}$, $B\cap C=\{2,3\}$, so:

$\{2,3,4\}\times \{2,3\}=\left\{\right(2,2),(2,3),(3,2),(3,3),(4,2),(4,3\left)\right\}$

(ii) $(A\times C)\cup (B\times C)$

Solution:
$A\times C=\left\{\right(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(4,1),(4,2),(4,3),(7,1),(7,2),(7,3\left)\right\}$
$B\times C=\left\{\right(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(4,1),(4,2),(4,3\left)\right\}$ (subset of $A\times C$)
Union is $A\times C$.

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QUESTION TWO

(a) Define:
(i) Binary operation
A function $\ast :S\times S\to S$ mapping pairs to elements of $S$.

(ii) Function
A relation assigning each element of domain exactly one element in codomain.

(b) Prove $\sqrt{3}$ irrational.

Solution:
Assume $\sqrt{3}=\frac{p}{q}$ (coprime integers). Then $p^2=3q^2$. So $p^2\equiv 0\left(mod3\right)$, thus $p\equiv 0\left(mod3\right)$. Let $p=3k$,

then $9k^2=3q^2$

$⟹q^2=3k^2$,

so $q\equiv 0\left(mod3\right)$. Contradiction (coprime).

(c) Simplify given $A\subset B$:
(i) $(A\cup B{)}^c\cup B^c$

Solution:
Since $A\subset B$, $A\cup B=B$, so $(A\cup B{)}^c=B^c$,

thus $B^c\cup B^c=B^c$.

(ii) $(A\cup B)\cap B^c$
Solution:
$A\cup B=B$, so $B\cap B^c=\varnothing$.

(d) Rationalize denominators:
(i) $\frac{\sqrt{2}-1}{\sqrt{3}}$

Solution:

$\frac{\sqrt{2}-1}{\sqrt{3}}X\frac{\sqrt{3}}{\sqrt{3}}$

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

$=\frac{\sqrt{6}-\sqrt{3}}{3}$

(ii) $\frac{1+\sqrt{2}}{1-3\sqrt{2}}$

Solution:

$\frac{1+\sqrt{2}}{1-3\sqrt{2}}X\frac{1+3\sqrt{2}}{1+3\sqrt{2}}$

$=\frac{(1+\sqrt{2})(1+3\sqrt{2})}{1-18}$

$=\frac{1X1+1X3\sqrt{2}+\sqrt{2}X1+\sqrt{2}X3\sqrt{2}}{-17}$

$=\frac{1+3\sqrt{2}+\sqrt{2}+6}{-17}$

$=\frac{7+4\sqrt{2}}{-17}$

(e) Express as $\frac{p}{q}$ (coprime):

(i) $12.3231$

Solution:

$12.3231=\frac{123231}{10000},\text{GCD}(123231,10000)=1.$

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

(ii) $-0.543$

Solution:

$-0.543=-\frac{543}{1000},\text{GCD}(543,1000)=1.$

Final Answer: $-\frac{543}{1000}$

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QUESTION THREE

(a) Define:
(i) Power set
Set of all subsets of a set.
(ii) Asymptote
A line that a curve approaches as it tends to infinity.

(b) Binary operation $a\ast b=ab+b-5$ on $Z$.
(i) Associative? Commutative?
Solution:

• Commutative? $a\ast b=ab+b-5$, $b\ast a=ba+a-5=ab+a-5$. Not equal (e.g., $a=1,b=0$: $1\ast 0=-5$, $0\ast 1=-4$). No.
• Associative? Check $(a\ast b)\ast c=a\ast (b\ast c)$:
  • $(a\ast b)\ast c=(ab+b-5)\ast c=(ab+b-5)c+c-5=abc+bc-5c+c-5=abc+bc-4c-5$
  • $a\ast (b\ast c)=a\ast (bc+c-5)=a(bc+c-5)+(bc+c-5)-5=abc+ac-5a+bc+c-10$
    Not equal (e.g., $a=1,b=1,c=1$: left = 1+1-4-5=-7, right=1+1-5+1+1-10=-11)). No.

(ii) $(4\ast 1)\ast 2$ and $4\ast (1\ast 2)$

Solution:

• $4\ast 1=4\cdot 1+1-5=0$, then $0\ast 2=0\cdot 2+2-5=-3$
• $1\ast 2=1\cdot 2+2-5=-1$, then $4\ast (-1)=4\cdot (-1)+(-1)-5=-10$
  Final Answer: $-3$ and $-10$

(c) Domain and range:
(i) $f\left(x\right)=\frac{7}{x-2}$

Solution:
Domain: $x\ne 2$ ($R\setminus \left\{2\right\}$), Range: $y\ne 0$ ($R\setminus \left\{0\right\}$).

(ii) $f\left(x\right)=3+\sqrt{5x-1}$
Solution:
Domain: $5x-1\ge 0⟹x\ge \frac{1}{5}$ ($[\frac{1}{5},\infty )$), Range: $[3,\infty )$.

(d) $f\left(x\right)=2x^2+6x+4$
(i) Complete the square

Solution:

$2(x^2+3x)+4=2(x^2+3x+{\left(\frac{3}{2}\right)}^2-\frac{9}{4})+4$

$\mathrm{}=2({(x+\frac{3}{2})}^2-\frac{9}{4})+4$

$=2{(x+\frac{3}{2})}^2-\frac{9}{2}+4$

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

(ii) Sketch graph

• Turning point: $(-1.5,-0.5)$
• $y$-intercept: $x=0$, $y=4$
• $x$-intercepts: $2x^2+6x+4=0⟹x^2+3x+2=0⟹(x+1)(x+2)=0$, so $(-1,0),(-2,0)$
• Line of symmetry: $x=-1.5$
  Graph Sketch:
• 

(e) Roots of $3x^2+2x-5=0$ are $\alpha ,\beta$. Find equations with roots:
(i) $\frac{\alpha }{\beta },\frac{\beta }{\alpha }$

Solution:
Sum: $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

$=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }$

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

$=\frac{\frac{4}{9}+\frac{10}{3}}{-\frac{5}{3}}$

$=\frac{\frac{34}{9}}{-\frac{5}{3}}$

$=-\frac{34}{15}$

Product: $\frac{\alpha }{\beta }\cdot \frac{\beta }{\alpha }=1$
Equation: $t^2-\left(\text{sum}\right)t+\text{product}=0$

$⟹t^2+\frac{34}{15}t+1=0$

$⟹15t^2+34t+15=0$.

(ii) $\alpha +2,\beta +2$

Solution:
Sum: $(\alpha +2)+(\beta +2)=\alpha +\beta +4$

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

$=\frac{10}{3}$

Product: $(\alpha +2)(\beta +2)=\alpha \beta +2(\alpha +\beta )+4$

$=-\frac{5}{3}+2(-\frac{2}{3})+4$

$=-\frac{5}{3}-\frac{4}{3}+4$

$=-3+4$

$=1$

Equation: $t^2-\frac{10}{3}t+1=0$

$⟹3t^2-10t+3=0$.

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QUESTION FOUR

(a) State:
(i) Factor theorem
If $f\left(c\right)=0$, then $(x-c)$ is a factor of polynomial $f\left(x\right)$.
(ii) Line symmetry of quadratic function
The axis of symmetry $x=-\frac{b}{2a}$ is the vertical line through the vertex.

(b) $f\left(x\right)=x^3+4x^2+x-6$
(i) Factorize
Solution:
Possible roots: $\pm 1,2,3,6$.
$f\left(1\right)=1+4+1-6=0⟹(x-1)$ factor.
Synthetic division:

  1 | 1  4  1  -6
       |    1  5   6
       -----------
         1  5  6   0

So $f\left(x\right)=(x-1)(x^2+5x+6)=(x-1)(x+2)(x+3)$.

(ii) Roots
Solution:
$f\left(x\right)=0⟹x=1,-2,-3$.

(iii) Sketch graph

• Roots: $(-3,0),(-2,0),(1,0)$
• $y$-intercept: $(0,-6)$
• Behavior: cubic, positive leading coefficient.
  Graph Sketch:

(iv) Solve $x^3+4x^2+x-6\le 0$

Solution:
From factors, sign changes at $x=-3,-2,1$. Test intervals:

• : $(-)(-)(-)=-\le 0$
• : $(-)(+)(-)=+\le ̸0$
• : $(-)(+)(+)=-\le 0$
• $x>1$: $(+)(+)(+)=+\le ̸0$
  Solution: $(-\infty ,-3]\cup [-2,1]$.

(c) $f\left(x\right)=\frac{x-4}{2x-1}$, $g\left(x\right)=3x-1$
(i) $f^{-1}\left(x\right)$

Solution:
Let $y=\frac{x-4}{2x-1}$. Solve:
$y(2x-1)=x-4$

$⟹2xy-y=x-4$

$⟹2xy-x=y-4$

$⟹x(2y-1)=y-4$

$⟹x=\frac{y-4}{2y-1}$
So $f^{-1}\left(x\right)=\frac{x-4}{2x-1}$.

(ii) $(f\circ g{)}^{-1}(x)$

Solution:
First, $f\circ g=f\left(g\right(x\left)\right)=f(3x-1)=\frac{(3x-1)-4}{2(3x-1)-1}=\frac{3x-5}{6x-3}$.
Now find inverse: let $y=\frac{3x-5}{6x-3}$. Solve:
$y(6x-3)=3x-5$

$⟹6xy-3y=3x-5$

$⟹6xy-3x=3y-5$

$⟹x(6y-3)=3y-5$

$⟹x=\frac{3y-5}{6y-3}$.

So $(f\circ g{)}^{-1}(x)=\frac{3x-5}{6x-3}$

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

(d) Sketch $f\left(x\right)=\mid 2x^2+5x+2\mid$

Solution:
Factor: $2x^2+5x+2=(2x+1)(x+2)$.

• Roots: $x=-\frac{1}{2},-2$
• Vertex: $x=-\frac{b}{2a}=-\frac{5}{4}$, $y=\mid 2\left(\frac{25}{16}\right)-\frac{25}{4}+2\mid =\mid -\frac{9}{8}\mid =\frac{9}{8}$
• $y$-intercept: $\mid 2\mid =2$
• Absolute value, so above $x$-axis.
  Graph Sketch:

(e) Prove $\frac{\sin A}{\sin A}+\frac{1+\cos A}{\sin A}=2\csc A$

Solution:
Left: $1+\frac{1+\cos A}{\sin A}=\frac{\sin A}{\sin A}+\frac{1+\cos A}{\sin A}=\frac{\sin A+1+\cos A}{\sin A}$.

Right: $2\csc A=\frac{2}{\sin A}$.
Not equal. Likely typo; intended:

$\frac{\sin A}{1-\cos A}+\frac{1+\cos A}{\sin A}=\frac{{\sin }^{2}A+(1-{\cos }^{2}A)}{(1-\cos A)\sin A}=\frac{{\sin }^{2}A+{\sin }^{2}A}{(1-\cos A)\sin A}=\frac{2{\sin }^{2}A}{\sin A(1-\cos A)}=\frac{2\sin A}{1-\cos A}$

Not matching. Assuming standard identity:

$\frac{\sin A}{1-\cos A}=\csc A+\cot A,\text{but not directly related.}$

Given identity is incorrect as stated.

(f) Differentiate $y=\ln \frac{2x-3}{3x+2}$

Solution:

$y=\ln (2x-3)-\ln (3x+2)$ $\frac{dy}{dx}=\frac{2}{2x-3}-\frac{3}{3x+2}$

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

$=\frac{6x+4-6x+9}{(2x-3)(3x+2)}$

$=\frac{13}{(2x-3)(3x+2)}$

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QUESTION FIVE

(a) Define:
(i) Radical function
Function involving roots, e.g., $f\left(x\right)=\sqrt{x}$.
(ii) Natural exponential function
$e^x$, where $e\approx 2.71828$.

(b) Sketch graphs:
(i) $f\left(x\right)=2+\sqrt{x+3}$

• Domain: $x\ge -3$
• $x$-intercept: $2+\sqrt{x+3}=0⟹\sqrt{x+3}=-2$ (none)
• $y$-intercept: $x=0$, $y=2+\sqrt{3}$
• Starts at $(-3,2)$, increases.
  Graph Sketch:

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

• Vertical asymptote: $x=\frac{2}{3}$
• Horizontal asymptote: $y=\frac{2}{3}$
• $x$-intercept: $2x-3=0⟹x=1.5$
• $y$-intercept: $x=0$, $y=\frac{-3}{-2}=1.5$
  Graph Sketch:

(c) Solve $2\tan x-\sec x=0$ on $[0^{\circ },360^{\circ }]$

Solution:

$2\frac{\sin x}{\cos x}-\frac{1}{\cos x}=0$

$⟹\frac{2\sin x-1}{\cos x}=0$

$⟹2\sin x-1=0$

$⟹\sin x=\frac{1}{2}$

Solutions: $x=30^{\circ },150^{\circ }$.

(d) Polynomial $2x^3-a{x}^2+bx+3$:
Remainder $-15$ when divided by $(x+1)$, remainder $-46$ when divided by $(x-3)$.
Solution:
By Remainder Theorem:

• $f(-1)=2(-1{)}^3-a(-1{)}^2+b(-1)+3=-2-a-b+3=-a-b+1=-15$
• $f\left(3\right)=2\left(27\right)-a\left(9\right)+b\left(3\right)+3=54-9a+3b+3=-9a+3b+57=-46$
  

Equations:

$-a-b=-16\text{(1)}$ $-9a+3b=-103\text{(2)}$

Multiply (1) by 3: $-3a-3b=-48$
Add to (2): $-12a=-151⟹a=\frac{151}{12}$?

Check:
From (1): $a+b=16$
From (2): $9a-3b=103$ (multiplied by -1)

Multiply (1) by 3: $3a+3b=48$
Add to (2): $12a=151⟹a=\frac{151}{12}$
Then $b=16-\frac{151}{12}=\frac{192-151}{12}=\frac{41}{12}$.

Final Answer: $a=\frac{151}{12},b=\frac{41}{12}$

(e) Partial fractions:
(i) $\frac{x-7}{(x+1)(x^2+1)}$

Solution:
Assume:

$\frac{x-7}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$

Multiply:
$x-7=A(x^2+1)+(Bx+C)(x+1)$
Set $x=-1$: $-1-7=A(1+1)$

$⟹-8=2A⟹A=-4$

Expand:

$x-7=-4(x^2+1)+(Bx+C)(x+1)=-4x^2-4+B{x}^2+Bx+Cx+C$

$=(B-4)x^2+(B+C)x+(C-4)$

Compare coefficients:

• $x^2$: $B-4=0⟹B=4$
• $x$: $B+C=1$
• $⟹4+C=1$
• $⟹C=-3$
• Constant: $C-4=-3-4=-7$ (matches)
  So:

$\frac{-4}{x+1}+\frac{4x-3}{x^2+1}$

(ii) $\frac{12}{(x-3)(x-4)}$

Solution:

$\frac{12}{(x-3)(x-4)}=\frac{A}{x-3}+\frac{B}{x-4}$

$12=A(x-4)+B(x-3)$
Set $x=3$: $12=A(-1)$

$⟹A=-12$

Set $x=4$: $12=B\left(1\right)$

$⟹B=12$

So:

$\frac{-12}{x-3}+\frac{12}{x-4}$

(f) Solve:

(i) ${\log }_2(x^2-x+2)=1+2{\log }_{2}x$

Solution:
RHS: $1+{\log }_2{x}^2={\log }_{2}2+{\log }_2{x}^2={\log }_2\left(2x^2\right)$
So:
$x^2-x+2=2x^2$

$⟹x^2+x-2=0$

$⟹(x+2)(x-1)=0$
Solutions: $x=1,x=-2$. But domain: $x>0$ and $x^2-x+2>0$ (always true). So $x=1$ (since $x>0$).

(ii) $3^{2x+1}+26\cdot 3^x-9=0$

Solution:
Let $u=3^x$:
$3\cdot (3^x{)}^2+26\cdot 3^x-9=0$

$⟹3u^2+26u-9=0$
Discriminant: $676+108=784=28^2$
$u=\frac{-26\pm 28}{6}$

• $u=\frac{2}{6}=\frac{1}{3}$
• $⟹3^x=\frac{1}{3}$
• $⟹x=-1$
• $u=\frac{-54}{6}=-9$ (invalid, since $3^x>0$)
  Final Answer: $x=-1$

@Dr. Microbiota

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END OF SOLUTIONS

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