INTRODUCTION TO MATHEMATICAL METHODS

SEMESTER ONE EXAMINATION JAN INTAKE 2022 

DATE : 30th MAY, 2022

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Question 1

QUESTION ONE

(a)
(i) Express $2x^2 + 7x + 5 = 0$ in the form $(x + a)^2 = k$.
Solution:

$$\begin{align*} 2x^2 + 7x + 5 &= 0 \\ x^2 + \frac{7}{2}x + \frac{5}{2} &= 0 \quad \text{(Divide by 2)} \\ x^2 + \frac{7}{2}x &= -\frac{5}{2} \quad \text{(Isolate $x$ terms)} \\ x^2 + \frac{7}{2}x + \left(\frac{7}{4}\right)^2 &= -\frac{5}{2} + \left(\frac{7}{4}\right)^2 \quad \text{(Complete the square)} \\ \left(x + \frac{7}{4}\right)^2 &= -\frac{5}{2} + \frac{49}{16} \\ \left(x + \frac{7}{4}\right)^2 &= -\frac{40}{16} + \frac{49}{16} \\ \left(x + \frac{7}{4}\right)^2 &= \frac{9}{16} \end{align*}$$

Answer: $\left(x + \frac{7}{4}\right)^2 = \frac{9}{16}$

(ii) Hence state the value of $a$ and $k$.
Solution:
From part (i), $a = \frac{7}{4}$, $k = \frac{9}{16}$.
Answer: $a = \frac{7}{4}$, $k = \frac{9}{16}$

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(b) Rationalize the denominator:
(i) $\frac{5}{2 + \sqrt{5}}$
Solution:
Multiply numerator and denominator by conjugate $2 - \sqrt{5}$:

$$\begin{array}{rl}{\displaystyle \frac{5}{2+\sqrt{5}}X\frac{2-\sqrt{5}}{2-\sqrt{5}}}& {\displaystyle =\frac{5(2-\sqrt{5})}{(2{)}^2-(\sqrt{5}{)}^2}}\\ {\displaystyle }& {\displaystyle }\end{array}$$$$\begin{array}{r}{\displaystyle =\frac{10-5\sqrt{5}}{4-5}}\\ {\displaystyle }& {\displaystyle }\end{array}$$$$\begin{array}{r}{\displaystyle =\frac{10-5\sqrt{5}}{-1}}\\ {\displaystyle }& {\displaystyle }\end{array}$$

$$\begin{align*} \frac{5}{2 + \sqrt{5}} \cdot \frac{2 - \sqrt{5}}{2 - \sqrt{5}} &= \frac{5(2 - \sqrt{5})}{(2)^2 - (\sqrt{5})^2} \\ &= \frac{10 - 5\sqrt{5}}{4 - 5} \\ &= \frac{10 - 5\sqrt{5}}{-1} \\ &= -10 + 5\sqrt{5} \end{align*}$$

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

(ii) $\frac{3 - \sqrt{5}}{1 + 3\sqrt{5}}$
Solution:
Multiply numerator and denominator by conjugate $1 - 3\sqrt{5}$:

$$\begin{array}{rl}{\displaystyle \frac{3-\sqrt{5}}{1+3\sqrt{5}}\cdot \frac{1-3\sqrt{5}}{1-3\sqrt{5}}}& {\displaystyle =\frac{(3-\sqrt{5})(1-3\sqrt{5})}{(1{)}^2-(3\sqrt{5}{)}^2}}\\ {\displaystyle }\end{array}$$

$$\begin{array}{rl}{\displaystyle \text{Numerator: }}& {\displaystyle 3X1=3}\\ {\displaystyle }& {\displaystyle \mathrm{}}\end{array}$$

$$\begin{array}{r}{\displaystyle 3X(-3\sqrt{5})=-9\sqrt{5}}\\ {\displaystyle }& {\displaystyle }\end{array}$$

$$\begin{array}{r}{\displaystyle (-\sqrt{5})X1=-\sqrt{5}}\\ {\displaystyle }& {\displaystyle }\end{array}$$

$$\begin{array}{r}{\displaystyle (-\sqrt{5})X(-3\sqrt{5})=15}\\ \end{array}$$

$$\begin{array}{rl}{\displaystyle }& {\displaystyle \text{Total: }3-9\sqrt{5}-\sqrt{5}+15=18-10\sqrt{5}}\\ {\displaystyle }\end{array}$$

$$\begin{array}{rl}{\displaystyle \text{Denominator: }}& {\displaystyle 1-9X5=1-45=-44}\end{array}$$

$$\begin{array}{r}{\displaystyle \mathrm{}}\\ {\displaystyle }& {\displaystyle \frac{18-10\sqrt{5}}{-44}}\end{array}$$

$$\begin{array}{r}{\displaystyle =\frac{-18+10\sqrt{5}}{44}}\end{array}$$

$$\begin{align*} \frac{3 - \sqrt{5}}{1 + 3\sqrt{5}} \cdot \frac{1 - 3\sqrt{5}}{1 - 3\sqrt{5}} &= \frac{(3 - \sqrt{5})(1 - 3\sqrt{5})}{(1)^2 - (3\sqrt{5})^2} \\ \text{Numerator: } & 3 \cdot 1 = 3 \\ & 3 \cdot (-3\sqrt{5}) = -9\sqrt{5} \\ & (-\sqrt{5}) \cdot 1 = -\sqrt{5} \\ & (-\sqrt{5}) \cdot (-3\sqrt{5}) = 15 \\ & \text{Total: } 3 - 9\sqrt{5} - \sqrt{5} + 15 = 18 - 10\sqrt{5} \\ \text{Denominator: } & 1 - 9 \cdot 5 = 1 - 45 = -44 \\ & \frac{18 - 10\sqrt{5}}{-44} = \frac{-18 + 10\sqrt{5}}{44} = \frac{-9 + 5\sqrt{5}}{22} \end{align*}$$

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

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(c) Express in form $\frac{a}{b}$ ($a, b$ integers, $b \neq 0$):
(i) $1.666\ldots$
Solution:
Let $x = 1.666\ldots$

$$\begin{array}{r}{\displaystyle 10x}{\displaystyle =16.666\dots }\\ {\displaystyle \mathrm{}}\end{array}$$

$$\begin{array}{r}{\displaystyle 10x-x}{\displaystyle =16.666\dots -1.666\dots }\\ {\displaystyle \mathrm{}}\end{array}$$

$$\begin{array}{r}{\displaystyle 9x}{\displaystyle =15}\\ {\displaystyle }\end{array}$$

$$\begin{align*} 10x &= 16.666\ldots \\ 10x - x &= 16.666\ldots - 1.666\ldots \\ 9x &= 15 \\ x &= \frac{15}{9} = \frac{5}{3} \end{align*}$$

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

(ii) $0.87$
Solution:

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

Answer: $\frac{87}{100}$

(iii) $-5.463$
Solution:

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

Answer: $-\frac{5463}{1000}$

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(d) Differentiate:
(i) $f(x) = (2x - x^4)(x^3 - x - 12)$
Solution:
Use product rule: $\frac{d}{dx}[uv] = u'v + uv'$
Let $u = 2x - x^4$, $v = x^3 - x - 12$

$$u' = 2 - 4x^3, \quad v' = 3x^2 - 1$$ $$f'(x) = (2 - 4x^3)(x^3 - x - 12) + (2x - x^4)(3x^2 - 1)$$

Expand:

$$\begin{array}{rl}{\displaystyle }& {\displaystyle (2-4x^3)(x^3-x-12)=2x^3-2x-24-4x^6+4x^4+48x^3}\end{array}$$

$$\begin{array}{r}{\displaystyle =-4x^6+4x^4+50x^3-2x-24}\\ {\displaystyle }& {\displaystyle }\end{array}$$

$$\begin{array}{r}{\displaystyle (2x-x^4)(3x^2-1)}\end{array}$$

$$\begin{array}{r}{\displaystyle =6x^3-2x-3x^6+x^4}\end{array}$$

$$\begin{array}{r}{\displaystyle =-3x^6+x^4+6x^3-2x}\\ {\displaystyle }\end{array}$$

$$\begin{array}{rl}{\displaystyle \text{Sum: }}& {\displaystyle (-4x^6+4x^4+50x^3-2x-24)+(-3x^6+x^4+6x^3-2x)}\end{array}$$

$$\begin{align*} &(2 - 4x^3)(x^3 - x - 12) = 2x^3 - 2x - 24 - 4x^6 + 4x^4 + 48x^3 = -4x^6 + 4x^4 + 50x^3 - 2x - 24 \\ &(2x - x^4)(3x^2 - 1) = 6x^3 - 2x - 3x^6 + x^4 = -3x^6 + x^4 + 6x^3 - 2x \\ \text{Sum: } & (-4x^6 + 4x^4 + 50x^3 - 2x - 24) + (-3x^6 + x^4 + 6x^3 - 2x) = -7x^6 + 5x^4 + 56x^3 - 4x - 24 \end{align*}$$

Answer: $f'(x) = -7x^6 + 5x^4 + 56x^3 - 4x - 24$

(ii) $y = \frac{3x - 1}{x^2 + 5x}$
Solution:
Use quotient rule: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}$
Let $u = 3x - 1$, $v = x^2 + 5x$

$$u' = 3, \quad v' = 2x + 5$$ $${}^{}$$

$$y' = \frac{3(x^2 + 5x) - (3x - 1)(2x + 5)}{(x^2 + 5x)^2}$$

Numerator:

$$\begin{array}{rl}{\displaystyle }& {\displaystyle 3x^2+15x-[(3x-1)(2x+5)]}\\ {\displaystyle }& {\displaystyle }\end{array}$$

$$\begin{array}{r}{\displaystyle (3x-1)(2x+5)=6x^2+15x-2x-5}\end{array}$$

$$\begin{array}{r}{\displaystyle =6x^2+13x-5}\\ {\displaystyle }& {\displaystyle \mathrm{}}\end{array}$$

$$\begin{array}{r}{\displaystyle \mathrm{THEREFORE \; 3}{x}^2+15x-(6x^2+13x-5)}\end{array}$$

$$\begin{array}{r}{\displaystyle =3x^2+15x-6x^2-13x+5}\end{array}$$

$$\begin{align*} &3x^2 + 15x - [(3x-1)(2x+5)] \\ &(3x-1)(2x+5) = 6x^2 + 15x - 2x - 5 = 6x^2 + 13x - 5 \\ &3x^2 + 15x - (6x^2 + 13x - 5) = 3x^2 + 15x - 6x^2 - 13x + 5 = -3x^2 + 2x + 5 \end{align*}$$

Answer: $y' = \frac{-3x^2 + 2x + 5}{(x^2 + 5x)^2}$

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(e) Solve simultaneously:

$$5^x \cdot 25^{2y} = 5, \quad 3^{2x} \cdot 9^{y-1} = \frac{1}{27}$$

Solution:
Rewrite using base 5 and 3:

$$25^{2y}=(5^2{)}^{2y}=5^{4y},{\mathrm{}}^{}$$

$$25^{2y} = (5^2)^{2y} = 5^{4y}, \quad 9^{y-1} = (3^2)^{y-1} = 3^{2(y-1)}$$

Equations become:

$$5^{x}X{5}^{4y}=5^1$$

$$⟹5^{x+4y}=5^1$$

$$⟹x+4y=1\text{EQUESTION(1)}$$

$$5^x \cdot 5^{4y} = 5^1 \implies 5^{x+4y} = 5^1 \implies x + 4y = 1 \quad \text{(1)}$$ $$3^{2x}X{3}^{2(y-1)}=3^{-3}(\text{since }\frac{1}{27}=3^{-3})$$

$$⟹3^{2x+2y-2}=3^{-3}$$

$$3^{2x} \cdot 3^{2(y-1)} = 3^{-3} \quad (\text{since } \frac{1}{27} = 3^{-3}) \implies 3^{2x + 2y - 2} = 3^{-3} \implies 2x + 2y - 2 = -3$$ $$\mathrm{}$$

$$2x + 2y = -1 \quad \text{(2)}$$

Solve system:
From (1): $x = 1 - 4y$
Substitute into (2):

$$2(1-4y)+2y=-1$$

$$⟹2-8y+2y=-1$$

$$⟹2-6y=-1$$

$$⟹-6y=-3$$

$$2(1 - 4y) + 2y = -1 \implies 2 - 8y + 2y = -1 \implies 2 - 6y = -1 \implies -6y = -3 \implies y = \frac{1}{2}$$ $$$$

$$x = 1 - 4(\frac{1}{2}) = 1 - 2 = -1$$

Answer: $x = -1$, $y = \frac{1}{2}$

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(f) Find range of $f(x) = 4x - 15$, domain $D_f = \{x : 15 < x < 250, x \in \mathbb{Z}\}$.
Solution:
Domain is integers from 16 to 249 inclusive.
$f$ is linear and increasing.
Min: $f(16) = 4(16) - 15 = 64 - 15 = 49$
Max: $f(249) = 4(249) - 15 = 996 - 15 = 981$
Range is all integers from 49 to 981 inclusive.
Answer: Range = $\{y \in \mathbb{Z} : 49 \leq y \leq 981\}$

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QUESTION TWO

(a) Given set $S = \{0.3, -2, 1.414, \frac{22}{7}, \sqrt{2}, \sqrt{3}, 0.75, \pi, \sqrt{5}, 2, \frac{2}{5}, -3\}$:

• $A = \{\text{integers}\} = \{-3, -2, 2\}$
• $B = \{\text{rationals}\} = \{0.3 = \frac{3}{10}, -2, \frac{22}{7}, 0.75 = \frac{3}{4}, 2, \frac{2}{5}\}$
• $C = \{\text{irrationals}\} = \{1.414 \approx \sqrt{2}? \text{ but } \sqrt{2} \text{ is listed separately}, \sqrt{2}, \sqrt{3}, \pi, \sqrt{5}\}$
  Note: $1.414$ is rational (terminating decimal), but $\sqrt{2} \approx 1.414$ is irrational. Since $\sqrt{2}$ is separately listed, $1.414$ is likely rational.

(i) $B \cap C$
Rational and irrational sets are disjoint, so $B \cap C = \emptyset$.
Answer: $\emptyset$

(ii) $A \cup B'$
$B' = S \setminus B = \{\sqrt{2}, \sqrt{3}, \pi, \sqrt{5}\}$ (since $1.414$ is rational)
$A \cup B' = \{-3, -2, 2\} \cup \{\sqrt{2}, \sqrt{3}, \pi, \sqrt{5}\} = \{-3, -2, 2, \sqrt{2}, \sqrt{3}, \pi, \sqrt{5}\}$
Answer: $\{-3, -2, 2, \sqrt{2}, \sqrt{3}, \pi, \sqrt{5}\}$

(iii) $B \cup C'$
$C' = S \setminus C = \{0.3, -2, \frac{22}{7}, 0.75, 2, \frac{2}{5}\}$ (rationals)
$B \cup C' = B$ (since $C' \subseteq B$)
Answer: $B = \{0.3, -2, \frac{22}{7}, 0.75, 2, \frac{2}{5}\}$

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

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

(ii) Inverse: Let $y = \frac{5}{3x-1}$. Solve for $x$:

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

$$⟹3xy-y=5$$

$$⟹3xy=y+5$$

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

So $f^{-1}(x) = \frac{x + 5}{3x}$
Answer: $f^{-1}(x) = \frac{x+5}{3x}$

(iii) Solve $\log g(x) = \frac{2}{x-1}$ where $g(x) = x-1$.
Equation is $\log(x-1) = \frac{2}{x-1}$. Let $u = x-1$:

$$\log u = \frac{2}{u} \implies u \log u = 2$$

By inspection, $u = 4$: $4 \log 4 = 4 \cdot 0.602 \approx 2.408 \neq 2$
$u = 2$: $2 \log 2 \approx 2 \cdot 0.301 = 0.602 \neq 2$
Trial: $u = 10$: $10 \log 10 \approx 10 \cdot 1 = 10 > 2$
$u = 5$: $5 \log 5 \approx 5 \cdot 0.699 = 3.495 > 2$
$u = 3$: $3 \log 3 \approx 3 \cdot 0.477 = 1.431 < 2$
No integer solution. Use numerical method or note that $u \log u = 2$ has solution $u \approx 4.32$ (but exact form not required).
Note: The problem likely expects an exact solution. Let $v = \log u$, then $u = 10^v$, so $10^v \cdot v = 2$. This is transcendental; no algebraic solution.
Clarify: The equation is written as "$\left( \log \right)(x) = \frac{2}{x-1}$", which is ambiguous. Assuming $\log$ is base 10 and applied to $g(x)$, so $\log_{10}(x-1) = \frac{2}{x-1}$.
Answer: No algebraic solution.

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

(i) Commutative? Check $a * b = b * a$:
$a * b = a^2 + b^2 - 2$, $b * a = b^2 + a^2 - 2 = a^2 + b^2 - 2$. Yes.
Answer: Yes

(ii) Associative? Check $(a * b) * c = a * (b * c)$:
Left: $(a * b) * c = (a^2 + b^2 - 2) * c = (a^2 + b^2 - 2)^2 + c^2 - 2$
Right: $a * (b * c) = a * (b^2 + c^2 - 2) = a^2 + (b^2 + c^2 - 2)^2 - 2$
Not equal in general (e.g., $a=1,b=1,c=1$:

left = $(1+1-2)^2 +1-2 = 0 +1-2=-1$,

right = $1 + (1+1-2)^2 -2=1+0-2=-1$ — equal,

but try $a=1,b=2,c=3$:
Left: $(1*2)*3 = (1+4-2)*3 = 3*3 = 9 + 9 - 2 = 16$
Right: $1*(2*3) = 1*(4+9-2)=1*11=1+121-2=120 \neq 16$. No.
Answer: No

(iii) $(3 * 2) * 5$:
$3 * 2 = 3^2 + 2^2 - 2 = 9 + 4 - 2 = 11$
$11 * 5 = 11^2 + 5^2 - 2 = 121 + 25 - 2 = 144$
Answer: $144$

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(d) Height $f(t) = 15t - 4t^2$ metres.

(i) Greatest height: Vertex at $t = -\frac{b}{2a} = -\frac{15}{2(-4)} = \frac{15}{8} = 1.875$ s

Height: $f(1.875)=15(1.875)-4(1.875{)}^2$

$=28.125-4(3.515625)$

$=28.125-14.0625$

$f(1.875) = 15(1.875) - 4(1.875)^2 = 28.125 - 4(3.515625) = 28.125 - 14.0625 = 14.0625$ m
Answer: $14.0625$ m

(ii) Time: $t = 1.875$ s
Answer: $1.875$ seconds

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(e) Solve:
(i) $\log_2 (x^2 + 5x - 2) = \log_2 (x^2 + 3x - 6) + \log_4 9$
Note $\log_4 9 = \frac{\log_2 9}{\log_2 4} = \frac{2 \log_2 3}{2} = \log_2 3$
Equation:

$${\log }_2(x^2+5x-2)={\log }_2(x^2+3x-6)+{\log }_{2}3$$

$$\log_2 (x^2 + 5x - 2) = \log_2 (x^2 + 3x - 6) + \log_2 3 = \log_2 [3(x^2 + 3x - 6)]$$

So:

$$x^2+5x-2=3(x^2+3x-6)=3x^2+9x-18$$

$$x^2 + 5x - 2 = 3(x^2 + 3x - 6) = 3x^2 + 9x - 18$$ $$0=2x^2+4x-16$$

$$0 = 2x^2 + 4x - 16 \implies x^2 + 2x - 8 = 0$$

Factors: $(x+4)(x-2)=0$, so $x=-4$ or $x=2$
Check domain: Arguments of log must be positive.
For $x=2$: $x^2+5x-2=4+10-2=12>0$, $x^2+3x-6=4+6-6=4>0$
For $x=-4$: $x^2+5x-2=16-20-2=-6<0$ (invalid)
Answer: $x=2$

(ii) $2^x + 2^{2-x} = 5$
Let $u = 2^x$, then $2^{2-x} = \frac{4}{2^x} = \frac{4}{u}$
Equation: $u + \frac{4}{u} = 5$
Multiply by $u$: $u^2 + 4 = 5u \implies u^2 - 5u + 4 = 0$
Factors: $(u-4)(u-1)=0$, so $u=4$ or $u=1$
Thus $2^x=4 \implies x=2$, or $2^x=1 \implies x=0$
Answer: $x=0$ or $x=2$

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QUESTION THREE

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

(i) $C\cap A=(-2,3]\cap [-1,7]$

$C \cap A = (-2, 3] \cap [-1,7] = [-1, 3]$
Interval: $[-1, 3]$
Number line:

-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10  
      [========]  

(ii) $(A \cap B) - C$
First, $A \cap B = (-2,3] \cap [0,5] = [0,3]$
Then $[0,3] - C = [0,3] \setminus [-1,7] = \emptyset$

(since $[0,3] \subseteq [-1,7]$)
Interval: $\emptyset$

(iii) $A \cap (A' \cup B)$
$A' = X \setminus A = [-3, -2] \cup (3, 10]$
$A^{\mathrm{\prime }}\cup B=[-3,-2]\cup (3,10]\cup [0,5]$

$A' \cup B = [-3, -2] \cup (3,10] \cup [0,5] = [-3,-2] \cup [0,10]$
$A\cap (A^{\mathrm{\prime }}\cup B)=(-2,3]\cap ([-3,-2]\cup [0,10])$

$A \cap (A' \cup B) = (-2,3] \cap ([-3,-2] \cup [0,10]) = [0,3]$
Interval: $[0,3]$

(iv) $X'$ (complement in universal set, but universal set is given as $X$, so likely in $\mathbb{R}$)
Answer: $(-\infty, -3) \cup (10, \infty)$

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(b) Quadratic $f(x) = 3x^2 - 4x - 5$.

(i) Vertex: $x = -\frac{b}{2a} = \frac{4}{6} = \frac{2}{3}$
$$

$f\left(\frac{2}{3}\right)=3{\left(\frac{2}{3}\right)}^2-4\left(\frac{2}{3}\right)-5$

$=3X\frac{4}{9}-\frac{8}{3}-5$

$=\frac{4}{3}-\frac{8}{3}-5$

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

$f\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) - 5 = 3 \cdot \frac{4}{9} - \frac{8}{3} - 5 = \frac{4}{3} - \frac{8}{3} - 5 = -\frac{4}{3} - 5 = -\frac{19}{3}$

Answer: $\left(\frac{2}{3}, -\frac{19}{3}\right)$

(ii) $x$-intercepts: Solve $3x^2 - 4x - 5 = 0$
$$

$x=\frac{4\pm \sqrt{16+60}}{6}$

$=\frac{4\pm \sqrt{76}}{6}$

$=\frac{4\pm 2\sqrt{19}}{6}$

$x = \frac{4 \pm \sqrt{16 + 60}}{6} = \frac{4 \pm \sqrt{76}}{6} = \frac{4 \pm 2\sqrt{19}}{6} = \frac{2 \pm \sqrt{19}}{3}$

Answer: $\frac{2 \pm \sqrt{19}}{3}$

(iii) $y$-intercept: $f(0) = -5$
Answer: $(0, -5)$

(iv) Sketch:

• Opens upwards (coefficient of $x^2 > 0$)
• Vertex $\left(\frac{2}{3} \approx 0.667, -\frac{19}{3} \approx -6.333\right)$
• $x$-intercepts $\frac{2 \pm \sqrt{19}}{3}$ ($\sqrt{19} \approx 4.36$, so $\approx 2.12$ and $-0.787$)
• $y$-intercept $(0,-5)$
• $$
• $(0, -5)$

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(c) Bottles: $x$ bottles at K150, $y$ at K170. Actual cost: $150x + 170y = 18800$
Hypothetical: Half at 150 and twice at 170:
$\frac{1}{2}x$ at K150, $2y$ at K170, cost: $150 \cdot \frac{x}{2} + 170 \cdot 2y = 75x + 340y = 19600$
System:

$$150x + 170y = 18800 \quad \text{(1)}$$ $$\mathrm{}$$

$$75x + 340y = 19600 \quad \text{(2)}$$

Multiply (1) by 2: $300x + 340y = 37600$
Subtract (2): $300x+340y-(75x+340y)=37600-19600$

$⟹225x=18000$

$300x + 340y - (75x + 340y) = 37600 - 19600 \implies 225x = 18000 \implies x = 80$

From (1): $150(80)+170y=18800$

$⟹12000+170y=18800$

$⟹170y=6800$

$150(80) + 170y = 18800 \implies 12000 + 170y = 18800 \implies 170y = 6800 \implies y = 40$
Total bottles: $x + y = 80 + 40 = 120$
Answer: 120 bottles

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(d) Roots $\alpha, \beta$ of $f(x) = 2 + 5x - 3x^2 = 0$ (or $3x^2 - 5x - 2 = 0$)

(i) Equation with roots $\alpha^3, \beta^3$.
Sum $\alpha + \beta = \frac{5}{3}$, product $\alpha\beta = -\frac{2}{3}$
Sum of new roots: $\alpha^3 + \beta^3 = (\alpha+\beta)(\alpha^2 - \alpha\beta + \beta^2) = (\alpha+\beta)[(\alpha+\beta)^2 - 3\alpha\beta]$

$$=\frac{5}{3}[{\left(\frac{5}{3}\right)}^2-3(-\frac{2}{3})]$$

$$=\frac{5}{3}[\frac{25}{9}+2]$$

$$=\frac{5}{3}[\frac{25}{9}+\frac{18}{9}]$$

$$=\frac{5}{3}X\frac{43}{9}$$

$$= \frac{5}{3} \left[ \left(\frac{5}{3}\right)^2 - 3\left(-\frac{2}{3}\right) \right] = \frac{5}{3} \left[ \frac{25}{9} + 2 \right] = \frac{5}{3} \left[ \frac{25}{9} + \frac{18}{9} \right] = \frac{5}{3} \cdot \frac{43}{9} = \frac{215}{27}$$

Product: $\alpha^3\beta^3 = (\alpha\beta)^3 = \left(-\frac{2}{3}\right)^3 = -\frac{8}{27}$
Equation: $t^2 - (\text{sum})t + \text{product} = 0$

$$t^2-\frac{215}{27}t-\frac{8}{27}=0$$

$$t^2 - \frac{215}{27}t - \frac{8}{27} = 0 \implies 27t^2 - 215t - 8 = 0$$

Answer: $27t^2 - 215t - 8 = 0$

(ii) $\frac{1}{\alpha^2 + 1} + \frac{1}{\beta^2 + 1} = \frac{(\beta^2 + 1) + (\alpha^2 + 1)}{(\alpha^2 + 1)(\beta^2 + 1)} = \frac{\alpha^2 + \beta^2 + 2}{\alpha^2\beta^2 + \alpha^2 + \beta^2 + 1}$
${}^{}$

${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta ={\left(\frac{5}{3}\right)}^2-2(-\frac{2}{3})$

$=\frac{25}{9}+\frac{4}{3}$

$=\frac{25}{9}+\frac{12}{9}$

$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \left(\frac{5}{3}\right)^2 - 2\left(-\frac{2}{3}\right) = \frac{25}{9} + \frac{4}{3} = \frac{25}{9} + \frac{12}{9} = \frac{37}{9}$
${}^{}$

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

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

Numerator: ${\alpha }^2+{\beta }^2+2=\frac{37}{9}+2$

$=\frac{37}{9}+\frac{18}{9}$

$\alpha^2 + \beta^2 + 2 = \frac{37}{9} + 2 = \frac{37}{9} + \frac{18}{9} = \frac{55}{9}$

Denominator: ${\alpha }^2{\beta }^2+{\alpha }^2+{\beta }^2+1=\frac{4}{9}+\frac{37}{9}+1$

$=\frac{41}{9}+1$

$\alpha^2\beta^2 + \alpha^2 + \beta^2 + 1 = \frac{4}{9} + \frac{37}{9} + 1 = \frac{41}{9} + 1 = \frac{50}{9}$

So $\frac{55\mathrm{/}9}{50\mathrm{/}9}$

$=\frac{55}{50}$

$\frac{55/9}{50/9} = \frac{55}{50} = \frac{11}{10}$

Answer: $\frac{11}{10}$

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(e) Derivative of $f(x) = (1 - 3x - 2x^2)^{-3}$ at $x=3$.
Let $u = 1 - 3x - 2x^2$, so $f = u^{-3}$
$f' = -3u^{-4} \cdot u'$, $u' = -3 - 4x$
At $x=3$:
$u = 1 - 9 - 18 = -26$
$u' = -3 - 12 = -15$

$f^{\mathrm{\prime }}=-3(-26{)}^{-4}X(-15)$

$=-3X\frac{1}{(-26{)}^4}X(-15)$

$=-3X\frac{1}{456976}X(-15)$

$=\frac{45}{456976}$

$f' = -3(-26)^{-4} \cdot (-15) = -3 \cdot \frac{1}{(-26)^4} \cdot (-15) = -3 \cdot \frac{1}{456976} \cdot (-15) = \frac{45}{456976} = \frac{5}{50775.11}$

Better: $(-26{)}^{-4}=\frac{1}{26^4}$

$(-26)^{-4} = \frac{1}{26^4} = \frac{1}{456976}$

So $f^{\mathrm{\prime }}=-3X\frac{1}{456976}X(-15)$

$=\frac{45}{456976}$

$f' = -3 \cdot \frac{1}{456976} \cdot (-15) = \frac{45}{456976} = \frac{5}{50775.111}$?

Simplify: $\frac{45}{456976} = \frac{5 \cdot 9}{50775.111 \cdot 9}$?
Divide numerator and denominator by 3: $\frac{15}{152325.333}$ — no.
GCD of 45 and 456976.
Compute numerically: $26^2=676$, $26^4=676^2=456976$
$f' = \frac{45}{456976} \approx 0.00009845$
But exact: $\frac{45}{456976} = \frac{5 \cdot 9}{5 \cdot 91395.2}$?
Factor: 45 = 9×5, 456976 ÷ 16? Better reduce fraction:
GCD of 45 and 456976. 45=3^2×5, 456976 even, not divisible by 5, so GCD=1?
Answer: $\frac{45}{456976}$ or simplify by dividing numerator and denominator by... 45 and 456976 share no common factors, so $\frac{45}{456976}$.

But check calculation:
$f'(x) = -3(1-3x-2x^2)^{-4} \cdot (-3-4x) = 3(3+4x)(1-3x-2x^2)^{-4}$
At $x=3$: $3(3+12)(1-9-18)^{-4} = 3 \cdot 15 \cdot (-26)^{-4} = 45 \cdot \frac{1}{456976}$
Yes.
Answer: $\frac{45}{456976}$

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QUESTION FOUR

(a) Divide $x^4 - 2x^3 - 7x^2 + 7x + 5$ by $x^2 + 2x - 1$.
Polynomial division:

               x^2 - 4x + 3  
x^2 + 2x - 1 | x^4 - 2x^3 - 7x^2 + 7x + 5  
               -(x^4 + 2x^3 - x^2)  
                      -4x^3 - 6x^2 + 7x  
                      -(-4x^3 - 8x^2 + 4x)  
                              2x^2 + 3x + 5  
                              -(2x^2 + 4x - 2)  
                                     -x + 7  

Quotient: $x^2 - 4x + 3$, Remainder: $-x + 7$
Answer: Quotient = $x^2 - 4x + 3$, Remainder = $-x+7$

$\mathrm{}$

$-x + 7$

✏️ Step-by-step Division:

We divide term-by-term:

Step 1: Divide leading terms

$\frac{x^4}{x^2}=x^2$

Multiply:

$x^2(x^2+2x-1)=x^4+2x^3-x^2$

Subtract:

$(x^4-2x^3-7x^2)-(x^4+2x^3-x^2)=-4x^3-6x^2$

Bring down $+7x$

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Step 2: Divide

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

Multiply:

$-4x(x^2+2x-1)=-4x^3-8x^2+4x$

Subtract:

$(-4x^3-6x^2+7x)-(-4x^3-8x^2+4x)=2x^2+3x$

Bring down $+5$

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Step 3: Divide

$\frac{2x^2}{x^2}=2$

Multiply:

$2(x^2+2x-1)=2x^2+4x-2$

Subtract:

$(2x^2+3x+5)-(2x^2+4x-2)=-x+7$

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✅ Final Answer:

$\overline{)x^2-4x+2+\frac{-x+7}{x^2+2x-1}}$

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(b) Cartesian product $A \times B$ has six ordered pairs. Given: $(2,1), (2,3), (4,1), (4,5)$.

(i) Elements: From pairs, $A = \{2,4\}$, $B = \{1,3,5\}$ (since 1,3,5 appear)
But total pairs should be $|A| \times |B| = 6$, so $|A|=2$, $|B|=3$ or vice versa. Given pairs show A has 2,4 and B has 1,3,5, so missing pairs: $(2,5), (4,3)$?
Existing: (2,1),(2,3),(4,1),(4,5) — so missing (2,5) and (4,3)? But (4,5) is given, (2,3) given.
B should be {1,3,5}, so pairs: (2,1),(2,3),(2,5),(4,1),(4,3),(4,5). Given four, missing (2,5) and (4,3).
Answer: $A = \{2,4\}$, $B = \{1,3,5\}$

(ii) Missing ordered pairs: $(2,5), (4,3)$
Answer: $(2,5), (4,3)$

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(c) $P(x) = 2x^3 + 3x^2 - 2x - 3$

(i) Factors: Possible rational roots $\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2}$
Try $x=1$: $2+3-2-3=0$, so $(x-1)$ is factor.
Divide:

        2x^2 + 5x + 3  
x-1 | 2x^3 + 3x^2 - 2x - 3  
      -(2x^3 - 2x^2)  
            5x^2 - 2x  
            -(5x^2 - 5x)  
                  3x - 3  
                  -(3x - 3)  
                        0  

So $P(x) = (x-1)(2x^2 + 5x + 3)$
Factor quadratic: $2x^2 + 5x + 3 = (2x+3)(x+1)$
Thus $P(x) = (x-1)(2x+3)(x+1)$
Answer: $(x-1)(x+1)(2x+3)$

(ii) $P(x) > 0$
Roots at $x=1, x=-1, x=-\frac{3}{2}$
Sign chart:

Interval: (-∞, -3/2) | (-3/2, -1) | (-1, 1) | (1, ∞)  
P(x):           -          +          -         +  

So $P(x) > 0$ for $x \in (-\frac{3}{2}, -1) \cup (1, \infty)$
Answer: $x \in \left(-\frac{3}{2}, -1\right) \cup (1, \infty)$

(iii) Sketch:

• Roots at $x=-\frac{3}{2}, x=-1, x=1$
• As $x \to \infty$, $P(x) \to \infty$; $x \to -\infty$, $P(x) \to -\infty$
• Multiplicity 1 at each root

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(d) Solve and graph inequalities:

(i) $\frac{x+3}{x-5} < 3$
Bring to zero: $\frac{x+3}{x-5}-3<0=\frac{x+3-3(x-5)}{x-5}$

$=\frac{x+3-3x+15}{x-5}$

$\frac{x+3}{x-5} - 3 < 0 = \frac{x+3 - 3(x-5)}{x-5} = \frac{x+3-3x+15}{x-5} = \frac{-2x+18}{x-5} < 0$
Critical points: $x=9, x=5$
Sign chart:

Interval: (-∞,5) | (5,9) | (9,∞)  
Numerator: + | + | -  
Denom: - | + | +  
Fraction: - | + | -  

So $\frac{-2x+18}{x-5} < 0$ when $x < 5$ or $x > 9$
But $x=5$ undefined, so solution: $x \in (-\infty, 5) \cup (9, \infty)$
Graph:

Number line:  
<-----o=====o----->  
      5     9  
Shade left of 5 and right of 9.  

(ii) $|2x+9| > 5$

Equivalent to: $2x+9 < -5$ or $2x+9 > 5$
First: $2x<-14$

$2x < -14 \implies x < -7$

Second: $2x>-4$

$2x > -4 \implies x > -2$

Solution: $x < -7$ or $x > -2$

Graph:

<======o---------o=====>  
      -7        -2  
Shade left of -7 and right of -2.  

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(e) Nature of roots of $2x^2 + \sqrt{3}x - 4 = 0$
Discriminant $d = b^2 - 4ac = (\sqrt{3})^2 - 4(2)(-4) = 3 + 32 = 35 > 0$, so two distinct real roots.
Answer: Two distinct real roots

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(f) Solve $\sqrt{3 - x} - \sqrt{7 + x} = 2$

Isolate: $\sqrt{3 - x} = 2 + \sqrt{7 + x}$

Square both sides:
$3 - x = 4 + 4\sqrt{7+x} + (7+x)$
$3 - x = 11 + x + 4\sqrt{7+x}$
$3 - x - 11 - x = 4\sqrt{7+x}$
$-8 - 2x = 4\sqrt{7+x}$

Divide by 2: $-4 - x = 2\sqrt{7+x}$

Square again:
$(-4 - x)^2 = 4(7 + x)$
$16 + 8x + x^2 = 28 + 4x$
$x^2 + 4x - 12 = 0$

Factors: $(x+6)(x-2)=0$, so $x=-6$ or $x=2$

Check:
For $x=2$: $\sqrt{3-2} - \sqrt{7+2} = 1 - 3 = -2 \neq 2$
For $x=-6$: $\sqrt{3-(-6)} - \sqrt{7-6} = \sqrt{9} - \sqrt{1} = 3 - 1 = 2$
Answer: $x = -6$

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QUESTION FIVE

(a) $f(x) = x^2 + 4$, domain $D_f = \{x: -2 \leq x \leq 2\}$

(i) Range: Min at $x=0$, $f(0)=4$; max at endpoints, $f(-2)=f(2)=4+4=8$
So range: $[4, 8]$
Answer: $[4, 8]$

(ii) One-to-one? $f(-1)=1+4=5$, $f(1)=1+4=5$, same output, not injective.
Answer: No

(iii) Inverse does not exist (not bijective).
Answer: Does not exist

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(b) Polynomial $x^3 + ax^2 + bx - 5$.
Remainder when divided by $x-1$ is $-1$: $f(1) = -1$
$1+a+b-5=-1$

$⟹a+b-4=-1$

$1 + a + b - 5 = -1 \implies a + b -4 = -1 \implies a + b = 3$
Remainder when divided by $x+1$ is $-5$: $f(-1) = -5$
$-1+a-b-5=-5$

$⟹a-b-6=-5$

$-1 + a - b - 5 = -5 \implies a - b -6 = -5 \implies a - b = 1$
Solve:
$a + b = 3$
$a - b = 1$
Add: $2a = 4 \implies a=2$
Then $b=1$
Answer: $a=2$, $b=1$

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(c) Solve $3^{2x} - 12 \cdot 3^x + 27 = 0$
Let $u = 3^x$, then $u^2 - 12u + 27 = 0$
Factors: $(u-3)(u-9)=0$, so $u=3$ or $u=9$
Thus $3^x=3$

$3^x=3 \implies x=1$, $3^x=9$

$3^x=9 \implies x=2$
Answer: $x=1$ or $x=2$

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(d) Evaluate $\left(2x^{1/3} y^{1/3}\right)^6 \times \left(\frac{1}{2} x^{1/4} y^{1/4}\right)^4$ at $x=2$, $y=3$
Simplify:
First part: $(2^6) (x^{1/3})^6 (y^{1/3})^6 = 64 x^{2} y^{2}$

Second part: $\left(\frac{1}{2}\right)^4 (x^{1/4})^4 (y^{1/4})^4 = \frac{1}{16} x y$

Product: $64 x^2 y^2 \cdot \frac{1}{16} x y = \frac{64}{16} x^{3} y^{3} = 4 (xy)^3$

At $x=2$, $y=3$: $\mathrm{}$

$4X(6{)}^3=4X216$

$4 \cdot (6)^3 = 4 \cdot 216 = 864$
Answer: 864

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(e) Line through $(4,-6)$ parallel to $6x + 2y + 12 = 0$
Slope of given line: $2y = -6x -12 \implies y = -3x -6$, slope $-3$
Parallel line has same slope.
Equation: $y - (-6) = -3(x - 4)$
$y +6 = -3x +12$
$y = -3x +6$
Answer: $y = -3x + 6$ or $3x + y - 6 = 0$

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(f) Solve $\frac{2x-1}{7} - \frac{3x-4}{14} = \frac{x+1}{2}$

Multiply through by 14:
$2(2x-1) - (3x-4) = 7(x+1)$
$4x - 2 - 3x + 4 = 7x + 7$
$x + 2 = 7x + 7$
$2 - 7 = 7x - x$
$-5 = 6x$
$x = -\frac{5}{6}$
Answer: $x = -\frac{5}{6}$

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(g) Differentiate $f(x) = 2x^2 - x$ using first principles.
Definition: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$$f(x+h)=2(x+h{)}^2-(x+h)$$

$$=2(x^2+2xh+h^2)-x-h$$

$$f(x+h) = 2(x+h)^2 - (x+h) = 2(x^2 + 2xh + h^2) - x - h = 2x^2 + 4xh + 2h^2 - x - h$$ $$$$

$$f(x+h)-f(x)=(2x^2+4xh+2h^2-x-h)-(2x^2-x)$$

$$=4xh+2h^2-h$$

$$f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - x - h) - (2x^2 - x) = 4xh + 2h^2 - h$$ $$\frac{f(x+h)-f(x)}{h}=\frac{4xh+2h^2-h}{h}$$

$$\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 - h}{h} = 4x + 2h - 1$$

Limit as $h \to 0$: $4x - 1$
Answer: $f'(x) = 4x - 1$

@Dr. Microbiota

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END OF SOLUTIONS