INTRODUCTION TO MATHEMATICAL METHODS SEMESTER ONE EXAMINATION JULY INTAKE 2017

DATE : 10th MAY, 2017

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QUESTION ONE

(a) Define the following:
(i) Singleton set
(ii) Product of two sets

Answer:
(i) Singleton set: A set containing exactly one element.
(ii) Product of two sets: The set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$, denoted $A \times B$.

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(b) Let $A = \{-3, -\frac{1}{5}, -\sqrt{5}, 0.251, \frac{1}{\sqrt{5}}, \sqrt{7}, \sqrt{2}, 4\}$ be a universal set. List the elements of A which are:
(i) Integers
(ii) Rational numbers
(iii) Irrational numbers

Answer:

• Integers: Elements with no fractional/decimal part.
  $\boxed{-3,\ 4}$
• Rational numbers: Numbers expressible as $\frac{p}{q}$ ($p, q$ integers, $q \neq 0$).
  $\boxed{-3,\ -\frac{1}{5},\ 0.251,\ 4}$
  Explanation:
  • $-3 = \frac{-3}{1}$, $4 = \frac{4}{1}$ (integers are rational).
  • $-\frac{1}{5}$ is a fraction.
  • $0.251 = \frac{251}{1000}$ (terminating decimal).
• Irrational numbers: Non-repeating, non-terminating decimals.
  $\boxed{-\sqrt{5},\ \frac{1}{\sqrt{5}},\ \sqrt{7},\ \sqrt{2}}$
  Explanation: $\sqrt{5}, \sqrt{7}, \sqrt{2}$ are irrational, and reciprocals/negatives of irrationals remain irrational.

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(c) Given $A = \{1, 2, 3\}$, $B = \{p, q\}, C = \{r\}$.
(i) Find $A \times B$
(ii) Show that $A \times B \neq B \times A$

Answer:
(i) $A \times B = \{(1, p),\ (1, q),\ (2, p),\ (2, q),\ (3, p),\ (3, q)\}$
(ii) $B \times A = \{(p, 1),\ (p, 2),\ (p, 3),\ (q, 1),\ (q, 2),\ (q, 3)\}$
Since $(1, p) \neq (p, 1)$, $A \times B \neq B \times A$.

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(d) Express 0.7142 in fractional form.

Answer:
Let $x = 0.7142$.
Multiply by 10000:
$10000x = 7142$
$x = \frac{7142}{10000} = \frac{3571}{5000}$
Explanation: Simplify by dividing numerator and denominator by 2.

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(e) Find $\frac{dy}{dx}$ of the following.
(i) $y = (x^3 + 6)e^{3x}$
(ii) $y = \frac{\cos x}{x}$

Answer:

(i) $y=(x^3+6)e^{3x}$

This is a product of two functions:

• $u=x^3+6$
• $v=e^{3x}$

Using the product rule:

$\frac{dy}{dx}=u^{\prime }v+u{v}^{\prime }$

• $u^{\prime }=\frac{d}{dx}(x^3+6)=3x^2$
• $v^{\prime }=\frac{d}{dx}\left(e^{3x}\right)=3e^{3x}$

So:

$\frac{dy}{dx}=3x^2\cdot e^{3x}+(x^3+6)\cdot 3e^{3x}$

Factor out $3e^{3x}$:

$\frac{dy}{dx}=3e^{3x}(x^2+x^3+6)=3e^{3x}(x^3+x^2+6)$

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(ii) $y=\cos x^x$

This one’s a bit trickier—it’s not clear whether you meant:

• Option A: $y=\cos \left(x^x\right)$
• Option B: $y=\frac{\cos x}{x}$

Let me solve both interpretations:

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Option A: $y=\cos \left(x^x\right)$

Let $u=x^x$. Then:

• First, find $\frac{du}{dx}$:

  $x^x=e^{x\ln x}\Rightarrow \frac{d}{dx}\left(x^x\right)=e^{x\ln x}\cdot (\ln x+1)=x^x(\ln x+1)$

• Now differentiate $y=\cos \left(u\right)$:

  $\frac{dy}{dx}=-\sin \left(x^x\right)\cdot \frac{d}{dx}\left(x^x\right)=-\sin \left(x^x\right)\cdot x^x(\ln x+1)$

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Option B: $y=\frac{\cos x}{x}$

Use the quotient rule:

$\frac{dy}{dx}=\frac{x(-\sin x)-\cos x\left(1\right)}{x^2}=\frac{-x\sin x-\cos x}{x^2}$

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(i) Product Rule:

. Differentiate $y=(x^3+6)e^{3x}$

This is a product of two functions:

• $u\left(x\right)=x^3+6$
• $v\left(x\right)=e^{3x}$

Using the product rule:

$\frac{dy}{dx}=u^{\prime }\left(x\right)v\left(x\right)+u\left(x\right)v^{\prime }\left(x\right)$

Compute the derivatives:

• $u^{\prime }\left(x\right)=3x^2$
• $v^{\prime }\left(x\right)=3e^{3x}$

Substitute into the product rule:

$\frac{dy}{dx}=3x^2\cdot e^{3x}+(x^3+6)\cdot 3e^{3x}$

Factor out $3e^{3x}$:

$\frac{dy}{dx}=3e^{3x}(x^2+x^3+6)=3e^{3x}(x^3+x^2+6)$

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(ii) Quotient Rule: $\frac{d}{dx}\left[\frac{u}{v}\right]=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$, where $u=\cos x$, $v=x$.

Differentiate $y=\frac{\cos \left(x\right)}{x}$

This is a quotient of two functions:

• $u\left(x\right)=\cos \left(x\right)$
• $v\left(x\right)=x$

Using the quotient rule:

$\frac{dy}{dx}=\frac{u^{\prime }\left(x\right)v\left(x\right)-u\left(x\right)v^{\prime }\left(x\right)}{\left[v\right(x){]}^2}$

Compute the derivatives:

• $u^{\prime }\left(x\right)=-\sin \left(x\right)$
• $v^{\prime }\left(x\right)=1$

Substitute into the quotient rule:

$\frac{dy}{dx}=\frac{(-\sin (x\left)\right)\cdot x-\cos \left(x\right)\cdot 1}{x^2}=\frac{-x\sin \left(x\right)-\cos \left(x\right)}{x^2}$

Factor out the negative sign:

$\frac{dy}{dx}=-\frac{x\sin \left(x\right)+\cos \left(x\right)}{x^2}$

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$$\frac{}{}$$

Question Two

(a) Define the following:
(i) Transitive relation
(ii) Injection function

Answer:
(i) Transitive relation: If $(a, b) \in \mathfrak{R}$ and $(b, c) \in \mathfrak{R}$, then $(a, c) \in \mathfrak{R}$.
(ii) Injection function: A function where distinct inputs map to distinct outputs ($f(a) = f(b) \implies a = b$).

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(b) Determine whether the relation $\mathfrak{R} = \{(x, y): y \text{ is divisible by } x\}$ in $A = \{1, 2, 3, 4, 5, 6\}$ is reflexive, symmetric, and transitive.

Answer:

• Reflexive: Yes, since $x \mid x$ (e.g., $(1,1), (2,2)$).
• Symmetric: No. Counterexample: $(2,4) \in \mathfrak{R}$ (4 divisible by 2), but $(4,2) \notin \mathfrak{R}$ (2 not divisible by 4).
• Transitive: Yes. If $x \mid y$ and $y \mid z$, then $x \mid z$ (e.g., $(2,4)$ and $(4,8)$ $\implies (2,8)$, but 8 not in $A$; consider $(1,2)$ and $(2,4) \implies (1,4)$).

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(c) A function $f : \mathbb{Z} \rightarrow \mathbb{Z}$ defined as:

$$f(x) = \begin{cases} 1 + x, & 1 \leq x < 2 \\ 2x - 11, & 2 \leq x < 4 \\ 3x - 101, & 4 \leq x < 6 \end{cases}$$

(i) Domain
(ii) Range
(iii) $f(4)$

Answer:
(i) Domain: $1 \leq x < 6$ (all integers in $[1, 6)$).
$\boxed{\{1, 2, 3, 4, 5\}}$
(ii) Range:

• For $x=1$: $f(1) = 1+1=2$
• $x=2$: $f(2) = 2(2)-11 = -7$
• $x=3$: $f(3) = 2(3)-11 = -5$
• $x=4$: $f(4) = 3(4)-101 = -89$
• $x=5$: $f(5) = 3(5)-101 = -86$
  Range: $\boxed{\{2, -7, -5, -89, -86\}}$
  (iii) $f(4) = 3(4) - 101 = \boxed{-89}$

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(d) Classify $f(x) = \frac{2x + 3}{x - 3}, \, x \neq 3$ as injection, surjection, or bijection.

Answer:

• Injection: Check if $f(a) = f(b) \implies a = b$. $$\frac{2a+3}{a-3}=\frac{2b+3}{b-3}$$
• $$\frac{2a+3}{a-3} = \frac{2b+3}{b-3} \implies (2a+3)(b-3) = (2b+3)(a-3)$$ Simplify: $2ab-6a+3b-9=2ab-6b+3a-9$

$⟹-6a+3b=-6b+3a$

$⟹-9a=-9b$

$2ab - 6a + 3b - 9 = 2ab - 6b + 3a - 9 \implies -6a + 3b = -6b + 3a \implies -9a = -9b \implies a = b$.
Yes, injection.

• Surjection: Check if range $= \mathbb{R}$ (codomain). As $x \to \infty$, $f(x) \to 2$, and

vertical asymptote at $x=3$. Range is $\mathbb{R} \setminus \{2\}$, not all $\mathbb{R}$. Not surjection.

• Bijection: No (since not surjection).

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(e) Let $f(x) = 2x - 3$, $g(x) = \frac{x + 3}{2}$. Show $fog(x) = gof(x)$.

Answer:
Compute:

• $fog(x)=f(g(x))=f\left(\frac{x+3}{2}\right)=2\left(\frac{x+3}{2}\right)-3$
• $fog(x) = f(g(x)) = f\left(\frac{x+3}{2}\right) = 2\left(\frac{x+3}{2}\right) - 3 = x + 3 - 3 = x$
• $gof(x)=g(f(x))=g(2x-3)=\frac{(2x-3)+3}{2}$
• $gof(x) = g(f(x)) = g(2x - 3) = \frac{(2x-3)+3}{2} = \frac{2x}{2} = x$
  Thus, $fog(x) = gof(x) = x$.

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Question Three

(a) State the following:
(i) The factor theorem
(ii) Natural exponential function

Answer:
(i) Factor theorem: If $p(a) = 0$, then $(x - a)$ is a factor of polynomial $p(x)$.
(ii) Natural exponential function: $e^x$, where $e \approx 2.71828$.

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(b) Given $\alpha, \beta$ are roots of $3x^{2} + x + 2 = 0$. Find $\alpha^{2} + \beta^{2}$.

Answer:
For quadratic $ax^2 + bx + c = 0$:

$$\alpha + \beta = -\frac{b}{a} = -\frac{1}{3}, \quad \alpha\beta = \frac{c}{a} = \frac{2}{3}$$ $${}^{}$$

$${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta ={(-\frac{1}{3})}^2-2\left(\frac{2}{3}\right)$$

$$=\frac{1}{9}-\frac{4}{3}$$

$$=\frac{1}{9}-\frac{12}{9}$$

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{1}{3}\right)^2 - 2\left(\frac{2}{3}\right) = \frac{1}{9} - \frac{4}{3} = \frac{1}{9} - \frac{12}{9} = \boxed{-\frac{11}{9}}$$

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(c) Polynomial $p(x) = 2x^{3} + qx^{2} + px + 6$ is divisible by $x-2$ and leaves remainder $-12$ when divided by $x+1$. Find $q, p$.

Answer:

• Divisible by $x-2$ $\implies p(2) = 0$: $$2(8) + q(4) + p(2) + 6 = 0 \implies 16 + 4q + 2p + 6 = 0 \implies 4q + 2p = -22 \quad \text{(1)}$$
• Remainder $-12$ when divided by $x+1$ $\implies p(-1) = -12$: $$2(-1)^3 + q(-1)^2 + p(-1) + 6 = -12 \implies -2 + q - p + 6 = -12 \implies q - p = -16 \quad \text{(2)}$$

Solve system:
From (2): $q = p - 16$.
Substitute into (1):

$$4(p-16)+2p=-22$$

$$⟹4p-64+2p=-22$$

$$⟹6p=42$$

$$4(p - 16) + 2p = -22 \implies 4p - 64 + 2p = -22 \implies 6p = 42 \implies p = 7$$

Then $q = 7 - 16 = -9$.
$\boxed{q = -9,\ p = 7}$

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(d) Factorise $f(x) = x^{3} - 2x^{2} - x + 2$, sketch graph, and solve $f(x) \leq 0$.

Answer:

• Factorization:
  Possible roots: $\pm1, \pm2$.
  $f(1) = 1 - 2 - 1 + 2 = 0$ $\implies (x-1)$ factor.
  Polynomial division: $$\frac{x^3 - 2x^2 - x + 2}{x-1} = x^2 - x - 2$$ Factor quadratic: $x^2 - x - 2 = (x-2)(x+1)$.
  Thus, $f(x) = (x-1)(x-2)(x+1)$.
• Graph:
  • Intercepts:
    • $x$-intercepts: $x = -1, 1, 2$.
    • $y$-intercept: $f(0) = 2$.
  • Behavior:
    • As $x \to \infty$, $f(x) \to \infty$.
    • As $x \to -\infty$, $f(x) \to -\infty$.
      Sketch: Cubic with roots at $-1, 1, 2$, passing through $(0,2)$.
      
• Solution to $f(x) \leq 0$:
  Sign chart:

  Interval	$x < -1$	$-1 < x < 1$	$1 < x < 2$	$x > 2$
  $(x+1)$	$-$	$+$	$+$	$+$
  $(x-1)$	$-$	$-$	$+$	$+$
  $(x-2)$	$-$	$-$	$-$	$+$
  $f(x)$	$-$	$+$	$-$	$+$
  $f(x) \leq 0$ when $x \in (-\infty, -1] \cup [1, 2]$.
  Solution set: $\boxed{(-\infty, -1] \cup [1, 2]}$

OR USE THIS BELOW

To solve the inequality

$f\left(x\right)=x^3-2x^2-x+2\le 0$

we use the factorized form:

$f\left(x\right)=(x-1)(x-2)(x+1)$

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🧮 Step-by-step Sign Analysis

Critical points (roots):

• $x=-1,1,2$

Intervals to test:

• $(-\infty ,-1)$
• $(-1,1)$
• $(1,2)$
• $(2,\infty )$

Sign of each factor in each interval:

Interval	$x+1$	$x-1$	$x-2$	$f\left(x\right)$ Sign
	−	−	−	−
	+	−	−	+
	+	+	−	−
$x>2$	+	+	+	+

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✅ Final Answer

$f\left(x\right)\le 0\text{on}(-\infty ,-1]\cup [1,2]$

This includes the points where $f\left(x\right)=0$, namely $x=-1,1,2$, and the intervals where the function is negative.

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(e) Differentiate from first principles:
(i) $f(x) = \frac{1}{x}$
(ii) $f(x) = x^{2} + 6x + 9$

Answer:

First principle: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

$${\mathrm{(i)\; f}}^{\mathrm{\prime }}(x)=\underset{h\to 0}{\lim }\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$$

$$=\underset{h\to 0}{\lim }\frac{\frac{x-(x+h)}{x(x+h)}}{h}$$

$$=\underset{h\to 0}{\lim }\frac{-h}{hx(x+h)}$$

$$=\underset{h\to 0}{\lim }\frac{-1}{x(x+h)}$$

$$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h} = \lim_{h \to 0} \frac{-h}{h x (x+h)} = \lim_{h \to 0} \frac{-1}{x(x+h)} = \boxed{-\frac{1}{x^2}}$$

$${\mathrm{(ii)\; f}}^{\mathrm{\prime }}(x)=\underset{h\to 0}{\lim }\frac{[(x+h{)}^2+6(x+h)+9]-[x^2+6x+9]}{h}$$

$$=\underset{h\to 0}{\lim }\frac{[x^2+2xh+h^2+6x+6h+9]-[x^2+6x+9]}{h}$$

$$=\underset{h\to 0}{\lim }\frac{2xh+h^2+6h}{h}$$

$$=\underset{h\to 0}{\lim }(2x+h+6)$$

$$f'(x) = \lim_{h \to 0} \frac{[(x+h)^2 + 6(x+h) + 9] - [x^2 + 6x + 9]}{h} = \lim_{h \to 0} \frac{[x^2 + 2xh + h^2 + 6x + 6h + 9] - [x^2 + 6x + 9]}{h} = \lim_{h \to 0} \frac{2xh + h^2 + 6h}{h} = \lim_{h \to 0} (2x + h + 6) = \boxed{2x + 6}$$

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Question Four

(a) Find exact values:
(i) $\sin 15^\circ$
(ii) $\cos \frac{7\pi}{12}$

Answer:
(i) $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$

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

$$= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \boxed{\frac{\sqrt{6} - \sqrt{2}}{4}}$$

(ii) $\cos \frac{7\pi}{12} = \cos\left(105^\circ\right) = \cos(60^\circ + 45^\circ) = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ$

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

$$=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$$

$$= \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \boxed{\frac{\sqrt{2} - \sqrt{6}}{4}}$$

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(b) For $y = 5\sin\left(x + \frac{\pi}{2}\right)$, find period, amplitude, phase shift. Sketch graph.

Answer:

• Amplitude: $|5| = 5$.
• Period: $\frac{2\pi}{1} = 2\pi$.
• Phase shift: $-\frac{\pi}{2}$ (left by $\frac{\pi}{2}$).
• Graph:
  • Standard sine wave shifted left by $\frac{\pi}{2}$.
  • Key points: At $x = -\frac{\pi}{2}$, $y = 5\sin(0) = 0$; at $x = 0$, $y = 5\sin(\frac{\pi}{2}) = 5$; etc.
    

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(c) Solve:

$${\log }_{10}(x-y)=1$$

$$\log_{10}(x - y) = 1$$ $$\log_{e} e^{x} + 2\log_{e} e^{y} = 4$$

Answer:
Simplify:

• First equation: $\log_{10}(x - y) = 1 \implies x - y = 10^1 = 10$.
• Second equation: $\log_e e^x = x$, $2\log_e e^y = 2y$ $\implies x + 2y = 4$.
  Solve system:

$$x - y = 10 \quad (1)$$ $$x + 2y = 4 \quad (2)$$

Subtract (1) from (2): $(x+2y)-(x-y)=4-10$

$⟹3y=-6$

$(x + 2y) - (x - y) = 4 - 10 \implies 3y = -6 \implies y = -2$.
From (1): $x-(-2)=10$

$⟹x+2=10$

$x - (-2) = 10 \implies x + 2 = 10 \implies x = 8$.

Check: $\log_{10}(8 - (-2)) = \log_{10}(10) = 1$, and $8 + 2(-2) = 8 - 4 = 4$.
Solution: $\boxed{{\displaystyle x=8,y=-2}}$

OR USE THIS BELOW

$\boxed{x = 8,\ y = -2}$

Let’s solve the system step by step:

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✏️ Given:

1. ${\log }_{10}(x-y)=1$
2. ${\log }_e{e}^x+2{\log }_e{e}^y=4$

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🔍 Step 1: Solve Equation 1

${\log }_{10}(x-y)=1\Rightarrow x-y=10^1=10$

So:

$x=y+10\text{(Equation A)}$

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🔍 Step 2: Simplify Equation 2

Recall:

• ${\log }_e{e}^x=x$
• ${\log }_e{e}^y=y$

So:

$x+2y=4\text{(Equation B)}$

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🔁 Step 3: Substitute Equation A into B

Substitute $x=y+10$ into Equation B:

$(y+10)+2y=4\Rightarrow 3y+10=4\Rightarrow 3y=-6\Rightarrow y=-2$

Then:

$x=y+10=-2+10=8$

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✅ Final Answer:

$\overline{)x=8,y=-2}$

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(d) Find domain, range, and inverse (if exists):
(i) $g(x) = \frac{4x+3}{6x-4}$
(ii) $g(x) = \sqrt{3x-5}$

Answer:
(i)

• Domain: Denominator $\neq 0$, so $6x - 4 \neq 0 \implies x \neq \frac{2}{3}$. Domain: $\mathbb{R} \setminus \{\frac{2}{3}\}$.
• Range: Solve for $x$: $y=\frac{4x+3}{6x-4}⟹y(6x-4)=4x+3$

$⟹6xy-4y=4x+3$

$⟹6xy-4x=4y+3$

$⟹x(6y-4)=4y+3$

$y = \frac{4x+3}{6x-4} \implies y(6x-4) = 4x+3 \implies 6xy - 4y = 4x + 3 \implies 6xy - 4x = 4y + 3 \implies x(6y - 4) = 4y + 3 \implies x = \frac{4y+3}{6y-4}$.
Inverse exists when denominator $\neq 0$, so range is $\mathbb{R} \setminus \{\frac{2}{3}\}$.

• Inverse: $g^{-1}(x) = \frac{4x+3}{6x-4}$ (same as original, but confirm with domain).
  Note: Function is its own inverse if $g(g(x)) = x$.

OR LETS DO THIS ONLY FOR LAZY STUDENTS LIKE YOU 😂🤣😆😹😄😜

$g\left(x\right)=\frac{4x+3}{6x-4}$

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🧭 Domain

The domain excludes values that make the denominator zero.

• Set denominator to zero:
  $6x-4=0\Rightarrow x=\frac{2}{3}$

✅ Domain:

$\overline{)x\in R,x\ne \frac{2}{3}}$

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📈 Range

To find the range, consider the horizontal asymptote and behavior of the function.

• As $x\to \pm \infty$, the leading terms dominate: $g\left(x\right)\approx \frac{4x}{6x}=\frac{2}{3}$
• So, horizontal asymptote: $y=\frac{2}{3}$

To confirm that $y=\frac{2}{3}$ is not in the range, solve:

$\frac{4x+3}{6x-4}=\frac{2}{3}\Rightarrow 3(4x+3)=2(6x-4)\Rightarrow 12x+9=12x-8\Rightarrow 9=-8\text{(False)}$

✅ Range:

$\overline{)y\in R,y\ne \frac{2}{3}}$

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🔄 Inverse Function

Let $y=\frac{4x+3}{6x-4}$. Solve for $x$:

$y(6x-4)=4x+3\Rightarrow 6xy-4y=4x+3\Rightarrow 6xy-4x=4y+3\Rightarrow x(6y-4)=4y+3\Rightarrow x=\frac{4y+3}{6y-4}$

✅ Inverse:

$g^{-1}\left(x\right)=\overline{)\frac{4x+3}{6x-4}}$

Interestingly, the inverse is identical to the original function. So:

✅ g(x) is self-inverse
That means:

$g\left(g\right(x\left)\right)=x$

(ii)

• Domain: $3x-5\ge 0$
• $3x - 5 \geq 0 \implies x \geq \frac{5}{3}$. Domain: $\left[\frac{5}{3}, \infty\right)$.
• Range: $\sqrt{\cdot} \geq 0$, so range: $[0, \infty)$.

Inverse: Solve $y=\sqrt{3x-5}$

$⟹y^2=3x-5$

$⟹3x=y^2+5$

$y = \sqrt{3x-5} \implies y^2 = 3x - 5 \implies 3x = y^2 + 5 \implies x = \frac{y^2 + 5}{3}$.

Thus, $g^{-1}(x) = \frac{x^2 + 5}{3}$, domain $x \geq 0$.

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(e) Integrate:
(i) $\int_{1}^{3} (2x + 6) \, dx$
(ii) $\int_{0}^{\infty} e^{-x} \, dx$

Answer:
(i)

$$\int (2x + 6) \, dx = x^2 + 6x + C$$

Evaluate:

$$\left[ x^2 + 6x \right]_{1}^{3} = (9 + 18) - (1 + 6) = 27 - 7 = \boxed{20}$$

(ii) Improper integral:

$${\int }_0^{\mathrm{\infty }}{e}^{-x}dx=\underset{b\to \mathrm{\infty }}{\lim }{\int }_0^b{e}^{-x}dx=\underset{b\to \mathrm{\infty }}{\lim }{[-e^{-x}]}_0^b=\underset{b\to \mathrm{\infty }}{\lim }(-e^{-b}-(-e^0))=\underset{b\to \mathrm{\infty }}{\lim }(-e^{-b}+1)=0+1=\boxed{{\displaystyle 1}}$$

$$\boxed{{\displaystyle \mathrm{}}}$$

$$\boxed{{\displaystyle \mathrm{OR\; WE\; DO\; THIS\; NIGERS}}}$$

🧮 (i) Definite Integral:

${\int }_1^3(2x+6)dx$

Step 1: Integrate the expression

$\int (2x+6)dx=x^2+6x+C$

Step 2: Apply limits from 1 to 3

$={[x^2+6x]}_1^3=(3^2+6\cdot 3)-(1^2+6\cdot 1)$

$=(9+18)-(1+6)$

$=27-7$

$\mathrm{}=\overline{)20}$

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🧮 (ii) Improper Integral:

${\int }_0^{\infty }{e}^{-x}dx$

This is a standard exponential decay integral.

Step 1: Integrate

$\int e^{-x}dx=-e^{-x}+C$

Step 2: Apply limits from 0 to ∞

$={[-e^{-x}]}_0^{\infty }=\underset{b\to \infty }{\lim }(-e^{-b})-(-e^0)$

$=0-(-1)$

$=\overline{)1}$

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✅ Final Answers:

Part	Integral Expression	Result
(i)	${\int }_1^3(2x+6)dx$	$\overline{)20}$
(ii)	${\int }_0^{\infty }{e}^{-x}dx$	$\overline{)1}$

$$\int_{0}^{\infty} e^{-x} \, dx = \lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx = \lim_{b \to \infty} \left[ -e^{-x} \right]_{0}^{b} = \lim_{b \to \infty} \left( -e^{-b} - (-e^{0}) \right) = \lim_{b \to \infty} ( -e^{-b} + 1 ) = 0 + 1 = \boxed{1}$$

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Question Five

(a) State the first principle of differentiation.

Answer:
The derivative of $f(x)$ is:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

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(b) Prove the identities:
(i) $\frac{1 + \sin x}{1 - \sin x} = (\sec x + \tan x)^2$

(ii) $\frac{\sin^2 x (\sec x - \csc x)}{\cos x \tan x} = \tan x - 1$

Answer:
(i) Right-hand side:

$$(\sec x + \tan x)^2 = \left( \frac{1}{\cos x} + \frac{\sin x}{\cos x} \right)^2 = \left( \frac{1 + \sin x}{\cos x} \right)^2 = \frac{(1 + \sin x)^2}{\cos^2 x}$$

Left-hand side:

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

Thus, LHS = RHS.

(ii) Simplify left-hand side:

$$\text{Numerator: }{\sin }^{2}x(\sec x-\csc x)={\sin }^{2}x(\frac{1}{\cos x}-\frac{1}{\sin x})$$

$$={\sin }^{2}x\cdot \frac{\sin x-\cos x}{\sin x\cos x}$$

$$=\sin xX(\sin x-\cos x)\mathrm{/}\cos x$$

$$\text{Numerator: } \sin^2 x (\sec x - \csc x) = \sin^2 x \left( \frac{1}{\cos x} - \frac{1}{\sin x} \right) = \sin^2 x \cdot \frac{\sin x - \cos x}{\sin x \cos x} = \sin x \cdot (\sin x - \cos x) / \cos x = \sin x \tan x - \sin x$$

Better:

$$\frac{{\sin }^{2}x(\sec x-\csc x)}{\cos x\tan x}=\frac{{\sin }^{2}x(\frac{1}{\cos x}-\frac{1}{\sin x})}{\cos xX\frac{\sin x}{\cos x}}$$

$$=\frac{{\sin }^{2}x\left(\frac{\sin x-\cos x}{\sin x\cos x}\right)}{\sin x}$$

$$=\frac{\sin x(\sin x-\cos x)}{\sin x}$$

$$\frac{\sin^2 x (\sec x - \csc x)}{\cos x \tan x} = \frac{\sin^2 x \left( \frac{1}{\cos x} - \frac{1}{\sin x} \right)}{\cos x \cdot \frac{\sin x}{\cos x}} = \frac{\sin^2 x \left( \frac{\sin x - \cos x}{\sin x \cos x} \right)}{\sin x} = \frac{\sin x (\sin x - \cos x)}{\sin x} = \sin x - \cos x$$

Mistake. Correct:
Denominator: $\cos x \tan x = \cos x \cdot \frac{\sin x}{\cos x} = \sin x$.
Numerator: $\sin^2 x \left( \frac{1}{\cos x} - \frac{1}{\sin x} \right) = \sin^2 x \cdot \frac{\sin x - \cos x}{\sin x \cos x} = \sin x \cdot \frac{\sin x - \cos x}{\cos x} = \sin x (\tan x - 1)$.
Thus,

$$\frac{\sin x (\tan x - 1)}{\sin x} = \tan x - 1$$

Proved.

LETS ALSO USE THIS METHOD 

✅ (i) Prove:

$\frac{1+\sin x}{1-\sin x}=(\sec x+\tan x{)}^2$

✏️ Step 1: Expand RHS

Recall:

• $\sec x=\frac{1}{\cos x}$
• $\tan x=\frac{\sin x}{\cos x}$

So:

$\mathrm{(sec}x+\tan x{)}^2={\left(\frac{1+\sin x}{\cos x}\right)}^2=\frac{(1+\sin x{)}^2}{{\cos }^{2}x}$

✏️ Step 2: Expand numerator

$(1+\sin x{)}^2=1+2\sin x+{\sin }^{2}x$

So RHS becomes:

$\frac{1+2\sin x+{\sin }^{2}x}{{\cos }^{2}x}$

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✏️ Step 3: Simplify LHS

Start with:

$\frac{1+\sin x}{1-\sin x}$

Multiply numerator and denominator by $1+\sin x$ (rationalizing):

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

Recall:

• $1-{\sin }^{2}x={\cos }^{2}x$

So LHS becomes:

$\frac{(1+\sin x{)}^2}{{\cos }^{2}x}$

✅ LHS = RHS, hence:

$\overline{)\frac{1+\sin x}{1-\sin x}=(\sec x+\tan x{)}^2}$

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✅ (ii) Prove:

$\frac{\sin 2x(\sec x-\csc x)}{\cos x\tan x}=\tan x-1$

✏️ Step 1: Use identities

• $\sin 2x=2\sin x\cos x$
• $\sec x=\frac{1}{\cos x}$
• $\csc x=\frac{1}{\sin x}$
• $\tan x=\frac{\sin x}{\cos x}$

Substitute into LHS:

$\frac{2\sin x\cos x(\frac{1}{\cos x}-\frac{1}{\sin x})}{\cos x\cdot \frac{\sin x}{\cos x}}$

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✏️ Step 2: Simplify numerator

$2\sin x\cos x(\frac{1}{\cos x}-\frac{1}{\sin x})=2\sin x\cos x\left(\frac{\sin x-\cos x}{\sin x\cos x}\right)=2(\sin x-\cos x)$

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✏️ Step 3: Simplify denominator

$\cos x\cdot \frac{\sin x}{\cos x}=\sin x$

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✏️ Step 4: Final simplification

$\frac{2(\sin x-\cos x)}{\sin x}=2(1-\frac{\cos x}{\sin x})=2(1-\cot x)$

But this does not match the RHS $\tan x-1$ unless there's a typo in the original identity.

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⚠️ Possible Issue:

The original expression:

$\frac{\sin 2x(\sec x-\csc x)}{\cos x\tan x}$

simplifies to:

$2(1-\cot x)$

not $\tan x-1$

So unless the RHS was meant to be $2(1-\cot x)$, the identity as written is not valid.

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(c) Solve $8 \cos^2 x + 2 \cos x - 1 = 0$ for $0 \leq x \leq 2\pi$.

Answer:
Let $u = \cos x$:

$$8u^2 + 2u - 1 = 0$$

Quadratic formula:

$$u=\frac{-2\pm \sqrt{4+32}}{16}$$

$$=\frac{-2\pm \sqrt{36}}{16}$$

$$u = \frac{ -2 \pm \sqrt{4 + 32} }{16} = \frac{ -2 \pm \sqrt{36} }{16} = \frac{ -2 \pm 6 }{16}$$

Solutions:

$$u = \frac{4}{16} = \frac{1}{4}, \quad u = \frac{-8}{16} = -\frac{1}{2}$$

Thus,

$\cos x = \frac{1}{4} \implies x = \cos^{-1}(0.25) \approx 1.318$ rad, and $x = 2\pi - 1.318 \approx 4.965$ rad.

$\cos x = -\frac{1}{2} \implies x = \frac{2\pi}{3}, \frac{4\pi}{3}$.
All solutions:

$$\boxed{x = \frac{2\pi}{3},\ \frac{4\pi}{3},\ \cos^{-1}(0.25),\ 2\pi - \cos^{-1}(0.25)}$$

Exact: $x = \frac{2\pi}{3}, \frac{4\pi}{3}, \arccos\left(\frac{1}{4}\right), 2\pi - \arccos\left(\frac{1}{4}\right)$.

OR LETS DO THIS GUYS

✅ Given:

$8{\cos }^{2}x+2\cos x-1=0\text{for}0\le x\le 2\pi$

Let:

$u=\cos x$

Then the equation becomes:

$8u^2+2u-1=0$

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✏️ Step 1: Solve the quadratic

Use the quadratic formula:

$u=\frac{-2\pm \sqrt{(2{)}^2-4(8\left)\right(-1)}}{2\left(8\right)}$

$=\frac{-2\pm \sqrt{4+32}}{16}$

$=\frac{-2\pm \sqrt{36}}{16}$ $$

$u=\frac{-2\pm 6}{16}$

So:

• $u=\frac{4}{16}=\frac{1}{4}$
• $u=\frac{-8}{16}=-\frac{1}{2}$

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✏️ Step 2: Solve for $x$

Case 1: $\cos x=\frac{1}{4}$

This occurs in Quadrants I and IV:

$x={\cos }^{-1}\left(\frac{1}{4}\right)\approx 1.3181\text{(radians)}$

$$ $x=2\pi -1.3181\approx 4.9651$

Case 2: $\cos x=-\frac{1}{2}$

This occurs in Quadrants II and III:

$x={\cos }^{-1}(-\frac{1}{2})=\frac{2\pi }{3}\approx 2.0944$

$\mathrm{}$ $x=\frac{4\pi }{3}\approx 4.1888$

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✅ Final Answer:

$\overline{)x\approx 1.3181,2.0944,4.1888,4.9651}\text{(in radians, within }[0,2\pi ]\text{)}$

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(d) Let $A = \{1, 2, 3, 4\}$, $f = \{(1,4), (2,1), (3,3), (4,2)\}$, $g = \{(1,3), (2,1), (3,2), (4,4)\}$. Find:
(i) $f \circ g$
(ii) $g \circ f$

Answer:
Composition: $(f \circ g)(x) = f(g(x))$.
(i) $f \circ g$:

• $g(1)=3 \to f(3)=3$
• $g(2)=1 \to f(1)=4$
• $g(3)=2 \to f(2)=1$
• $g(4)=4 \to f(4)=2$
  So, $f \circ g = \{(1,3), (2,4), (3,1), (4,2)\}$.

(ii) $g \circ f$:

• $f(1)=4 \to g(4)=4$
• $f(2)=1 \to g(1)=3$
• $f(3)=3 \to g(3)=2$
• $f(4)=2 \to g(2)=1$
  So, $g \circ f = \{(1,4), (2,3), (3,2), (4,1)\}$.

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(e) Integrate:
(i) $\int \cos x \cdot e^{\sin x} \, dx$
(ii) $\int x e^x \, dx$

Answer:
(i) Substitution: Let $u = \sin x$, $du = \cos x \, dx$.

$$\int \cos x \cdot e^{\sin x} \, dx = \int e^u \, du = e^u + C = \boxed{e^{\sin x} + C}$$

(ii) Integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x$, $dv = e^x dx$ $\implies du = dx$, $v = e^x$.

$$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C=\boxed{{\displaystyle e^x(x-1)+C}}$$

$$\boxed{{\displaystyle }}$$

$$\boxed{{\displaystyle \mathrm{OR\; WE\; DO\; THIS\; WAY}}}$$

✅ (i) $\int \cos x\cdot e^{\sin x}dx$

Let’s use substitution:

Let:

$u=\sin x\Rightarrow \frac{du}{dx}=\cos x\Rightarrow du=\cos xdx$

So the integral becomes:

$\int e^{\sin x}\cdot \cos xdx=\int e^{u}du=e^u+C=e^{\sin x}+C$

✅ Answer:

$\overline{)\int \cos x\cdot e^{\sin x}dx=e^{\sin x}+C}$

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✅ (ii) $\int x{e}^{x}dx$

Use integration by parts:

Let:

• $u=x\Rightarrow du=dx$
• $dv=e^{x}dx\Rightarrow v=e^x$

Apply the formula:

$\int udv=uv-\int vdu$

So:

$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C$

✅ Answer:

$\overline{)\int x{e}^{x}dx=(x-1)e^x+C}$

$$\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C = \boxed{e^x (x - 1) + C}$$

@Dr. Microbiota

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END OF SOLUTIONS