INTRODUCTION TO MATHEMATICAL METHODS

SEMESTER ONE EXAMINATION JULY INTAKE 2016

DATE : 9th NOVEMBER, 2016

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Question One

(a) Define the following
i. Cardinality of a set: The number of elements in a set, denoted as $|A|$ for a set $A$.
ii. Domain of a function: The set of all input values (independent variables) for which the function is defined.

(b) Given that $X \subset Y$ are any sets. Simplify the following
i. $(X \cup Y) \cap (X \cup Y)$
Answer: $X \cup Y$
Explanation:

• Intersection of a set with itself is the set: $(A \cap A = A)$.
• Given $X \subset Y$, $X \cup Y = Y$.
• Thus, $(X \cup Y) \cap (X \cup Y) = X \cup Y = Y$.

ii. $Y \cap (Y - X)^c$
Answer: $X$
Explanation:

• $Y - X = Y \cap X^c$.
• $(Y - X)^c = (Y \cap X^c)^c = Y^c \cup X$ (De Morgan's Law).
• $Y \cap (Y^c \cup X) = (Y \cap Y^c) \cup (Y \cap X) = \emptyset \cup (Y \cap X)$.
• Since $X \subset Y$, $Y \cap X = X$.
• Thus, $Y \cap (Y - X)^c = X$.

(c) Prove the following identity
i. $\sqrt{\frac{1 + \sin x}{1 - \sin x}} = \sec x + \tan x$
Proof:

• Let’s prove the identity:

  $\sqrt{\frac{1+\sin x}{1-\sin x}}=\sec x+\tan x$

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  ✏️ Step 1: Express RHS in terms of sine and cosine

  Recall:

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

  So:

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

  Now square both sides:

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

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  ✏️ Step 2: Simplify LHS

  Start with:

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

  Square both sides:

  ${\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)}^2=\frac{1+\sin x}{1-\sin x}$

  Now rationalize the denominator:

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

  Recall:

  • $1-{\sin }^{2}x={\cos }^{2}x$

  So:

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

  Which matches the squared RHS.

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  ✅ Final Step: Take square root of both sides

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

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  📌 Proven:

  $\overline{)\sqrt{\frac{1+\sin x}{1-\sin x}}=\sec x+\tan x}$

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

• Left-Hand Side (LHS):

$$\sec x - \csc x = \frac{1}{\cos x} - \frac{1}{\sin x} = \frac{\sin x - \cos x}{\sin x \cos x}.$$ $${}^{}$$

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

• Denominator: $\cos x \tan x = \cos x \cdot \frac{\sin x}{\cos x} = \sin x$.
• Thus,

$$\frac{\sin x \tan x - \sin x}{\sin x} = \tan x - 1.$$

• LHS = RHS.

(d) Solve $8 \cos^2 x + 2 \cos x - 1 = 0$ for $0 \leq x \leq 2\pi$
Answer: $x = \cos^{-1}\left(\frac{1}{4}\right), \quad 2\pi - \cos^{-1}\left(\frac{1}{4}\right), \quad \frac{2\pi}{3}, \quad \frac{4\pi}{3}$
Explanation:

• Let $u = \cos x$: $8u^2 + 2u - 1 = 0$.
• Quadratic formula: $u = \frac{-2 \pm \sqrt{4 + 32}}{16} = \frac{-2 \pm 6}{16}$.
• Solutions: $u = \frac{1}{4}$ or $u = -\frac{1}{2}$.
• Case 1: $\cos x = \frac{1}{4}$:
  • $x = \cos^{-1}\left(\frac{1}{4}\right)$ (in quadrant I, approx 1.318 rad).
  • $x = 2\pi - \cos^{-1}\left(\frac{1}{4}\right)$ (in quadrant IV).
• Case 2: $\cos x = -\frac{1}{2}$:
  • $x = \frac{2\pi}{3}$ (quadrant II).
  • $x = \frac{4\pi}{3}$ (quadrant III).
• Solutions: $x \approx 1.318, \quad 4.965, \quad \frac{2\pi}{3}, \quad \frac{4\pi}{3}$ (or exact forms).

FOR UNDERSTANDING

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

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✏️ Step 1: Let $u=\cos x$

This transforms the equation into a quadratic in $u$:

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

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✏️ Step 2: Solve the quadratic

Use the quadratic formula:

$u=\frac{-2\pm \sqrt{2^2-4\left(8\right)(-1)}}{2\left(8\right)}=\frac{-2\pm \sqrt{4+32}}{16}=\frac{-2\pm \sqrt{36}}{16}$ $u=\frac{-2\pm 6}{16}$

So:

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

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✏️ Step 3: Solve for $x$

Now solve $\cos x=\frac{1}{4}$ and $\cos x=-\frac{1}{2}$

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:

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

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✅ Final Answer:

$\overline{)x\approx 1.3181,4.9651,\frac{2\pi }{3},\frac{4\pi }{3}}$

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Question Two

(a) Define the following
i. Composite function of two functions: For functions $f$ and $g$, the composite $f \circ g$ is defined as $(f \circ g)(x) = f(g(x))$, where $g(x)$ is in the domain of $f$.
ii. Equality of two sets: Sets $A$ and $B$ are equal ($A = B$) if they contain exactly the same elements, i.e., $A \subseteq B$ and $B \subseteq A$.

(b) Given that $X \subset Y$ are sets. Prove the following
i. $(X \cup Y) \cap (X \cup Y) = X$
Note: The given equation is incorrect. Counterexample: $X = \{1\}, Y = \{1,2\}$, then $(X \cup Y) \cap (X \cup Y) = \{1,2\} \neq X$.
Correction: Likely intended as $(X \cap Y) \cup (X \cap Y) = X \cap Y = X$ (since $X \subset Y$).

ii. $X \cap (Y - X) = \emptyset$
Proof:

• $Y - X = Y \cap X^c$.
• $X \cap (Y \cap X^c) = (X \cap X^c) \cap Y = \emptyset \cap Y = \emptyset$.

(c) Given $f(x) = \frac{1}{x+2}$ ($x \neq -2$) and $g(x) = x + 2$. Find
i. $f \circ g(x)$
Answer: $\frac{1}{x+4}$
Explanation:

• $f \circ g(x) = f(g(x)) = f(x+2) = \frac{1}{(x+2)+2} = \frac{1}{x+4}$.
• Domain: $x \neq -4$.

ii. $f \circ g(1)$
Answer: $\frac{1}{5}$
Explanation:

• $f \circ g(1) = \frac{1}{1+4} = \frac{1}{5}$.

iii. Existence of inverse of $f \circ g(x)$
Answer: Inverse exists.
Explanation:

• $f \circ g(x) = \frac{1}{x+4}$ is one-to-one (strictly decreasing for $x > -4$ and $x < -4$).
• Bijective from $\mathbb{R} \setminus \{-4\}$ to $\mathbb{R} \setminus \{0\}$, so inverse exists.

(d) Use chain rule to differentiate
i. $y = (x^3 + 2x)^6$
Answer: $\frac{dy}{dx} = 6(x^3 + 2x)^5 (3x^2 + 2)$
Explanation:

• Let $u = x^3 + 2x$, $y = u^6$.
• $\frac{dy}{dx} = 6u^5 \cdot (3x^2 + 2) = 6(x^3 + 2x)^5 (3x^2 + 2)$.

ii. $f(x) = (6x - 3x^3 - 2)^{5/6}$
Answer: $f'(x) = \frac{5}{2}(2 - 3x^2)(6x - 3x^3 - 2)^{-1/6}$
Explanation:

• Let $u = 6x - 3x^3 - 2$, $f(x) = u^{5/6}$.
• $f'(x) = \frac{5}{6}u^{-1/6} \cdot (6 - 9x^2) = \frac{5}{6}(6 - 9x^2)(6x - 3x^3 - 2)^{-1/6}$.
• Simplify: $\frac{5}{6} \cdot 3(2 - 3x^2) u^{-1/6} = \frac{5}{2}(2 - 3x^2)(6x - 3x^3 - 2)^{-1/6}$.

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Question Three

(a) State the following
i. Remainder theorem: If polynomial $f(x)$ is divided by $x - c$, the remainder is $f(c)$.
ii. De Morgan’s law: For sets, $(X \cup Y)' = X' \cap Y'$ and $(X \cap Y)' = X' \cup Y'$.

(b)
i. Remainder of $f(x) = x^3 + 3x^2 + x + 10$ divided by $x + 2$
Answer: 12
Explanation:

• $x + 2 = x - (-2)$, so remainder $f(-2) = (-2)^3 + 3(-2)^2 + (-2) + 10 = -8 + 12 - 2 + 10 = 12$.

ii. Show $(X \cup Y)' = X' \cap Y'$
Proof:

• Let $a \in (X \cup Y)'$ → $a \notin X \cup Y$ → $a \notin X$ and $a \notin Y$ → $a \in X'$ and $a \in Y'$ → $a \in X' \cap Y'$.
• Let $a \in X' \cap Y'$ → $a \in X'$ and $a \in Y'$ → $a \notin X$ and $a \notin Y$ → $a \notin X \cup Y$ → $a \in (X \cup Y)'$.
• Thus, $(X \cup Y)' = X' \cap Y'$.

(c) Given $f(x) = x^3 - x^2 - 4x + 4$
i. Factorise completely
Answer: $f(x) = (x-1)(x-2)(x+2)$
Explanation:

• Roots: Possible rational roots $\pm1, \pm2, \pm4$.
• $f(1) = 1 - 1 - 4 + 4 = 0$, so $x-1$ is a factor.
• Synthetic division:

  1 | 1  -1  -4  4
        1    0  -4
    -----------------
        1   0  -4 | 0
  

• Quotient $x^2 - 4 = (x-2)(x+2)$.
• Thus, $f(x) = (x-1)(x-2)(x+2)$.

ii. Solutions to $f(x) = 0$
Answer: $x = -2, 1, 2$
Explanation:

• Set factors to zero: $x-1=0$, $x-2=0$, $x+2=0$.

iii. Sketch graph

• Description:
  • Roots at $x = -2, 1, 2$, y-intercept $f(0) = 4$.
  • Behavior:
    • $x < -2$: $f(x) < 0$ (since all factors negative).
    • $-2 < x < 1$: $f(x) > 0$ (even number of negative factors).
    • $1 < x < 2$: $f(x) < 0$ (odd number of negative factors).
    • $x > 2$: $f(x) > 0$ (all factors positive).
  • Cubic with positive leading coefficient: $\to -\infty$ as $x \to -\infty$, $\to \infty$ as $x \to \infty$.
  • Sketch crosses x-axis at $(-2,0)$, $(1,0)$, $(2,0)$, and y-axis at $(0,4)$.

iv. Solution to $x^3 - x^2 - 4x + 4 \geq 0$
Answer: $[-2, 1] \cup [2, \infty)$
Explanation:

• Sign analysis:
  • $f(x) \geq 0$ for $x \in [-2, 1] \cup [2, \infty)$.

(d) Use quotient rule to differentiate
i. $y = \frac{\sqrt{x}}{x^3 + 1} = \frac{x^{1/2}}{x^3 + 1}$
Answer: $y' = \frac{1 - 5x^3}{2\sqrt{x}(x^3 + 1)^2}$
Explanation:

• Quotient rule: $y' = \frac{u'v - uv'}{v^2}$, $u = x^{1/2}$, $v = x^3 + 1$.
• $u' = \frac{1}{2}x^{-1/2}$, $v' = 3x^2$.
• Numerator: $\frac{1}{2}{x}^{-1\mathrm{/}2}(x^3+1)-x^{1\mathrm{/}2}X3x^2$
• $=\frac{1}{2}{x}^{5\mathrm{/}2}+\frac{1}{2}{x}^{-1\mathrm{/}2}-3x^{5\mathrm{/}2}$
• $\frac{1}{2}x^{-1/2}(x^3 + 1) - x^{1/2} \cdot 3x^2 = \frac{1}{2}x^{5/2} + \frac{1}{2}x^{-1/2} - 3x^{5/2} = -\frac{5}{2}x^{5/2} + \frac{1}{2}x^{-1/2}$.
• Simplify: $\frac{1}{2}x^{-1/2}(1 - 5x^3)$.
• $y^{\mathrm{\prime }}=\frac{\frac{1}{2}{x}^{-1\mathrm{/}2}(1-5x^3)}{(x^3+1{)}^2}$
• $y' = \frac{\frac{1}{2}x^{-1/2}(1 - 5x^3)}{(x^3 + 1)^2} = \frac{1 - 5x^3}{2\sqrt{x}(x^3 + 1)^2}$.

ii. $g(x) = \frac{x^2 + 3}{x^3 + x}$
Answer: $g'(x) = \frac{-x^4 - 8x^2 - 3}{(x^3 + x)^2}$
Explanation:

• Quotient rule: $u = x^2 + 3$, $v = x^3 + x$.
• $u' = 2x$, $v' = 3x^2 + 1$.
• Numerator: $2x(x^3+x)-(x^2+3)(3x^2+1)$

$=2x^4+2x^2-(3x^4+x^2+9x^2+3)$

$=2x^4+2x^2-3x^4-10x^2-3$

$2x(x^3 + x) - (x^2 + 3)(3x^2 + 1) = 2x^4 + 2x^2 - (3x^4 + x^2 + 9x^2 + 3) = 2x^4 + 2x^2 - 3x^4 - 10x^2 - 3 = -x^4 - 8x^2 - 3$.

• $g'(x) = \frac{-x^4 - 8x^2 - 3}{(x^3 + x)^2}$.

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Question Four

(a) Define the following
i. Many to one function: A function where two or more distinct domain elements map to the same range element (e.g., $f(x_1) = f(x_2)$ for $x_1 \neq x_2$).
ii. Exponential function: A function of the form $f(x) = a^x$ or $f(x) = b \cdot a^{kx}$ ($a > 0$, $a \neq 1$).

(b) Determine if function is one-to-one, many-to-one, or neither. Find inverse if exists.
i. $f(x) = \frac{3}{3 + 2x}, \, x \neq -\frac{3}{2}$
Answer: One-to-one; inverse $f^{-1}(x) = \frac{3(1-x)}{2x}$
Explanation:

• One-to-one: Assume $f(x_1) = f(x_2)$:

$$\frac{3}{3+2x_1}=\frac{3}{3+2x_2}$$

$$⟹3+2x_1=3+2x_2$$

$$\frac{3}{3+2x_1} = \frac{3}{3+2x_2} \implies 3+2x_1 = 3+2x_2 \implies x_1 = x_2.$$

• Inverse: Solve $y = \frac{3}{3+2x}$ for $x$:

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

$$⟹2xy=3-3y$$

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

• Thus, $f^{-1}(x) = \frac{3(1-x)}{2x}$, domain $x \neq 0$.

ii. $g(x) = x^2 - 4$
Answer: Many-to-one; no inverse (unless domain restricted).
Explanation:

• Many-to-one: $g(2) = g(-2) = 0$.
• No inverse over $\mathbb{R}$ (not bijective).
• If domain restricted to $x \geq 0$, inverse $g^{-1}(x) = \sqrt{x+4}$; if $x \leq 0$, $g^{-1}(x) = -\sqrt{x+4}$.

(c) Spread of virus: $y = \frac{1000}{1 + 999e^{-0.8t}}$, $t \geq 0$
i. Infected after 4 days
Answer: 24 students
Explanation:

• $t = 4$: $y = \frac{1000}{1 + 999e^{-3.2}}$.
• $e^{-3.2} \approx 0.04076$, so $999 \times 0.04076 \approx 40.719$.
• $1 + 40.719 = 41.719$, $y = \frac{1000}{41.719} \approx 23.97 \approx 24$.

ii. Campus closes when 40% infected (y=400). Find t.
Answer: 8.13 days
Explanation:

• Solve $400 = \frac{1000}{1 + 999e^{-0.8t}}$:

$$400(1+999e^{-0.8t})=1000$$

$$⟹999e^{-0.8t}=\frac{1000}{400}-1=1.5$$

$$⟹e^{-0.8t}=\frac{1.5}{999}$$

$$400(1 + 999e^{-0.8t}) = 1000 \implies 999e^{-0.8t} = \frac{1000}{400} - 1 = 1.5 \implies e^{-0.8t} = \frac{1.5}{999} = \frac{1}{666}.$$

• $-0.8t = \ln(1/666) = -\ln(666) \implies t = \frac{\ln(666)}{0.8}$.
• $\ln(666) \approx 6.501$, so $t \approx \frac{6.501}{0.8} = 8.126 \approx 8.13$ days.

(d) Use product rule to differentiate
i. $g(x) = (x^2 + 3)(x^3 - 4x)$
Answer: $g'(x) = 5x^4 - 3x^2 - 12$
Explanation:

• Product rule: $u = x^2 + 3$, $v = x^3 - 4x$.
• $u' = 2x$, $v' = 3x^2 - 4$.

$g^{\mathrm{\prime }}(x)=2x(x^3-4x)+(x^2+3)(3x^2-4)$

$=2x^4-8x^2+3x^4+5x^2-12$

$g'(x) = 2x(x^3 - 4x) + (x^2 + 3)(3x^2 - 4) = 2x^4 - 8x^2 + 3x^4 + 5x^2 - 12 = 5x^4 - 3x^2 - 12$.

ii. $g(x) = (6x + 3)(x^3 - 2x)$
Answer: $g'(x) = 24x^3 + 9x^2 - 24x - 6$
Explanation:

• Product rule: $u = 6x + 3$, $v = x^3 - 2x$.
• $u' = 6$, $v' = 3x^2 - 2$.

$g^{\mathrm{\prime }}(x)=6(x^3-2x)+(6x+3)(3x^2-2)$

$=6x^3-12x+18x^3+9x^2-12x-6$

$g'(x) = 6(x^3 - 2x) + (6x + 3)(3x^2 - 2) = 6x^3 - 12x + 18x^3 + 9x^2 - 12x - 6 = 24x^3 + 9x^2 - 24x - 6$.

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Question Five

(a) State when
i. Quadratic $ax^2 + bx + c = 0$ has real roots: Discriminant $D = b^2 - 4ac \geq 0$.
ii. $x - k$ is a factor of $p(x)$: $p(k) = 0$.

(b) Given $2\log_9 x + \log_3 y = 2$, show $xy = 9$. Solve with $\log_{10}(x+y) = 1$.
Proof for $xy = 9$:

• $2\log_9 x = 2 \cdot \frac{\log_3 x}{\log_3 9} = 2 \cdot \frac{\log_3 x}{2} = \log_3 x$.
• Thus, $\log_3 x + \log_3 y = \log_3(xy) = 2$, so $xy = 3^2 = 9$.
  Solve system:
• $\log_{10}(x+y) = 1 \implies x + y = 10$.
• $xy = 9$, so $x$ and $y$ are roots of $t^2 - 10t + 9 = 0$.
• Factors: $(t-1)(t-9) = 0$, so $t = 1$ or $t = 9$.
• Solutions: $(x,y) = (1,9)$ or $(9,1)$.

(c) Nature of roots without solving
i. $-x^2 + 8x + 16 = 0$
Answer: Two distinct real roots
Explanation:

• $D = b^2 - 4ac = 64 - 4(-1)(16) = 64 + 64 = 128 > 0$.

ii. $2x^2 + 7x - 5 = 0$
Answer: Two distinct real roots
Explanation:

• $D = 49 - 4(2)(-5) = 49 + 40 = 89 > 0$.

(d) First principles differentiation
i. $f(x) = x^3 - 2x$
Answer: $f'(x) = 3x^2 - 2$
Explanation:

• $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{[(x+h)^3 - 2(x+h)] - [x^3 - 2x]}{h}$.
• Expand: $\frac{[x^3+3x^{2}h+3x{h}^2+h^3-2x-2h]-x^3+2x}{h}$
• $=\frac{3x^{2}h+3x{h}^2+h^3-2h}{h}$
• $\frac{[x^3 + 3x^2h + 3xh^2 + h^3 - 2x - 2h] - x^3 + 2x}{h} = \frac{3x^2h + 3xh^2 + h^3 - 2h}{h} = 3x^2 + 3xh + h^2 - 2$.
• As $h \to 0$, $f'(x) = 3x^2 - 2$.

ii. $f(x) = \frac{1}{x}$
Answer: $f'(x) = -\frac{1}{x^2}$
Explanation:

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

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

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

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

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

(e) For $y = 5\sin\left(x + \frac{\pi}{2}\right)$
Answer:

• Amplitude: 5
• Period: $2\pi$
• Phase shift: $\frac{\pi}{2}$ left
• Graph sketch:
  • Since $\sin\left(x + \frac{\pi}{2}\right) = \cos x$, $y = 5\cos x$.
  • Key points:
    • $x = 0$: $y = 5$
    • $x = \frac{\pi}{2}$: $y = 0$
    • $x = \pi$: $y = -5$
    • $x = \frac{3\pi}{2}$: $y = 0$
    • $x = 2\pi$: $y = 5$
  • Cosine wave with amplitude 5, period $2\pi$, starting at max.
  • 

@Dr. Microbiota

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