INTRODUCTION TO MATHEMATICAL METHODS FINAL EXAM JULY INTAKE 2023

DATE : 17th November, 2023

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QUESTION ONE

a) Define
(i) A Power set
(ii) Disjoint set

Answer:
(i) The power set of a set $S$ is the set of all possible subsets of $S$, including the empty set and $S$ itself. It is denoted by $P(S)$.
(ii) Two sets are disjoint if they have no elements in common, i.e., their intersection is the empty set.

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b) Let $U = [-4, 12]$ be the universal set and set $A = [-2, 7)$, $B = (5, 10)$ and $C = [2, 8]$. Find
(i) $(A \cup B) \cap C^c$
(ii) $B - A$

Answer:
(i)

• $A \cup B = [-2, 10)$
• $C^c = [-4, 2) \cup (8, 12]$
• $(A \cup B) \cap C^c = [-2, 2) \cup (8, 10)$
  

(ii)

• $B - A = [7, 10)$

Explanation:
(i)

• Union $A \cup B$: $A = [-2, 7)$, $B = (5, 10)$. The union combines intervals: from $-2$ to $10$,

excluding $10$ (since $7$ is excluded in $A$, but $B$ includes values up to but not including $10$,

and $5$ is included via $A$). Thus, $A \cup B = [-2, 10)$.

• Complement $C^c$: $C = [2, 8]$, so $C^c$ in $U = [-4, 12]$ is $[-4, 2) \cup (8, 12]$.
• Intersection: $(A \cup B) \cap C^c$ is the overlap between $[-2, 10)$ and $[-4, 2) \cup (8, 12]$. This gives $[-2, 2) \cup (8, 10)$.

(ii)

• Set difference $B - A = B \cap A^c$.
• $A^c = [-4, -2) \cup [7, 12]$ (complement of $A$ in $U$).
• $B = (5, 10)$, so $B \cap A^c = (5, 10) \cap [7, 12] = [7, 10)$ (since $5$ and $10$ are excluded in $B$, but $7$ is included).

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c) Given that $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, Let set $A = \{1, 3, 4, 5, 7\}$ and $B = \{2, 3, 4, 6, 8\}$ and $C = \{1, 2, 3\}$ find
(i) $(A \cap B) \cup C^c$
(ii) $(B \cap A)^c \cap C$

Answer:
(i) $(A \cap B) \cup C^c = \{3, 4, 5, 6, 7, 8, 9, 10\}$
(ii) $(B \cap A)^c \cap C = \{1, 2\}$

Explanation:
(i)

• $A \cap B = \{3, 4\}$ (common elements).
• $C^c = U - C = \{4, 5, 6, 7, 8, 9, 10\}$.
• Union: $\{3, 4\} \cup \{4, 5, 6, 7, 8, 9, 10\} = \{3, 4, 5, 6, 7, 8, 9, 10\}$.

(ii)

• $B \cap A = \{3, 4\}$ (same as $A \cap B$).
• $(B \cap A)^c = U - \{3, 4\} = \{1, 2, 5, 6, 7, 8, 9, 10\}$.
• Intersection with $C = \{1, 2, 3\}$: $\{1, 2, 5, 6, 7, 8, 9, 10\} \cap \{1, 2, 3\} = \{1, 2\}$.

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d) For each of the following
(i) Write the following in roster form. $A = \{x \mid x \text{ is an integer and } x^2 - x - 6 \text{ is an odd number}\}$
(ii) Let set $A = \{1, 5\}$ and $B = \{2, 3\}$, find $P(A \times B)$
(iii) Let set $A = \{1, 2\}$ and $B = \{3\}$, verify $P(A) \cup P(B) \subset P(A \cup B)$

Answer:
(i) $A = \emptyset$ (empty set).

(ii) $P(A\times B)=\{\varnothing ,\{(1,2)\},\{(1,3)\},\{(5,2)\},\{(5,3)\},\{(1,2),(1,3)\},\{(1,2),(5,2)\},\{(1,2),(5,3)\},\{(1,3),$

$(5,2)\},\{(1,3),(5,3)\},\{(5,2),(5,3)\},\{(1,2),(1,3),(5,2)\},\{(1,2),(1,3),(5,3)\},\{(1,2),(5,2),(5,3)\},$

$\left\{\right(1,3),(5,2),(5,3\left)\right\},\left\{\right(1,2),(1,3),(5,2),(5,3\left)\right\}\}$.

(iii) $P(A) \cup P(B) \subset P(A \cup B)$ holds true.

Explanation:
(i)

• For any integer $x$, $x^2 - x = x(x-1)$ is even (product of consecutive integers),

so $x^2 - x - 6$ is even. Thus, no integers satisfy the condition, and $A = \emptyset$.

(ii)

• $A \times B = \{(1,2), (1,3), (5,2), (5,3)\}$.
• Power set $P(A \times B)$ has $2^4 = 16$ subsets, listed above.

(iii)

• $A \cup B = \{1, 2, 3\}$.
• $P(A) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$, $P(B) = \{\emptyset, \{3\}\}$, so $P(A) \cup P(B) = \{\emptyset, \{1\}, \{2\}, \{1,2\}, \{3\}\}$.
• $P(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$.
• All elements of $P(A) \cup P(B)$ are in $P(A \cup B)$, so the subset relation holds.

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e) Write the given expression in terms of $\log 2$, $\log 5$ and $\log 7$ to any base
$\log \left( \frac{\sqrt[3]{16} \times \sqrt[4]{125}}{\sqrt[8]{49}} \right)$

Answer:
$\frac{4}{3} \log 2 + \frac{3}{4} \log 5 - \frac{1}{4} \log 7$

Explanation:

• Simplify radicals:
  • $\sqrt[3]{16} = 16^{1/3} = (2^4)^{1/3} = 2^{4/3}$.
  • $\sqrt[4]{125} = 125^{1/4} = (5^3)^{1/4} = 5^{3/4}$.
  • $\sqrt[8]{49} = 49^{1/8} = (7^2)^{1/8} = 7^{1/4}$.
• Expression becomes $\log \left( \frac{2^{4/3} \cdot 5^{3/4}}{7^{1/4}} \right) = \log \left( 2^{4/3} \cdot 5^{3/4} \cdot 7^{-1/4} \right)$.
• Apply logarithm properties:
  $\log( a \cdot b \cdot c ) = \log a + \log b + \log c, \quad \log(a^b) = b \log a$
  $$
• $= \frac{4}{3} \log 2 + \frac{3}{4} \log 5 - \frac{1}{4} \log 7$.

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QUESTION TWO

a) Define
(i) A binary operation
(ii) An even function

Answer:
(i) A binary operation on a set $S$ is a function $* : S \times S \to S$ that assigns to each ordered pair $(a, b) \in S \times S$ an element $a * b \in S$.
(ii) A function $f$ is even if $f(-x) = f(x)$ for all $x$ in its domain. The graph is symmetric about the y-axis.

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b) For each of the following
(i) Simplify $[(A \cup B)^c \cap (A \cup B^c)]^c$
(ii) Use the Venn diagram to represent the sets $(A - B)^c \cap C$

Answer:
(i) $A \cup B$
(ii) Shade the region corresponding to $C$ except the part in $A$ but not in $B$.

Explanation:
(i)

• Apply set identities:
  $[(A \cup B)^c \cap (A \cup B^c)]^c = [ (A^c \cap B^c) \cap (A \cup B^c) ]^c$
  • Distribute: $A^c \cap B^c \cap A = \emptyset$, and $A^c \cap B^c \cap B^c = A^c \cap B^c$, so $\emptyset \cup (A^c \cap B^c) = A^c \cap B^c$.
  • Then $(A^c \cap B^c)^c = A \cup B$ (De Morgan's law).
• Simplified form: $A \cup B$.

(ii)

• $A - B = A \cap B^c$, so $(A - B)^c = A^c \cup B$.
• $(A - B)^c \cap C = (A^c \cup B) \cap C = (A^c \cap C) \cup (B \cap C)$.
• In a Venn diagram for sets $A, B, C$, shade:
  • All of $C$ not in $A$ (i.e., $C \cap A^c$).
  • All of $C$ in $B$ (i.e., $B \cap C$).
• This is equivalent to shading $C$ except the region where $C$ is in $A$ but not in $B$ (i.e., $C \cap A \cap B^c$).

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c) Rationalize the denominators for each of the following
(i) $\frac{1 + \sqrt{5}}{1 - 3\sqrt{5}}$

(ii) $\frac{\sqrt{7} + 1}{\sqrt{7} - 2}$

Answer:
(i) $-\frac{4 + \sqrt{5}}{11}$

(ii) $3 + \sqrt{7}$

Explanation:
(i)

• Multiply numerator and denominator by conjugate $1 + 3\sqrt{5}$:

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

$=\frac{1x1+1x3\sqrt{5}+\sqrt{5}x1+\sqrt{5}x3\sqrt{5}}{1-45}$

$=\frac{1+3\sqrt{5}+\sqrt{5}+15}{-44}$

$\frac{(1 + \sqrt{5})(1 + 3\sqrt{5})}{(1)^2 - (3\sqrt{5})^2} = \frac{1 \cdot 1 + 1 \cdot 3\sqrt{5} + \sqrt{5} \cdot 1 + \sqrt{5} \cdot 3\sqrt{5}}{1 - 45} = \frac{1 + 3\sqrt{5} + \sqrt{5} + 15}{-44} = \frac{16 + 4\sqrt{5}}{-44}$
$$

$=-\frac{4(4+\sqrt{5})}{44}$

$= -\frac{4(4 + \sqrt{5})}{44} = -\frac{4 + \sqrt{5}}{11}$.

(ii)

• Multiply numerator and denominator by conjugate $\sqrt{7} + 2$:

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

$=\frac{\sqrt{7}x\sqrt{7}+\sqrt{7}x2+\mathrm{1\; x }\sqrt{7}+1x2}{7-4}$

$=\frac{7+2\sqrt{7}+\sqrt{7}+2}{3}$

$=\frac{9+3\sqrt{7}}{3}$

$\frac{(\sqrt{7} + 1)(\sqrt{7} + 2)}{(\sqrt{7})^2 - (2)^2} = \frac{\sqrt{7} \cdot \sqrt{7} + \sqrt{7} \cdot 2 + 1 \cdot \sqrt{7} + 1 \cdot 2}{7 - 4} = \frac{7 + 2\sqrt{7} + \sqrt{7} + 2}{3} = \frac{9 + 3\sqrt{7}}{3} = 3 + \sqrt{7}$.

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d) Express the following numbers in the form $\frac{p}{q}$ where p and q are integers with no common factors, $q \neq 0$.
(i) $2.03\overline{3}$ (repeating decimal)
(ii) $-0.\overline{3}$ (repeating decimal)

Answer:
(i) $\frac{61}{30}$
(ii) $-\frac{1}{3}$

Explanation:
(i)

• Let $x = 2.03333\ldots$
• Then $10x = 20.3333\ldots$ and $100x = 203.3333\ldots$.
• Subtract: $100x-10x=203.333-20.333$

so $90x = 183$, $$

$x=\frac{183}{90}$

$x = \frac{183}{90} = \frac{61}{30}$.

• GCD of 61 and 30 is 1, so $\frac{61}{30}$.

(ii)

• Let $x = -0.3333\ldots$
• Then $10x = -3.3333\ldots$.
• Subtract: $10x - x = -3.333\ldots - (-0.333\ldots) = -3$,
• so $9x = -3$, $$
• $x=-\frac{3}{9}$
• $x = -\frac{3}{9} = -\frac{1}{3}$.

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e) Find the domain of $f(x) = \frac{\sqrt{2x+3}}{x-3}$. Hence sketch its graph show all intercepts and asymptotic lines.

Answer:

• Domain: $\left[ -\frac{3}{2}, 3 \right) \cup (3, \infty)$.
• Intercepts: x-intercept at $\left( -\frac{3}{2}, 0 \right)$, y-intercept at $\left( 0, -\frac{\sqrt{3}}{3} \right)$.
• Asymptotes: Vertical asymptote at $x = 3$, horizontal asymptote at $y = 0$.
• Graph: See sketch description below.

Explanation:

• Domain:
  • Square root defined when $2x+3\ge 0$

  $2x + 3 \geq 0 \Rightarrow x \geq -\frac{3}{2}$.

  • Denominator nonzero when $x \neq 3$.
  • Thus, domain: $\left[ -\frac{3}{2}, 3 \right) \cup (3, \infty)$.
• Intercepts:
  • x-intercept: Set numerator to zero: $\sqrt{2x+3} = 0 \Rightarrow 2x+3=0 \Rightarrow x = -\frac{3}{2}$. Point $\left( -\frac{3}{2}, 0 \right)$.
  • y-intercept: $f(0) = \frac{\sqrt{3}}{-3} = -\frac{\sqrt{3}}{3}$. Point $\left( 0, -\frac{\sqrt{3}}{3} \right)$.
• Asymptotes:
  • Vertical asymptote at $x = 3$ because as $x \to 3^{\pm}$, denominator $\to 0$ and numerator $\to \sqrt{9} = 3$, so $f(x) \to \pm\infty$ (specifically, $x \to 3^-$, $f(x) \to -\infty$; $x \to 3^+$, $f(x) \to +\infty$).
  • Horizontal asymptote: As $x \to \pm\infty$, $f(x) \approx \frac{\sqrt{2x}}{x} = \sqrt{\frac{2}{x}} \to 0$, so $y = 0$.
• Graph Sketch:
  • Behavior:
    • As $x \to -\frac{3}{2}^+$, $f(x) \to 0$.
    • For $x \in \left[ -\frac{3}{2}, 3 \right)$, $f(x) < 0$, decreases from 0 to $-\infty$ as $x \to 3^-$.
    • For $x > 3$, $f(x) > 0$, decreases from $+\infty$ to 0 as $x \to \infty$.
  • Key points:
    • $x = 0$: $f(0) = -\frac{\sqrt{3}}{3} \approx -0.577$.
    • $x = 1$: $f(1) = \frac{\sqrt{5}}{-2} \approx -1.118$.
    • $x = 2$: $f(2) = \frac{\sqrt{7}}{-1} \approx -2.646$.
    • $x = 4$: $f(4) = \frac{\sqrt{11}}{1} \approx 3.317$.
    • $x = 5$: $f(5) = \frac{\sqrt{13}}{2} \approx 1.803$.
  • Sketch:
    • Start at $\left( -\frac{3}{2}, 0 \right)$, curve down to $(0, -0.577)$, continue down to $(2, -2.646)$, then plunge to $-\infty$ as $x \to 3^-$.
    • For $x > 3$, start at $+\infty$ at $x = 3^+$, curve down through $(4, 3.317)$, $(5, 1.803)$, approaching $y = 0$ as $x \to \infty$.
  • Asymptotes: Vertical line $x = 3$, horizontal line $y = 0$.

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QUESTION THREE

a) Define
(i) Power set
(ii) Asymptotic line of a curve

Answer:
(i) The power set of a set $S$ is the set of all subsets of $S$, including the empty set and $S$ itself.
(ii) An asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both coordinates tend to infinity.

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b) Let * be a binary operation on a given set.
(i) Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation, give justification for your answer on Z+, define * by a * b = a - b + 3.
(ii) Let * be the binary operation on N given by a * b = H.C.F of a and b. Find $(20*75)*35$.

Answer:
(i) Not a binary operation. Justification: Result not always in $\mathbb{Z}^+$ (e.g., $1 * 5 = -1 \notin \mathbb{Z}^+$).
(ii) 5

Explanation:
(i)

• For $a, b \in \mathbb{Z}^+$ (positive integers), $a * b = a - b + 3$.
• Example: $a = 1, b = 5$, $1 * 5 = 1 - 5 + 3 = -1 \notin \mathbb{Z}^+$.
• Since the result is not always in $\mathbb{Z}^+$, it is not a binary operation.

(ii)

• $20 * 75 = \text{HCF}(20, 75) = 5$ (since $20 = 2^2 \times 5$, $75 = 3 \times 5^2$).
• Then $5 * 35 = \text{HCF}(5, 35) = 5$.
• Thus, $(20 * 75) * 35 = 5$.

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c) For the following
(i) In the quadratic equation, $kx^2 + (2k + 3)x - 2 = 0$, find the values of k for which this equation has two distinct real roots.
(ii) Given that the roots of the equation $2x^2 + 12x + 4 = 0$ are α and β. Find an equation whose roots are $\frac{1}{\alpha}$ and $-\frac{1}{\beta}$.

Answer:
(i) $k < -\frac{9}{2}$ or $k > -\frac{1}{2}$ (i.e., $k \in (-\infty, -4.5) \cup (-0.5, \infty)$).
(ii) One possible equation is $x^2 + \sqrt{7}x - \frac{1}{2} = 0$ (or $2x^2 + 2\sqrt{7}x - 1 = 0$).

Explanation:
(i)

• Discriminant $D=b^2-4ac=(2k+3{)}^2-4(k)(-2)$

$=4k^2+12k+9+8k$

$D = b^2 - 4ac = (2k+3)^2 - 4(k)(-2) = 4k^2 + 12k + 9 + 8k = 4k^2 + 20k + 9$.

• For two distinct real roots, $D > 0$:
  $4k^2 + 20k + 9 > 0$
  • Roots of equality: $k=\frac{-20\pm \sqrt{400-144}}{8}$

  $k = \frac{ -20 \pm \sqrt{400 - 144} }{8} = \frac{ -20 \pm 16}{8}$,

so $k = -\frac{4}{8} = -0.5$,     $k = -\frac{36}{8} = -4.5$.

• Parabola opens upwards, so $D > 0$ when $k < -4.5$ or $k > -0.5$.

(ii)

• Simplify equation: $2x^2+12x+4=0$

$2x^2 + 12x + 4 = 0 \Rightarrow x^2 + 6x + 2 = 0$.

• Roots α, β: Sum $\alpha + \beta = -6$, product $\alpha \beta = 2$.
• New roots: $r_1 = \frac{1}{\alpha}$,   $r_2 = -\frac{1}{\beta}$.
• Product $r_1{r}_2=\left(\frac{1}{\alpha }\right)(-\frac{1}{\beta })$

$=-\frac{1}{\alpha \beta }$

$r_1 r_2 = \left( \frac{1}{\alpha} \right) \left( -\frac{1}{\beta} \right) = -\frac{1}{\alpha \beta} = -\frac{1}{2}$.

• Sum $r_1+r_2=\frac{1}{\alpha }-\frac{1}{\beta }$

$=\frac{\beta -\alpha }{\alpha \beta }$

$r_1 + r_2 = \frac{1}{\alpha} - \frac{1}{\beta} = \frac{\beta - \alpha}{\alpha \beta} = \frac{\beta - \alpha}{2}$.

• $(\beta -\alpha {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$

$(\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha\beta = 36 - 8 = 28$,

so $\beta - \alpha = \pm 2\sqrt{7}$, and sum $= \frac{\pm 2\sqrt{7}}{2} = \pm \sqrt{7}$.

• Depending on labeling, sum is $\sqrt{7}$ or $-\sqrt{7}$, but product is fixed. For example,

with $\alpha = -3 + \sqrt{7}$, $\beta = -3 - \sqrt{7}$, sum is $-\sqrt{7}$.

• Equation: $x^2 - (\text{sum})x + \text{product} = 0$,

so one possibility is $x^2 + \sqrt{7}x - \frac{1}{2} = 0$.

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d) Solve the following without using calculator or log table.
(i) $\ln(3x - 2) + \ln(x + 1) = \ln(3x^2 + 1)$
(ii) $\log_3(2x - 1) = 2\log_9(x + 1)$

Answer:
(i) $x = 3$
(ii) $x = 2$

Explanation:
(i)

• Combine logs: $\ln[(3x-2)(x+1)] = \ln(3x^2 + 1)$.
• So $(3x-2)(x+1) = 3x^2 + 1$.
• Expand: $3x^2+3x-2x-2$

$\; \; \; \; \; \; \; \; \; \; \; =3x^2+x-2$

$3x^2 + 3x - 2x - 2 = 3x^2 + x - 2 = 3x^2 + 1$.

• Solve: $x-2=1$

$x - 2 = 1 \Rightarrow x = 3$.

• Check domain: $3x-2 > 0$ and $x+1 > 0$, so $x > \frac{2}{3}$. $x=3$ valid.

(ii)

• Simplify: $2 \log_9(x+1) = \log_9((x+1)^2)$.
• Convert base: ${\log }_9((x+1{)}^2)=\frac{{\log }_3((x+1{)}^2)}{{\log }_{3}9}$

$\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\frac{2{\log }_3(x+1)}{2}$

$\log_9((x+1)^2) = \frac{\log_3((x+1)^2)}{\log_3 9} = \frac{2 \log_3(x+1)}{2} = \log_3(x+1)$.

• So ${\log }_3(2x-1)={\log }_3(x+1)$

$\; \; \; \; \; \; \; \; \; \Rightarrow 2x-1=x+1$

$\log_3(2x-1) = \log_3(x+1) \Rightarrow 2x-1 = x+1 \Rightarrow x = 2$.

• Check domain: $2x-1 > 0$ and $x+1 > 0$, so $x > \frac{1}{2}$. $x=2$ valid.

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e) Determine the amplitude, phase shift, vertical shift and period for the following function and sketch the graph in the interval given
$f(x) = 2 - 3 \cos \left( 2x - \frac{\pi}{6} \right) \quad \left[ -\frac{\pi}{6}, 2\pi \right]$

Answer:

• Amplitude: 3
• Period: $\pi$
• Phase shift: $\frac{\pi}{12}$ to the right
• Vertical shift: 2
• Graph: See sketch description below.

Explanation:

• Standard form: $a + b \cos(c(x - \phi))$, here $f(x) = 2 + (-3) \cos(2x - \pi/6)$.
• Amplitude: $|b| = 3$.
• Period: $\frac{2\pi}{|c|} = \frac{2\pi}{2} = \pi$.
• Phase shift: Rewrite as $\cos\left(2\left(x - \frac{\pi}{12}\right)\right)$, so $\phi = \frac{\pi}{12}$ (right shift).
• Vertical shift: $a = 2$.
• Graph Sketch:
  • Range: $f(x) \in [-1, 5]$ (since $-3 \leq -3\cos(\cdot) \leq 3$, then add 2).
  • Key points (x, f(x)):
    • $x = -\pi/6$: $\theta = 2(-\pi/6) - \pi/6 = -\pi/2$, $f = 2 - 3\cos(-\pi/2) = 2 - 0 = 2$.
    • $x = \pi/12$: $\theta = 0$, $f = 2 - 3(1) = -1$.
    • $x = \pi/3$: $\theta = \pi/2$, $f = 2 - 0 = 2$.
    • $x = 7\pi/12$: $\theta = \pi$, $f = 2 - 3(-1) = 5$.
    • $x = 5\pi/6$: $\theta = 3\pi/2$, $f = 2 - 0 = 2$.
    • $x = 13\pi/12$: $\theta = 2\pi$, $f = -1$.
    • $x = 4\pi/3$: $\theta = 5\pi/2$, $f = 2$.
    • $x = 19\pi/12$: $\theta = 3\pi$, $f = 5$.
    • $x = 11\pi/6$: $\theta = 7\pi/2$, $f = 2$.
    • $x = 2\pi$: $\theta = 23\pi/6$, $\cos(23\pi/6) = \cos(-\pi/6) = \sqrt{3}/2$, $f = 2 - 3(\sqrt{3}/2) \approx -0.598$.
  • Behavior: Periodic with period $\pi$, oscillates between -1 and 5.
  • Shape: Starts at $(-\pi/6, 2)$, decreases to $(\pi/12, -1)$, increases to $(7\pi/12, 5)$, decreases to $(13\pi/12, -1)$, etc.
  • In interval $[-\pi/6, 2\pi]$, plot points and connect smoothly.

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QUESTION FOUR

a) State
(i) The remainder theorem
(ii) The discriminant of quadratic function

Answer:
(i) If a polynomial $f(x)$ is divided by $x - c$, the remainder is $f(c)$.
(ii) For $ax^2 + bx + c = 0$, the discriminant is $b^2 - 4ac$, which determines the nature of the roots.

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b) Given that $f(x) = x^3 + x^2 - 14x - 24$ is a polynomial of degree three.
(i) Factorise completely for which $f(x)$.
(ii) Find roots of $f(x) = 0$.
(iii) Sketch the graph of $f(x)$.

Answer:
(i) $f(x) = (x-4)(x+2)(x+3)$
(ii) Roots: $x = -3, -2, 4$
(iii) Graph: Cubic with roots at -3, -2, 4; y-intercept at (0, -24); local max and min.

Explanation:
(i)

• Possible rational roots: factors of 24. Test $x=4$: $f\left(4\right)=64+16-56-24=0$, so $(x-4)$ is a factor.
• Synthetic division:
• 
  
• Quotient: $x^2+5x+6=(x+2)(x+3)$.
• Thus, $f\left(x\right)=(x-4)(x+2)(x+3)$.

OR LETS DO THIS DOCTORS 

$\mathrm{🔹\; Given:}$

$f\left(x\right)=x^3+x^2-14x-24$

We aim to factorize completely.

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✅ Step 1: Group terms

Group the polynomial in pairs:

$f\left(x\right)=(x^3+x^2)+(-14x-24)$

Factor each group:

$=x^2(x+1)-2(7x+12)$

This doesn’t help directly, so we’ll try Rational Root Theorem.

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✅ Step 2: Use Rational Root Theorem

Try rational roots of the form:

$\text{Factors of constant term }(-24)÷\text{Factors of leading coefficient }\left(1\right)\Rightarrow \pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24$

Try $x=-2$:

$f(-2)=(-2{)}^3+(-2{)}^2-14(-2)-24=-8+4+28-24=0\Rightarrow x=-2\text{ is a root}$

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✅ Step 3: Use synthetic division to divide by $x+2$

We divide:

$f\left(x\right)÷(x+2)$

Synthetic division:

-2 |  1   1   -14   -24
    |      -2   2    24
    -------------------
      1  -1   -12    0

So:

$f\left(x\right)=(x+2)(x^2-x-12)$

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✅ Step 4: Factor the quadratic

$x^2-x-12=(x-4)(x+3)$

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✅ Final Factorization:

$f\left(x\right)=(x+2)(x-4)(x+3)$

---

✅ Final Answer:

$\overline{)f\left(x\right)=(x+2)(x-4)(x+3)}$

(ii)

• Roots: $x - 4 = 0$, $x + 2 = 0$, $x + 3 = 0$ ⇒ $x = 4, -2, -3$.

(iii)

• y-intercept: $f(0) = -24$.
• Behavior: As $x \to \infty$, $f(x) \to \infty$; as $x \to -\infty$, $f(x) \to -\infty$.
• Local extrema: $f'(x) = 3x^2 + 2x - 14 = 0$. Discriminant $4 + 168 = 172$, roots $\frac{ -1 \pm \sqrt{43} }{3} \approx -2.52, 1.85$.
  • $f''(x) = 6x + 2$: at $x \approx -2.52$, $f'' < 0$ (local max); at $x \approx 1.85$, $f'' > 0$ (local min).
• Sketch: Crosses x-axis at -3, -2, 4; y-axis at (0, -24); local max near (-2.52, 20.6), local min near (1.85, -39.7).

---

c) Let the functions $f(x) = \frac{4x}{x-1}$ and $g(x) = x - 4$. Find
(i) $f^{-1}(x)$
(ii) $(f \circ g)^{-1}(x)$

Answer:
(i) $f^{-1}(x) = \frac{x}{x-4}$
(ii) $(f \circ g)^{-1}(x) = \frac{5x - 16}{x - 4}$

Explanation:

(i)

(i) Set $y=\frac{4x}{x-1}$.

Solve for $x$: $y(x-1)=4x$

$\Rightarrow yx-y=4x$

$\Rightarrow yx-4x=y$

$\Rightarrow x(y-4)=y$

$\Rightarrow x=\frac{y}{y-4}$.

Thus, $f^{-1}\left(x\right)=\frac{x}{x-4}$.

(ii)

$$

(i)

Set $y = \frac{4x}{x-1}$.

Solve for $x$: $y(x-1) = 4x \Rightarrow yx - y = 4x \Rightarrow yx - 4x = y \Rightarrow x(y-4) = y \Rightarrow x = \frac{y}{y-4}$.

Thus, $f^{-1}(x) = \frac{x}{x-4}$.

(ii) f∘g(x)=f(g(x))

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

$f \circ g (x) = f(g(x)) = f(x-4) = \frac{4(x-4)}{(x-4)-1} = \frac{4(x-4)}{x-5}$.

Set $y = \frac{4(x-4)}{x-5}$.

Solve for $x$: $y(x-5)=4x-16$

$\Rightarrow yx-5y=4x-16$

$\Rightarrow yx-4x=5y-16$

$\Rightarrow x(y-4)=5y-16$

$y(x-5) = 4x - 16 \Rightarrow yx - 5y = 4x - 16 \Rightarrow yx - 4x = 5y - 16 \Rightarrow x(y-4) = 5y - 16 \Rightarrow x = \frac{5y - 16}{y-4}$.

Thus, $(f \circ g)^{-1}(x) = \frac{5x - 16}{x - 4}$.

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d) Given that $f(x) = 2x^2 - 12x + 3$ quadratic function.
(i) Complete the square of quadratic functions.
(ii) Hence sketch the graph of the function $f(x) = |2x^2 - 12x + 3|$, showing clearly the y – intercepts, turning point and line of symmetry.

Answer:
(i) $f(x) = 2(x - 3)^2 - 15$
(ii)

• y-intercept: (0, 3)
• x-intercepts: $\left( \frac{6 - \sqrt{30}}{2}, 0 \right)$, $\left( \frac{6 + \sqrt{30}}{2}, 0 \right)$ (approx (0.26, 0), (5.74, 0))
• Turning points: Minima at x-intercepts (y=0), maximum at (3, 15)
• Line of symmetry: $x = 3$
• Graph: W-shaped curve.

Explanation:

(i)

$\mathrm{(i)\; f}(x)=2(x^2-6x)+3$

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

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

$=2(x-3{)}^2-18+3$

$f(x) = 2(x^2 - 6x) + 3 = 2(x^2 - 6x + 9 - 9) + 3 = 2((x-3)^2 - 9) + 3 = 2(x-3)^2 - 18 + 3 = 2(x-3)^2 - 15$.

(ii)

• Original function: parabola opening upwards, vertex (3, -15).
• $|f(x)|$: Reflect parts below x-axis (where $f(x) < 0$) above x-axis.
• f(x) < 0 between roots of $2x^2 - 12x + 3 = 0$:
  

$\mathrm{<span><math><semantics><mrow><mo>=</mo><mfrac><mrow><mn>12</mn><mo>\pm </mo><msqrt><mn>120</mn></msqrt></mrow><mn>4</mn></mfrac><mo></mo></mrow></semantics></math></span>}$

$\mathrm{<span><math><semantics><mrow><mo>=</mo><mfrac><mrow><mn>12</mn><mo>\pm </mo><mn>2</mn><msqrt><mn>30</mn></msqrt></mrow><mn>4</mn></mfrac><mo></mo></mrow></semantics></math></span>}$

• y-intercept: $f(0) = 3$, so $|f(0)| = 3$, point (0, 3).
• Turning points:
  • Minima at x-intercepts (y=0).
  • Maximum at vertex of original parabola, but reflected: (3, | -15 |) = (3, 15).
• Line of symmetry: Original vertex at x=3, and roots symmetric about x=3, so symmetry line $x = 3$.
• Sketch:
  • From left, decreases to (0.26, 0), rises to (3, 15), falls to (5.74, 0), then rises.
  • For $x < 0.26$ and $x > 5.74$, same as original parabola.

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e) Prove the identity $\frac{1}{\cos x+1} + \frac{1}{1-\cos x} = 2 \csc^2 x$

Answer:

Left side:
$\frac{1}{\cos x+1} + \frac{1}{1-\cos x} = \frac{1}{\cos x+1} - \frac{1}{\cos x-1}$ 

Common denominator: $(\cos x+1)(\cos x-1) = \cos^2 x - 1 = -\sin^2 x$.

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

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

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

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

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QUESTION FIVE

a) State the following
(i) The remainder theorem
(ii) The polynomial of degree 8 and give your own example

Answer:
(i) If a polynomial $f(x)$ is divided by $x - c$, the remainder is $f(c)$.
(ii) A polynomial of degree 8 has the highest power of x as 8. Example: $x^8 + 3x^2 - 5$.

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b) Express the following as partial fractions
(i) $\frac{5x-1}{(x+1)(x-2)}$
(ii) $\frac{x^2 - 2x - 1}{(x-1)^2(x^2+1)}$

Answer:
(i) $\frac{2}{x+1} + \frac{3}{x-2}$

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

Explanation:
(i)

Assume $\frac{5x-1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$.

Then $5x-1 = A(x-2) + B(x+1)$.

Set $x = -1$: $5(-1)-1=A(-3)$

$\Rightarrow -6=-3A$

$5(-1)-1 = A(-3) \Rightarrow -6 = -3A \Rightarrow A = 2$.

Set $x = 2$: $5(2)-1=B(3)$

$\Rightarrow 9=3B$

$5(2)-1 = B(3) \Rightarrow 9 = 3B \Rightarrow B = 3$.

Thus, $\frac{2}{x+1} + \frac{3}{x-2}$.

(ii)  Assume $\frac{x^2 - 2x - 1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$.

Then $x^2 - 2x - 1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$.

Set $x = 1$: $1-2-1=B(1+1)$

$\Rightarrow -2=2B$

$1 - 2 - 1 = B(1+1) \Rightarrow -2 = 2B \Rightarrow B = -1$.

Compare coefficients:

Let $x = 0$: $-1=A(-1)(1)+(-1)(1)+D(1)$

$-1 = A(-1)(1) + (-1)(1) + D(1) \Rightarrow -1 = -A -1 + D \Rightarrow D = A$.

Let $x = 2$: $4-4-1=A(1)(5)+(-1)(5)+(2C+D)(1)$

$4 - 4 - 1 = A(1)(5) + (-1)(5) + (2C+D)(1) \Rightarrow -1 = 5A -5 + 2C + D$.

With $D = A$, $-1=6A+2C-5$

$\Rightarrow 6A+2C=4$

$-1 = 6A + 2C -5 \Rightarrow 6A + 2C = 4 \Rightarrow 3A + C = 2$     equestion(1).

Let $x = -1$: $1+2-1=A(-2)(2)+(-1)(2)+(-C+D)(4)$

$1 + 2 - 1 = A(-2)(2) + (-1)(2) + (-C+D)(4) \Rightarrow 2 = -4A -2 -4C + 4D$.
With $D = A$, $2=-4A-2-4C+4A$

$\Rightarrow 2=-2-4C$

$2 = -4A -2 -4C + 4A \Rightarrow 2 = -2 -4C \Rightarrow C = -1$.

From equestion(1): $3A-1=2$

$\Rightarrow 3A=3$

$3A - 1 = 2 \Rightarrow 3A = 3 \Rightarrow A = 1$, then $D = 1$.

Thus, $\frac{1}{x-1} - \frac{1}{(x-1)^2} + \frac{-x+1}{x^2+1} = \frac{1}{x-1} - \frac{1}{(x-1)^2} + \frac{1-x}{x^2+1}$.

---

c) Given that $\log_2 x + 2\log_4 y = 4$,
(i) Show that $xy = 16$.
(ii) Hence solve for $x$ and $y$ given that $\log_{10}(x+y) = 1$.

Answer:
(i) Shown below.
(ii) Solutions: $(x,y) = (2,8)$ or $(8,2)$.

Explanation:

(i)

$2{\log }_{4}y={\log }_4(y^2)$

$=\frac{{\log }_2(y^2)}{{\log }_{2}4}$

$=\frac{2{\log }_{2}y}{2}$

$2 \log_4 y = \log_4 (y^2) = \frac{\log_2 (y^2)}{\log_2 4} = \frac{2 \log_2 y}{2} = \log_2 y$.

So ${\log }_{2}x+{\log }_{2}y={\log }_2(xy)=4$

$\log_2 x + \log_2 y = \log_2 (xy) = 4 \Rightarrow xy = 2^4 = 16$.

(ii)

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

$\log_{10}(x+y) = 1 \Rightarrow x+y = 10^1 = 10$.

Solve $xy = 16$, $x+y = 10$: Quadratic equation $t^2 - 10t + 16 = 0$.

Factors: $(t-2)(t-8) = 0$, so $t = 2$ or $8$.

Solutions: $(x,y) = (2,8)$ or $(8,2)$.

OR LETS DO THIS DOCTORS 

🔹 Given:

${\log }_{2}x+2{\log }_{4}y=4$

---

✅ Step 1: Convert all logs to the same base

We’ll convert ${\log }_{4}y$ to base 2 using the identity:

${\log }_{b}a=\frac{{\log }_{k}a}{{\log }_{k}b}$

So:

${\log }_{4}y=\frac{{\log }_{2}y}{{\log }_{2}4}=\frac{{\log }_{2}y}{2}$

Now substitute into the original equation:

${\log }_{2}x+2\cdot \left(\frac{{\log }_{2}y}{2}\right)=4\Rightarrow {\log }_{2}x+{\log }_{2}y=4$

---

✅ Step 2: Combine logs

${\log }_{2}x+{\log }_{2}y={\log }_2\left(xy\right)$

So:

${\log }_2\left(xy\right)=4\Rightarrow xy=2^4=\overline{)16}$

---

🔹 Part (ii): Solve for $x$ and $y$ given:

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

---

✅ Step 3: Convert log to exponential form

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

---

✅ Step 4: Use system of equations

We now have:

1. $xy=16$
2. $x+y=10$

Let’s solve this system using substitution or quadratic method.

---

✅ Step 5: Use quadratic equation

Let’s assume $x$ and $y$ are roots of a quadratic:

$t^2-(x+y)t+xy=0\Rightarrow t^2-10t+16=0$

Solve using quadratic formula:

$t=\frac{10\pm \sqrt{(-10{)}^2-4(1\left)\right(16)}}{2\left(1\right)}=\frac{10\pm \sqrt{100-64}}{2}=\frac{10\pm \sqrt{36}}{2}=\frac{10\pm 6}{2}$

So:

• $t=\frac{10+6}{2}=8$
• $t=\frac{10-6}{2}=2$

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✅ Final Answer:

• $x=8,y=2$ or $x=2,y=8$
• So: $\overline{)(x,y)=(8,2)\text{ or }(2,8)}$

---

d) Find the value of the following trigonometric ratios without using a calculator
(i) $\sin 75^\circ$
(ii) $\sec \frac{\pi}{12}$

Answer:
(i) $\frac{\sqrt{6} + \sqrt{2}}{4}$
(ii) $\sqrt{6} - \sqrt{2}$

Explanation:
(i)

$\sin 75^{\circ }=\sin (45^{\circ }+30^{\circ })$

$=\sin 45^{\circ }\cos 30^{\circ }+\cos 45^{\circ }\sin 30^{\circ }$

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

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

$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

(ii)

$\sec \frac{\pi}{12} = \frac{1}{\cos \frac{\pi}{12}}$.

$\cos \frac{\pi }{12}=\cos 15^{\circ }$

$=\cos (45^{\circ }-30^{\circ })$

$=\cos 45^{\circ }\cos 30^{\circ }+\sin 45^{\circ }\sin 30^{\circ }$

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

$\cos \frac{\pi}{12} = \cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

$\sec \frac{\pi}{12} = \frac{4}{\sqrt{6} + \sqrt{2}}$. Rationalize:

$\frac{4}{\sqrt{6}+\sqrt{2}}X\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

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

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

$\frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \frac{4(\sqrt{6} - \sqrt{2})}{6 - 2} = \frac{4(\sqrt{6} - \sqrt{2})}{4} = \sqrt{6} - \sqrt{2}$.

LETS DO THIS DOCTORS ONLY FOR SLOW LEARNERS LIKE YOU 😄😂🤣

🔹 Part (i): Find $\sin 75^{\circ }$

We’ll use the angle addition identity:

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

Let’s write $75^{\circ }=45^{\circ }+30^{\circ }$

So:

$\sin 75^{\circ }=\sin (45^{\circ }+30^{\circ })=\sin 45^{\circ }\cos 30^{\circ }+\cos 45^{\circ }\sin 30^{\circ }$

Now recall exact values:

Angle	$\sin \theta$	$\cos \theta$
$30^{\circ }$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$
$45^{\circ }$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$

Substitute:

$\sin 75^{\circ }=(\frac{\sqrt{2}}{2}x\frac{\sqrt{3}}{2})+(\frac{\sqrt{2}}{2}x\frac{1}{2})$

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

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

---

✅ Final Answer:

$\overline{)\sin 75^{\circ }=\frac{\sqrt{6}+\sqrt{2}}{4}}$

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🔹 Part (ii): Find $\sec \left(\frac{\pi }{12}\right)$

Step 1: Convert to degrees

$\frac{\pi }{12}\text{ radians}=\frac{180^{\circ }}{12}=15^{\circ }$

So:

$\sec \left(\frac{\pi }{12}\right)=\sec 15^{\circ }=\frac{1}{\cos 15^{\circ }}$

---

Step 2: Use angle subtraction identity

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

Let $15^{\circ }=45^{\circ }-30^{\circ }$

So:

$\cos 15^{\circ }=\cos (45^{\circ }-30^{\circ })=\cos 45^{\circ }\cos 30^{\circ }+\sin 45^{\circ }\sin 30^{\circ }$

Use known values:

• $\cos 45^{\circ }=\frac{\sqrt{2}}{2}$
• $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$
• $\sin 45^{\circ }=\frac{\sqrt{2}}{2}$
• $\sin 30^{\circ }=\frac{1}{2}$

Substitute:

$\cos 15^{\circ }=(\frac{\sqrt{2}}{2}x\frac{\sqrt{3}}{2})+(\frac{\sqrt{2}}{2}x\frac{1}{2})$

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

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

$cos\; 15^\backslash circ\; =\; \backslash left(\; \backslash frac\{\backslash sqrt\{2\}\}\{2\}\; \backslash cdot\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}\; \backslash right)\; +\; \backslash left(\; \backslash frac\{\backslash sqrt\{2\}\}\{2\}\; \backslash cdot\; \backslash frac\{1\}\{2\}\; \backslash right)\; =\; \backslash frac\{\backslash sqrt\{6\}\}\{4\}\; +\; \backslash frac\{\backslash sqrt\{2\}\}\{4\}\; =\; \backslash frac\{\backslash sqrt\{6\}\; +\; \backslash sqrt\{2$So:

$\sec 15^{\circ }=\frac{1}{\cos 15^{\circ }}=\frac{1}{\frac{\sqrt{6}+\sqrt{2}}{4}}$

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

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✅ Final Answer:

$\overline{)\sec \left(\frac{\pi }{12}\right)=\frac{4}{\sqrt{6}+\sqrt{2}}}$

BUT Dr. Microbiota wa last 😄😂🤣

---

e) Solve the following equations
(i) $\sin A - \sqrt{3} \cos A = 0$
(ii) $2\cos A \sin A + 2 \sin A + \cos A + 1 = 0$

Answer:
(i) $A = \frac{\pi}{3} + k\pi$, $k \in \mathbb{Z}$
(ii) $A = \pi + 2k\pi$, $\frac{7\pi}{6} + 2k\pi$, $\frac{11\pi}{6} + 2k\pi$, $k \in \mathbb{Z}$

Explanation:
(i)

• $\sin A = \sqrt{3} \cos A \Rightarrow \tan A = \sqrt{3}$.
• Solutions: $A = \frac{\pi}{3} + k\pi$ (since period of tan is π).

(ii)

• Factor:
  $2 \cos A \sin A + 2 \sin A + \cos A + 1 = (2 \sin A + 1)(\cos A + 1) = 0$
  (Grouping: $(2 \cos A \sin A + \cos A) + (2 \sin A + 1) = \cos A (2 \sin A + 1) + 1 \cdot (2 \sin A + 1)$).
• Case 1: $2 \sin A + 1 = 0 \Rightarrow \sin A = -\frac{1}{2} \Rightarrow A = \frac{7\pi}{6} + 2k\pi$, $\frac{11\pi}{6} + 2k\pi$.
• Case 2: $\cos A + 1 = 0 \Rightarrow \cos A = -1 \Rightarrow A = \pi + 2k\pi$.

🔹 Part (i): Solve

$\sin A-\sqrt{3}\cos A=0$

Step 1: Rearrange the equation

$\sin A=\sqrt{3}\cos A$

Step 2: Divide both sides by $\cos A$ (assuming $\cos A\ne 0$)

$\frac{\sin A}{\cos A}=\sqrt{3}\Rightarrow \tan A=\sqrt{3}$

Step 3: Solve for $A$

We know:

$\tan A=\sqrt{3}\Rightarrow A=60^{\circ }\text{ or }A=240^{\circ }$

Because $\tan A=\sqrt{3}$ in Quadrant I and III

---

✅ Final Answer:

$\overline{)A=60^{\circ }\text{ or }240^{\circ }}$

---

🔹 Part (ii): Solve

$2\cos A\sin A+2\sin A+\cos A+1=0$

---

Step 1: Group terms

Let’s write it as:

$2\sin A\cos A+2\sin A+\cos A+1=0$

Group terms with $\sin A$:

$2\sin A(\cos A+1)+(\cos A+1)=0$

Factor out $(\cos A+1)$:

$(2\sin A+1)(\cos A+1)=0$

---

Step 2: Solve each factor

Case 1: $2\sin A+1=0$

$\sin A=-\frac{1}{2}\Rightarrow A=210^{\circ }\text{ or }330^{\circ }$

Because $\sin A=-\frac{1}{2}$ in Quadrant III and IV

---

Case 2: $\cos A+1=0$

$\cos A=-1\Rightarrow A=180^{\circ }$

---

✅ Final Answer:

$\overline{)A=180^{\circ },210^{\circ },\text{ or }330^{\circ }}$

@Dr. Microbiota

END OF SOLUTIONS