INTRODUCTION TO MATHEMATICAL METHODS TEST TWO JAN INTAKE 2024

DATE : 14th May, 2024

INSTRUCTIONS

1. Duration: 2 Hours.

---

QUESTION ONE

(a) Express the following as partial fractions
(i) $\frac{x-8}{(x^2+1)(x-2)}$ [3]

(ii) $\frac{9}{(x+1)^2(x-2)}$ [5]

(b) Given that $f(x) = 2x^2 - 2x - 5$ quadratic function.
(i) Complete the square of quadratic functions. [3]
(ii) Hence sketch the graph of the function, showing clearly the y – intercepts, turning point and line of symmetry. [3]

(c) Given that the roots of the equation $3x^2 + 10x - 6 = 0$ are $\alpha$ and $\beta$.
(i) Find an equation whose roots are $\alpha$ and $-\beta$ [4]
(ii) Hence determine the nature of its roots. [2]

---

Answer:

(a)(i)
The expression is $\frac{x-8}{(x^2+1)(x-2)}$.
Set up partial fractions:

$$\frac{x-8}{(x^2+1)(x-2)} = \frac{Ax + B}{x^2+1} + \frac{C}{x-2}$$

Multiply both sides by $(x^2+1)(x-2)$:

$$x - 8 = (Ax + B)(x - 2) + C(x^2 + 1)$$

Expand and equate coefficients:

$$(Ax+B)(x-2)=A{x}^2-2Ax+Bx-2B,C(x^2+1)=C{x}^2+C$$

$$(Ax + B)(x - 2) = Ax^2 - 2Ax + Bx - 2B, \quad C(x^2 + 1) = Cx^2 + C$$ $$x - 8 = (A + C)x^2 + (B - 2A)x + (-2B + C)$$

Equate coefficients:

• $x^2$: $A + C = 0$
• $x$: $B - 2A = 1$
• Constant: $-2B + C = -8$
  Solve the system:
  From $A + C = 0$, $C = -A$.
  Substitute into $-2B + C = -8$:

$$-2B - A = -8 \quad \text{(Equation 3')}$$

From $B - 2A = 1$:

$$B = 2A + 1$$

Substitute into Equation 3':

$$-2(2A+1)-A=-8$$

$$⟹-4A-2-A=-8$$

$$⟹-5A=-6$$

$$-2(2A + 1) - A = -8 \implies -4A - 2 - A = -8 \implies -5A = -6 \implies A = \frac{6}{5}$$

Then:

$$C=-\frac{6}{5},B=2\left(\frac{6}{5}\right)+1$$

$$C = -\frac{6}{5}, \quad B = 2 \left( \frac{6}{5} \right) + 1 = \frac{12}{5} + \frac{5}{5} = \frac{17}{5}$$

So:

$$\frac{x-8}{(x^2+1)(x-2)}=\frac{\frac{6}{5}x+\frac{17}{5}}{x^2+1}+\frac{-\frac{6}{5}}{x-2}$$

$$=\frac{6x+17}{5(x^2+1)}-\frac{6}{5(x-2)}$$

$$\frac{x-8}{(x^2+1)(x-2)} = \frac{\frac{6}{5}x + \frac{17}{5}}{x^2+1} + \frac{-\frac{6}{5}}{x-2} = \frac{6x + 17}{5(x^2 + 1)} - \frac{6}{5(x - 2)}$$ $$\boxed{\dfrac{6x + 17}{5(x^{2} + 1)} - \dfrac{6}{5(x - 2)}}$$

(a)(ii)
The expression is $\frac{9}{(x+1)^2(x-2)}$.
Set up partial fractions:

$$\frac{9}{(x+1)^2(x-2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-2}$$

Multiply both sides by $(x+1)^2(x-2)$:

$$9 = A(x+1)(x-2) + B(x-2) + C(x+1)^2$$

Expand and equate coefficients:

$$A(x+1)(x-2)=A(x^2-x-2)=A{x}^2-Ax-2A$$

$$A(x+1)(x-2) = A(x^2 - x - 2) = Ax^2 - Ax - 2A$$ $$B(x-2)=Bx-2B,$$

$$C(x+1{)}^2=C(x^2+2x+1)$$

$$\; \; \; \; \; \; \; \; =C{x}^2+2Cx+C$$

$$$$

$$B(x-2) = Bx - 2B, \quad C(x+1)^2 = C(x^2 + 2x + 1) = Cx^2 + 2Cx + C$$ $$9 = (A + C)x^2 + (-A + B + 2C)x + (-2A - 2B + C)$$

Equate coefficients:

• $x^2$: $A + C = 0$
• $x$: $-A + B + 2C = 0$
• Constant: $-2A - 2B + C = 9$
  Solve the system:
  From $A + C = 0$, $C = -A$.
  Substitute into $-A + B + 2C = 0$:

$$-A + B - 2A = 0 \implies B = 3A$$

Substitute into $-2A - 2B + C = 9$:

$$-2A-2(3A)-A=9$$

$$⟹-2A-6A-A=9$$

$$⟹-9A=9$$

$$-2A - 2(3A) - A = 9 \implies -2A - 6A - A = 9 \implies -9A = 9 \implies A = -1$$

Then:

$$B = 3(-1) = -3, \quad C = -(-1) = 1$$

So:

$$\frac{9}{(x+1{)}^2(x-2)}=\frac{-1}{x+1}+\frac{-3}{(x+1{)}^2}+\frac{1}{x-2}$$

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

$$\frac{9}{(x+1)^2(x-2)} = \frac{-1}{x+1} + \frac{-3}{(x+1)^2} + \frac{1}{x-2} = -\frac{1}{x+1} - \frac{3}{(x+1)^2} + \frac{1}{x-2}$$ $$\boxed{\dfrac{1}{x-2} - \dfrac{1}{x+1} - \dfrac{3}{(x+1)^2}}$$

(b)(i)
Given $f(x) = 2x^2 - 2x - 5$.
Complete the square:

$$f(x)=2(x^2-x)-5$$

$$=2(x^2-x+\frac{1}{4}-\frac{1}{4})-5$$

$$=2({(x-\frac{1}{2})}^2-\frac{1}{4})-5$$

$$f(x) = 2(x^2 - x) - 5 = 2\left(x^2 - x + \frac{1}{4} - \frac{1}{4}\right) - 5 = 2\left(\left(x - \frac{1}{2}\right)^2 - \frac{1}{4}\right) - 5$$ $$=2{(x-\frac{1}{2})}^2-\frac{2}{4}-5$$

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

$$=2{(x-\frac{1}{2})}^2-\frac{11}{2}$$

$$= 2\left(x - \frac{1}{2}\right)^2 - \frac{2}{4} - 5 = 2\left(x - \frac{1}{2}\right)^2 - \frac{1}{2} - 5 = 2\left(x - \frac{1}{2}\right)^2 - \frac{11}{2}$$ $$\boxed{f(x) = 2\left(x - \dfrac{1}{2}\right)^2 - \dfrac{11}{2}}$$

(b)(ii)
Sketch the graph of $f(x) = 2x^2 - 2x - 5$:

• y-intercept: When $x = 0$, $f(0) = -5$, so $(0, -5)$.
• Turning point: From completed square, vertex at $\left( \frac{1}{2}, -\frac{11}{2} \right) = (0.5, -5.5)$.
• Line of symmetry: $x = \frac{1}{2} = 0.5$.
• Shape: Parabola opens upwards (coefficient of $x^2$ is positive).
• x-intercepts (optional): Solve $2x^2 - 2x - 5 = 0$: $$x=\frac{2\pm \sqrt{(-2{)}^2-4(2)(-5)}}{4}$$
• $$\frac{\mathrm{}}{}$$
• $$=\frac{2\pm \sqrt{44}}{4}$$
• $$=\frac{2\pm 2\sqrt{11}}{4}$$
• $$x = \frac{2 \pm \sqrt{(-2)^2 - 4(2)(-5)}}{4} = \frac{2 \pm \sqrt{44}}{4} = \frac{2 \pm 2\sqrt{11}}{4} = \frac{1 \pm \sqrt{11}}{2}$$ Approximately $x \approx -1.158, \, x \approx 2.158$.

Sketch:

• Graph passes through $(0, -5)$, vertex at $(0.5, -5.5)$, symmetric about $x = 0.5$.
• Opens upwards, x-intercepts at $\left( \frac{1 - \sqrt{11}}{2}, 0 \right)$ and $\left( \frac{1 + \sqrt{11}}{2}, 0 \right)$.

(c)(i)
Given roots $\alpha, \beta$ of $3x^2 + 10x - 6 = 0$.
Sum: $\alpha +\beta =-\frac{b}{a}$

$\alpha + \beta = -\frac{b}{a} = -\frac{10}{3}$.

Product: $\alpha \beta =\frac{c}{a}$

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

$\alpha \beta = \frac{c}{a} = \frac{-6}{3} = -2$.

New roots: $\alpha$ and $-\beta$.
Sum: $\alpha + (-\beta) = \alpha - \beta$.

Product: $\alpha \cdot (-\beta) = -\alpha \beta = -(-2) = 2$.

The equation is $x^2 - (\text{sum})x + \text{product} = 0$, so:

$$x^2-(\alpha -\beta )x+2=0$$

$$x^2 - (\alpha - \beta)x + 2 = 0$$ $$\boxed{x^{2} - (\alpha - \beta)x + 2 = 0}$$

(c)(ii)
Discriminant of the new equation:

$$(\alpha -\beta {)}^2-4(1)(2)$$

$$(\alpha - \beta)^2 - 4(1)(2) = (\alpha - \beta)^2 - 8$$

Compute $(\alpha - \beta)^2$:

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

$$={(-\frac{10}{3})}^2-4(-2)$$

$$=\frac{100}{9}+8$$

$$=\frac{100}{9}+\frac{72}{9}$$

$$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \left(-\frac{10}{3}\right)^2 - 4(-2) = \frac{100}{9} + 8 = \frac{100}{9} + \frac{72}{9} = \frac{172}{9}$$

So discriminant:

$$\frac{172}{9}-\mathrm{8\; >0}$$

$$=\frac{172}{9}-\frac{72}{9}>0$$

$$\frac{172}{9} - 8 = \frac{172}{9} - \frac{72}{9} = \frac{100}{9} > 0$$

Since discriminant positive, roots are real and distinct.

$$\boxed{\text{real and distinct}}$$

---

QUESTION TWO

(a) State the following
(i) The factor theorem [2]
(ii) The polynomial of degree 4 and give your own example [2]

(b) Given that $f(x) = x^4 - 6x^3 - 3x^2 + 16x + 12$ is a polynomial of degree four.

(Note: Corrected from "degree three" as per the polynomial given.)

(i) Factorise completely for which $f(x)$. [5]
(ii) Find all solutions for which $f(x) = 0$. [3]
(iii) Sketch the graph of $f(x)$. [4]

(c) Find the quotient and remainder of the polynomial  $f\left(x\right)=6x^3-5x^2-3x+2$      

when it is divide by $x + 2$ by using synthetic division. [4]

---

Answer:

(a)(i)
The factor theorem states that for a polynomial $f(x)$, if $f(c) = 0$, then $(x - c)$ is a factor of $f(x)$,

and conversely, if $(x - c)$ is a factor, then $f(c) = 0$.

$$\boxed{\text{If } f(c) = 0 \text{ for a polynomial } f(x), \text{ then } (x - c) \text{ is a factor of } f(x), \text{ and vice versa.}}$$

(a)(ii)
A polynomial of degree 4 is a polynomial where the highest power of $x$ is 4.
Example: $x^4 + 2x^3 - x^2 + 5x - 6$.

$$\boxed{{\displaystyle \text{A polynomial of degree 4 has the highest exponent 4.}}}$$

$$\boxed{\text{A polynomial of degree 4 has the highest exponent 4. Example: } x^{4} + 2x^{3} - x^{2} + 5x - 6}$$

(b)(i)
Given $f(x) = x^4 - 6x^3 - 3x^2 + 16x + 12$.
Find rational roots using possible factors of constant term (12) over leading coefficient (1): $\pm 1, 2, 3, 4, 6, 12$.

• Test $x = -1$: $f(-1)=(-1{)}^4-6(-1{)}^3-3(-1{)}^2+16(-1)+12$

$f(-1) = (-1)^4 - 6(-1)^3 - 3(-1)^2 + 16(-1) + 12 = 1 + 6 - 3 - 16 + 12 = 0$, so $(x + 1)$ is a factor.

Synthetic division with root $-1$:

$$\begin{array}{r|rrrrr} & 1 & -6 & -3 & 16 & 12 \\ -1 & & -1 & 7 & -4 & -12 \\ \hline & 1 & -7 & 4 & 12 & 0 \\ \end{array}$$

Quotient: $x^3 - 7x^2 + 4x + 12$.
Now factor $x^3 - 7x^2 + 4x + 12$.

• Test $x = 2$: $f(2) = 8 - 28 + 8 + 12 = 0$, so $(x - 2)$ is a factor.
  Synthetic division with root $2$:

$$\begin{array}{r|rrrr} & 1 & -7 & 4 & 12 \\ 2 & & 2 & -10 & -12 \\ \hline & 1 & -5 & -6 & 0 \\ \end{array}$$

Quotient: $x^2 - 5x - 6$.
Factor: $x^2 - 5x - 6 = (x - 6)(x + 1)$.
So:

$$f(x)=(x+1)(x-2)(x-6)(x+1)$$

$$=(x+1{)}^2(x-2)(x-6)$$

$$f(x) = (x + 1)(x - 2)(x - 6)(x + 1) = (x + 1)^2 (x - 2)(x - 6)$$ $$\boxed{f(x) = (x + 1)^{2} (x - 2)(x - 6)}$$

(b)(ii)
Solve $f(x) = 0$:

$$(x + 1)^2 (x - 2)(x - 6) = 0$$

So:

$$x = -1 \quad (\text{double root}), \quad x = 2, \quad x = 6$$

Solutions: $x = -1, 2, 6$.

$$\boxed{x = -1, \quad x = 2, \quad x = 6}$$

(b)(iii)
Sketch the graph of $f(x) = x^4 - 6x^3 - 3x^2 + 16x + 12$:

• Degree 4, positive leading coefficient: as $x \to \pm \infty$, $y \to \infty$.
• Roots: $x = -1$ (multiplicity 2, touches x-axis), $x = 2$, $x = 6$ (cross x-axis).
• y-intercept: $f(0) = 12$, so $(0, 12)$.
• Behavior:
  • As $x \to -\infty$, $y \to \infty$.
  • Decreases to local minimum at $x \approx -1$ (since multiplicity 2, touches at $(-1, 0)$).
  • Increases to local maximum at $x \approx 1$ (since $f(1) = 1 - 6 - 3 + 16 + 12 = 20 > 0$).
  • Decreases to $(2, 0)$, crosses.
  • Continues decreasing to local minimum in $(2, 6)$ (since $f(3) = 81 - 162 - 27 + 48 + 12 = -48 < 0$).
  • Increases to $(6, 0)$, crosses.
  • Increases to $\infty$.
• Key points: $(-1, 0)$, $(0, 12)$, $(1, 20)$, $(2, 0)$, $(3, -48)$, $(6, 0)$.

Sketch description:

• Graph touches x-axis at $(-1, 0)$, crosses at $(2, 0)$ and $(6, 0)$.
• Local minimum at $x = -1$, local maximum near $x = 1$, local minimum in $(2, 6)$.
• y-intercept at $(0, 12)$.

(c)
Divide $f(x) = 6x^3 - 5x^2 - 3x + 2$ by $x + 2$ (root $-2$) using synthetic division:

$$\begin{array}{r|rrrr} & 6 & -5 & -3 & 2 \\ -2 & & -12 & 34 & -62 \\ \hline & 6 & -17 & 31 & -60 \\ \end{array}$$

Quotient: $6x^2 - 17x + 31$, Remainder: $-60$.

$$\boxed{{\displaystyle \text{Quotient: }6x^2-17x+31,\text{Remainder: }-60}}$$

🧮 Synthetic Division of

$f\left(x\right)=6x^3-5x^2-3x+2$
by $x+2$ (which means using −2 in synthetic division):

Setup:

-2 |   6   -5   -3    2
    |       -12  34  -62
    ---------------------
      6  -17   31  -60

✅ Final Result:

• Quotient: $6x^2-17x+31$
• Remainder: $-60$

So,

$\frac{6x^3-5x^2-3x+2}{x+2}=6x^2-17x+31-\frac{60}{x+2}$

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

$$\boxed{\text{Quotient: } 6x^{2} - 17x + 31, \quad \text{Remainder: } -60}$$

---

QUESTION THREE

(a) Sketch the following logarithmic and exponential graphs on the same diagram.
$f(x)=3^x+3\text{and}g(x)={\log }_3(x-3)$

$f(x) = 3^x + 3 \quad \text{and} \quad g(x) = \log_3 (x - 3)$

(b) Solve the following without using calculator or log table.
(i) $\log_2 (5x + 1) = 2\log_4 (2x + 8)$
(ii) $2\cos^2 x + \sin x - 1 = 0$

(c) Prove the following trigonometric identities
$\tan 2x = \frac{\tan x}{1-\tan^2 x}$
(Note: The identity as stated is incorrect; it should be $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$. Proof for the correct identity is provided.)

(d) Find the value of the following trigonometric ratios without using a calculator $\cot(-300^\circ)$

(e) Sketch the following function by determining the phase shift, vertical shift, period and amplitude. Sketch within the range given to the right of each function
$f(x) = 2 + \csc x$
$[-2\pi, 2\pi]$

THE END

---

Answer:

(a)
Sketch $f(x) = 3^x + 3$ and $g(x) = \log_3 (x - 3)$ on the same diagram:

• Exponential function $y = 3^x + 3$:
  • As $x \to -\infty$, $y \to 3^+$ (horizontal asymptote $y = 3$).
  • Passes through $(0, 4)$, $(-1, \frac{10}{3} \approx 3.333)$, $(1, 6)$.
  • Increases for all $x$.
• Logarithmic function $y = \log_3 (x - 3)$:
  • Domain: $x > 3$.
  • As $x \to 3^+$, $y \to -\infty$.
  • Passes through $(4, 0)$, $(12, 2)$.
  • Increases for $x > 3$.
• Intersection: Not required, but may intersect.
• Sketch:
  • $y = 3^x + 3$: Starts near $y = 3$ for large negative $x$, increases through $(0, 4)$, $(1, 6)$.
  • $y = \log_3 (x - 3)$: Vertical asymptote at $x = 3$, increases through $(4, 0)$, $(12, 2)$.
• 

(b)(i)
Solve $\log_2 (5x + 1) = 2 \log_4 (2x + 8)$:
First, simplify right side. Note $\log_4 a = \frac{\log_2 a}{\log_2 4} = \frac{\log_2 a}{2}$, so:

$$2{\log }_4(2x+8)=\mathrm{2\; x}\frac{{\log }_2(2x+8)}{2}$$

$$2 \log_4 (2x + 8) = 2 \cdot \frac{\log_2 (2x + 8)}{2} = \log_2 (2x + 8)$$

Equation is:

$$\log_2 (5x + 1) = \log_2 (2x + 8)$$

Arguments equal:

$$5x+1=2x+8$$

$$⟹3x=7$$

$$5x + 1 = 2x + 8 \implies 3x = 7 \implies x = \frac{7}{3}$$

Check domain: $5x + 1 > 0 \implies x > -\frac{1}{5}$, $2x + 8 > 0 \implies x > -4$, both satisfied.
Verify:
Left: $\log_2 \left(5 \cdot \frac{7}{3} + 1\right) = \log_2 \left(\frac{38}{3}\right)$
Right: $\log_2 \left(2 \cdot \frac{7}{3} + 8\right) = \log_2 \left(\frac{38}{3}\right)$, equal.

$$\boxed{x = \dfrac{7}{3}}$$

(b)(ii)
Solve $2 \cos^2 x + \sin x - 1 = 0$:
Use identity $\cos^2 x = 1 - \sin^2 x$:

$$2(1-{\sin }^{2}x)+\sin x-1=0$$

$$⟹2-2{\sin }^{2}x+\sin x-1=0$$

$$2(1 - \sin^2 x) + \sin x - 1 = 0 \implies 2 - 2\sin^2 x + \sin x - 1 = 0 \implies -2\sin^2 x + \sin x + 1 = 0$$

Multiply by $-1$:

$$2\sin^2 x - \sin x - 1 = 0$$

Let $u = \sin x$:

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

Discriminant $d = 1 + 8 = 9$, so:

$$u = \frac{1 \pm 3}{4} \implies u = 1 \quad \text{or} \quad u = -\frac{1}{2}$$

So $\sin x = 1$ or $\sin x = -\frac{1}{2}$:

• $\sin x = 1$: $x = \frac{\pi}{2} + 2k\pi$, $k \in \mathbb{Z}$.
• $\sin x = -\frac{1}{2}$: $x = \frac{7\pi}{6} + 2k\pi$ or $x = \frac{11\pi}{6} + 2k\pi$, $k \in \mathbb{Z}$.
  Solutions:

$$\boxed{x = \dfrac{\pi}{2} + 2k\pi, \quad x = \dfrac{7\pi}{6} + 2k\pi, \quad x = \dfrac{11\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}}$$

(c)
Prove the identity. Note: The given identity $\tan 2x = \frac{\tan x}{1 - \tan^2 x}$ is incorrect.

The correct identity is $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$. Proof for the correct identity:

$$\tan 2x=\frac{\sin 2x}{\cos 2x}$$

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

Divide numerator and denominator by $\cos^2 x$:

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

So:

$$\boxed{{\displaystyle \tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}}}$$

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

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

$$\boxed{{\displaystyle \frac{\mathrm{OR\; YOU\; CAN\; USE\; THIS\; WAY}}{}}}$$

$$\boxed{\tan 2x = \dfrac{2 \tan x}{1 - \tan^{2} x}}$$

The correct identity is:

$\tan \left(2x\right)=\frac{2\tan \left(x\right)}{1-{\tan }^2\left(x\right)}$

Let’s go step-by-step to prove this identity using known trigonometric formulas.

---

✅ Step-by-Step Proof:

We start with the double angle identity for tangent:

$\tan \left(2x\right)=\frac{2\tan \left(x\right)}{1-{\tan }^2\left(x\right)}$

We’ll derive this from the angle addition formula:

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

Let $A=x$ and $B=x$, so:

$\tan \left(2x\right)=\tan (x+x)=\frac{\tan x+\tan x}{1-\tan x\cdot \tan x}$

Simplify numerator and denominator:

• Numerator: $\tan x+\tan x=2\tan x$
• Denominator: $1-{\tan }^{2}x$

So we get:

$\tan \left(2x\right)=\frac{2\tan x}{1-{\tan }^{2}x}$

🎉 Final Result:

$\overline{)\tan \left(2x\right)=\frac{2\tan x}{1-{\tan }^{2}x}}$

This identity is especially useful in simplifying expressions and solving trigonometric equations involving double angles.

$\frac{}{}$

---

(d)
Find $\cot(-300^\circ)$:
$\cot$ is odd: $\cot(-\theta) = -\cot \theta$, and periodic with $180^\circ$: $\cot(\theta + 180^\circ k) = \cot \theta$ for integer $k$.

$$\cot (-300^{\circ })=\cot (-300^{\circ }+360^{\circ })=\cot (60^{\circ })$$

$$=\frac{1}{\tan 60^{\circ }}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$$

$$\cot(-300^\circ) = \cot(-300^\circ + 360^\circ) = \cot(60^\circ) = \frac{1}{\tan 60^\circ} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$ $$\boxed{{\displaystyle \frac{\sqrt{3}}{3}}}$$

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

$$\boxed{\dfrac{\sqrt{3}}{3}}$$

✅ Step 1: Use Cotangent Identity

Recall that:

$\cot \left(\theta \right)=\frac{1}{\tan \left(\theta \right)}$

So we need to find:

$\cot (-300^{\circ })=\frac{1}{\tan (-300^{\circ })}$

---

✅ Step 2: Use Periodicity and Reference Angle

Tangent and cotangent are odd functions, meaning:

$\tan (-\theta )=-\tan \left(\theta \right),\cot (-\theta )=-\cot \left(\theta \right)$

So:

$\cot (-300^{\circ })=-\cot \left(300^{\circ }\right)$

Now let’s simplify $\cot \left(300^{\circ }\right)$.

---

✅ Step 3: Find Reference Angle

• $300^{\circ }$ lies in the fourth quadrant
• Reference angle: $360^{\circ }-300^{\circ }=60^{\circ }$

In the fourth quadrant, cotangent is negative, so:

$\cot \left(300^{\circ }\right)=-\cot \left(60^{\circ }\right)$

Thus:

$\cot (-300^{\circ })=-(-\cot (60^{\circ }\left)\right)=\cot \left(60^{\circ }\right)$

---

✅ Step 4: Use Known Trig Value

From standard trigonometric ratios:

$\cot \left(60^{\circ }\right)=\frac{1}{\tan \left(60^{\circ }\right)}=\frac{1}{\sqrt{3}}$

$\backslash cot(60^\backslash circ)\; =\; \backslash frac\{1\}\{\backslash tan(60^\backslash circ)\}\; =\; \backslash frac\{1\}\{\backslash sqrt\{$

🎉 Final Answer:

$\overline{)\cot (-300^{\circ })=\frac{1}{\sqrt{3}}}$

---

(e)
Sketch $f(x) = 2 + \csc x$ in $[-2\pi, 2\pi]$:

• $\csc x = \frac{1}{\sin x}$, period $2\pi$.
• Vertical asymptotes: Where $\sin x = 0$, i.e., $x = -2\pi, -\pi, 0, \pi, 2\pi$.
• Vertical shift: +2, so horizontal reference line $y = 2$.
• Amplitude: None (cosecant has no amplitude).
• Phase shift: None.
• Key points:
  • Where $\sin x = 1$, $\csc x = 1$, $f(x) = 3$: at $x = -\frac{3\pi}{2}, \frac{\pi}{2}$.
  • Where $\sin x = -1$, $\csc x = -1$, $f(x) = 1$: at $x = -\frac{\pi}{2}, \frac{3\pi}{2}$.
• USUBehavior:
  • $[-2\pi, -\pi)$: As $x \to -2\pi^+$, $f \to \infty$; decreases to local max at $x = -\frac{3\pi}{2}$, $f = 3$; decreases to $-\infty$ as $x \to -\pi^-$.
  • $(-\pi, 0)$: As $x \to -\pi^+$, $f \to -\infty$; increases to local max at $x = -\frac{\pi}{2}$, $f = 1$; decreases to $-\infty$ as $x \to 0^-$.
  • $(0, \pi)$: As $x \to 0^+$, $f \to \infty$; decreases to local min at $x = \frac{\pi}{2}$, $f = 3$; increases to $\infty$ as $x \to \pi^-$.
  • $(\pi, 2\pi)$: As $x \to \pi^+$, $f \to -\infty$; increases to local max at $x = \frac{3\pi}{2}$, $f = 1$; decreases to $-\infty$ as $x \to 2\pi^-$.
• Local extrema:
  • Max: $\left(-\frac{3\pi}{2}, 3\right)$, $\left(-\frac{\pi}{2}, 1\right)$, $\left(\frac{3\pi}{2}, 1\right)$.
  • Min: $\left(\frac{\pi}{2}, 3\right)$.

Sketch description:

• Vertical asymptotes at $x = -2\pi, -\pi, 0, \pi, 2\pi$.
• Graph approaches $\pm\infty$ at asymptotes.
• Local maxima and minima as above.
• Range: $(-\infty, 1] \cup [3, \infty)$.

@Dr. Microbiota

---

END OF SOLUTIONS