INTRODUCTION TO MATHEMATICAL METHODS TEST TWO JULY INTAKE 2023

DATE : 12th OCTOBER 2023

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QUESTION ONE

(a) Express the following as partial fractions

(i)

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

Answer:

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

SOLUTION:
Assume:

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

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

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

Expand:

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

Equate coefficients:

• $x$ terms: $A + B = 5$
• Constant terms: $-2A + B = -1$
  Solve the system:
  From $A + B = 5$ and $-2A + B = -1$, subtract the second equation from the first:

$$(A+B)-(-2A+B)$$

$$=5-(-1)$$

$$⟹3A=6$$

$$(A + B) - (-2A + B) = 5 - (-1) \implies 3A = 6 \implies A = 2$$

Substitute $A = 2$ into $A + B = 5$:

$$2+B=5$$

$$2 + B = 5 \implies B = 3$$

Thus:

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

(ii)

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

Answer:

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

Explanation:
Assume:

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

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

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

Substitute $x = 1$:

$$(1{)}^2-2(1)-1=B(1^2+1)$$

$$⟹-2=2B$$

$$(1)^2 - 2(1) - 1 = B(1^2 + 1) \implies -2 = 2B \implies B = -1$$

Expand and equate coefficients:
Right-hand side:

$$A(x^3 - x^2 + x - 1) + B(x^2 + 1) + (Cx + D)(x^2 - 2x + 1)$$

Substitute $B = -1$:

$$= A(x^3 - x^2 + x - 1) - (x^2 + 1) + (Cx + D)(x^2 - 2x + 1)$$

Expand $(Cx + D)(x^2 - 2x + 1)$:

$$Cx^3 - 2Cx^2 + Cx + Dx^2 - 2Dx + D = Cx^3 + (-2C + D)x^2 + (C - 2D)x + D$$

Combine all:

$$[A x^3 - A x^2 + A x - A] + [-x^2 - 1] + [C x^3 + (-2C + D)x^2 + (C - 2D)x + D]$$ $$= (A + C)x^3 + (-A - 1 - 2C + D)x^2 + (A + C - 2D)x + (-A - 1 + D)$$

Equate to left-hand side $x^2 - 2x - 1$ (i.e., $0x^3 + 1x^2 - 2x - 1$):

• $x^3$: $A + C = 0$  equestion(1)
• $x^2$: $-A - 1 - 2C + D = 1$ equestion (2)
• $x$: $A + C - 2D = -2$ equestion (3)
• Constant: $-A - 1 + D = -1$ equestion (4)
  From (1): $C = -A$
  From (4): $-A + D = 0 \implies D = A$
  Substitute into equestion(3):

$$A + (-A) - 2(A) = -2 \implies -2A = -2 \implies A = 1$$

Then $C = -1$, $D = 1$
Verify equestion(2):

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

Thus:

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

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(b) Given that $f(x) = 2x^2 - 12x + 3$ quadratic function.

(i) Complete the square of quadratic functions. [3]

Answer:

$$f(x) = 2(x - 3)^2 - 15$$

Explanation:

$$f(x) = 2x^2 - 12x + 3 = 2(x^2 - 6x) + 3$$

Complete the square:

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

So:

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

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

$$f(x) = 2[(x - 3)^2 - 9] + 3 = 2(x - 3)^2 - 18 + 3 = 2(x - 3)^2 - 15$$

(ii) Hence sketch the graph of the function, showing clearly the y – intercepts, turning point and line of symmetry. [3]
Solution:

• Y-intercept: $(0, 3)$
• X-intercepts: $\left( \frac{6 - \sqrt{30}}{2}, 0 \right)$, $\left( \frac{6 + \sqrt{30}}{2}, 0 \right)$
• Turning point (vertex): $(3, -15)$
• Line of symmetry: $x = 3$
  Explanation:
• Y-intercept: $f(0) = 2(0)^2 - 12(0) + 3 = 3$, so $(0, 3)$.
• X-intercepts: Solve $2x^2 - 12x + 3 = 0$: $$x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 2 \cdot 3}}{4} = \frac{12 \pm \sqrt{120}}{4} = \frac{12 \pm 2\sqrt{30}}{4} = \frac{6 \pm \sqrt{30}}{2}$$
• Vertex: From $f(x) = 2(x - 3)^2 - 15$, vertex is $(3, -15)$.
• Line of symmetry: Vertical line through vertex, $x = 3$.
• Graph is a parabola opening upwards (since coefficient of $x^2$ is positive).

Sketch:

• Vertex at $(3, -15)$.
• Y-intercept at $(0, 3)$.
• X-intercepts approximately at $(0.261, 0)$ and $(5.739, 0)$ (since $\sqrt{30} \approx 5.477$).
• Line of symmetry $x = 3$.
  

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(c) Given that $\log_2 x + 2\log_4 y = 4$,

(i) Show that $xy = 16$. [2]

ANSWER:

$$\log_2 x + 2 \cdot \frac{\log_2 y}{2} = \log_2 x + \log_2 y = \log_2 (xy) = 4$$

Thus:

$$xy = 2^4 = 16$$

Explanation:

$$\log_4 y = \frac{\log_2 y}{\log_2 4} = \frac{\log_2 y}{2}$$

So:

$${\log }_{2}x+2X\left(\frac{{\log }_{2}y}{2}\right)$$

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

$$\log_2 x + 2 \cdot \left( \frac{\log_2 y}{2} \right) = \log_2 x + \log_2 y = \log_2 (xy)$$

Given equal to 4:

$${\log }_2(xy)=4$$

$$⟹xy=2^4$$

$$\log_2 (xy) = 4 \implies xy = 2^4 = 16$$

(ii) Hence solve for $x$ and $y$ given that

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

ANSWER:

$$(x, y) = (2, 8) \quad \text{or} \quad (8, 2)$$

Explanation:
From $\log_{10}(x+y) = 1$:

$$x + y = 10^1 = 10$$

From (i): $xy = 16$.
So $x$ and $y$ are roots of:

$$t^2-(x+y)t+xy=0$$

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

Factorize:

$$(t-2)(t-8)=0$$

$$(t - 2)(t - 8) = 0 \implies t = 2 \quad \text{or} \quad t = 8$$

So solutions: $(x, y) = (2, 8)$ or $(8, 2)$.
Check domain:

• Original logs require $x > 0$, $y > 0$, satisfied.
• $\log_{10}(x+y)$ defined for $x+y > 0$, satisfied.

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QUESTION TWO

(a) State the following

(i) The remainder theorem [2]
Solution:
When a polynomial $f(x)$ is divided by a linear divisor $(x - a)$, the remainder is $f(a)$.

(ii) The polynomial of degree 8 and give your own example [2]
Solution:
A polynomial of degree 8 has the highest power of the variable equal to 8.
Example: $f(x) = 3x^8 - 2x^5 + x - 7$.

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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 $f(x)$. [4]
Solution:

$$f(x) = (x + 3)(x + 2)(x - 4)$$

Explanation:
Possible rational roots: $\pm 1, 2, 3, 4, 6, 8, 12, 24$.
Test $x = -2$:

$$f(-2) = (-2)^3 + (-2)^2 - 14(-2) - 24 = -8 + 4 + 28 - 24 = 0$$

So $(x + 2)$ is a factor.
Synthetic division ($x = -2$):

$$\begin{array}{r|rrrr} & 1 & 1 & -14 & -24 \\ -2 & & -2 & 2 & 24 \\ \hline & 1 & -1 & -12 & 0 \\ \end{array}$$

Quotient: $x^2 - x - 12$.
Factor: $x^2 - x - 12 = (x - 4)(x + 3)$.
Thus:

$$f(x)=(x+2)(x-4)(x+3)$$

$$$$

$$f(x) = (x + 2)(x - 4)(x + 3)$$

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:

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

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✅ Final Answer:

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

(ii) Find all solutions for which $f(x) = 0$. [3]
Solution:

$$x = -3, \quad x = -2, \quad x = 4$$

Explanation:
Set $f(x) = 0$:

$$(x + 3)(x + 2)(x - 4) = 0$$

So $x = -3$, $x = -2$, or $x = 4$.

(iii) Sketch the graph of $f(x)$. [3]
Solution:

• Zeros: $x = -3$, $x = -2$, $x = 4$.
• Y-intercept: $f(0) = -24$, so $(0, -24)$.
• Behavior: As $x \to \infty$, $f(x) \to \infty$; as $x \to -\infty$, $f(x) \to -\infty$.
• Sign changes:
  • For $x < -3$: negative (e.g., $f(-4) = -16$)
  • $-3 < x < -2$: positive (e.g., $f(-2.5) = 1.875$)
  • $-2 < x < 4$: negative (e.g., $f(0) = -24$)
  • $x > 4$: positive (e.g., $f(5) = 36$)
    Graph crosses x-axis at each root.
    (Sketch described: cubic curve passing through (-3,0), (-2,0), (4,0), and (0,-24), with behavior as above.)

(iv) Hence or otherwise find the solution set for which $x^3 + x^2 - 14x - 24 \leq 0$. [2]
Solution:

$$x \in (-\infty, -3] \cup [-2, 4]$$

Explanation:
From sign analysis:

• $f(x) \leq 0$ when $x \in (-\infty, -3] \cup [-2, 4]$.
  Includes zeros and intervals where negative.

OR LETS DO THIS DOCTORS 

The inequality is:

$(x-4)(x+2)(x+3)\le 0.$

The roots are $x=-3$, $x=-2$, and $x=4$. These roots divide the real number line into intervals: $(-\infty ,-3)$, $(-3,-2)$, $(-2,4)$, and $(4,\infty )$. Test the sign of the expression in each interval.

• For $x\in (-\infty ,-3)$, choose $x=-4$: The product is $(-)\cdot (-)\cdot (-)=-$ (negative).
• For $x\in (-3,-2)$, choose $x=-2.5$: The product is $(-)\cdot (-)\cdot (+)=+$ (positive).
• For $x\in (-2,4)$, choose $x=0$: The product is $(-)\cdot (+)\cdot (+)=-$ (negative).
• For $x\in (4,\infty )$, choose $x=5$: $(5-4)=1>0,(5+2)=7>0,(5+3)=8>0.$ The product is $(+)\cdot (+)\cdot (+)=+$ (positive).

The expression is less than or equal to zero where it is negative or zero. It is negative in $(-\infty ,-3)$ and $(-2,4)$, and zero at $x=-3$, $x=-2$, and $x=4$. Including these roots, the solution set is:

$ANSWER\; :\; (-\infty ,-3]\cup [-2,4].$

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(c) Find the remainder of the polynomial $p(x) = x^4 - 3x^2 + x - 10$ when it is divide by $x + 2$ by using synthetic division [4]
Solution:
Remainder = $-8$
Explanation:
Synthetic division (divisor $x + 2$, so root $x = -2$):
Coefficients: $1$ (x⁴), $0$ (x³), $-3$ (x²), $1$ (x), $-10$ (constant).

$$\begin{array}{r|rrrrr} & 1 & 0 & -3 & 1 & -10 \\ -2 & & -2 & 4 & -2 & 2 \\ \hline & 1 & -2 & 1 & -1 & -8 \\ \end{array}$$

Remainder is $-8$.

We’re dividing by $x+2$ with coefficients [1, 0, -3, 1, -10]. The root is $-2$.

Here’s the synthetic division:

Final answer: The remainder is $-8$. Use synthetic division with root $x=-2$ (since dividing by $x+2$). Include the missing $x^3$ term as 0.

Coefficients: $[1,0,-3,1,-10]$

-2   |  1    0    -3     1    -10
     |      -2     4    -2      2
     -----------------------------
       1   -2     1    -1     -8

Final answer: remainder $=-8$.

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QUESTION THREE

(a) Define the following

(i) Discriminate of quadratic equation [2]
Solution:
For $ax^2 + bx + c = 0$, the discriminant is $D = b^2 - 4ac$. It determines the nature of roots:

• $D > 0$: two distinct real roots.
• $D = 0$: one real root (repeated).
• $D < 0$: no real roots (complex).

(ii) An exponential function [2]
Solution:
A function of the form $f(x) = a \cdot b^x$, where $a$ is a constant, $b > 0$, $b \neq 1$, and $x$ is the exponent.

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(b) Solve the following without using calculator or log table.

(i) 

$$\ln(3x-2) + \ln(x+1) = \ln(3x^2+1)$$

Solution:

$$x = 3$$
Combine logs:

$$\ln[(3x-2)(x+1)] = \ln(3x^2+1)$$

Equate arguments (since $\ln$ one-to-one):

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

Expand:

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

$$⟹3x^2+x-2$$

$$3x^2 + 3x - 2x - 2 = 3x^2 + 1 \implies 3x^2 + x - 2 = 3x^2 + 1$$

Simplify:

$$x-2=1$$

$$x - 2 = 1 \implies x = 3$$

Check domain: $3x-2 > 0 \implies x > 2/3$, $x+1 > 0 \implies x > -1$, and $x=3 > 2/3$, valid.

(ii)

$$\log_3 (2x-1) = 2\log_9 (x+1)$$

Solution:

$$x = 2$$

Note:   $\log_9 = \frac{\log_3}{\log_3 9} = \frac{\log_3}{2}$, so:

$$2{\log }_9(x+1)$$

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

$$2 \log_9 (x+1) = 2 \cdot \frac{\log_3 (x+1)}{2} = \log_3 (x+1)$$

Thus:

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

Equate arguments:

$$2x-1=x+1$$

$$\mathrm{}\text{<mo>2x − x = 1+ 1</mo><mn></mn>\hspace{0.25em}\hspace{0.05em}}$$

$$2x - 1 = x + 1 \implies x = 2$$

Check domain: $2x-1 > 0 \implies x > 0.5$, $x+1 > 0 \implies x > -1$, and $x=2 > 0.5$, valid.

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(c) Write the given expression in terms of $\log 2$, $\log 3$ and $\log 7$ to any base

$$\log \left( \frac{\sqrt[3]{16} \times \sqrt[4]{125}}{\sqrt[8]{49}} \right)$$

Note: The original expression has $\sqrt[4]{125}$, but 125 = 5³ introduces $\log 5$, which is not allowed. Assuming a typo and $\sqrt[4]{81}$ is intended (as 81 = 3⁴), to match required logs.

Solution (with $\sqrt[4]{81}$):

$$\log \left( \frac{\sqrt[3]{16} \times \sqrt[4]{81}}{\sqrt[8]{49}} \right) = \frac{4}{3} \log 2 + \log 3 - \frac{1}{4} \log 7$$

Explanation (assuming $\sqrt[4]{81}$):

• $\sqrt[3]{16} = 16^{1/3} = (2^4)^{1/3} = 2^{4/3}$
• $\sqrt[4]{81} = 81^{1/4} = (3^4)^{1/4} = 3^1 = 3$
• $\sqrt[8]{49} = 49^{1/8} = (7^2)^{1/8} = 7^{2/8} = 7^{1/4}$
  So:

$$\frac{2^{4/3} \times 3}{7^{1/4}} = 2^{4/3} \cdot 3^1 \cdot 7^{-1/4}$$

Take log:

$$\log(2^{4/3} \cdot 3 \cdot 7^{-1/4}) = \frac{4}{3} \log 2 + \log 3 - \frac{1}{4} \log 7$$

If original $\sqrt[4]{125}$ is kept, it would require $\log 5$, but problem specifies only $\log 2, \log 3, \log 7$.

OR MAYBE  LETS DO THIS DOCTORS 

We are given the expression:

$\log \left(\frac{\sqrt[3]{16}\times \sqrt[4]{125}}{\sqrt[8]{49}}\right)$

and asked to express it in terms of $\log 2$, $\log 3$, and $\log 7$.

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✅ Step 1: Rewrite each term using exponents

• $\sqrt[3]{16}=16^{1/3}=(2^4{)}^{1/3}=2^{4/3}$
• $\sqrt[4]{125}=125^{1/4}=(5^3{)}^{1/4}=5^{3/4}$
• $\sqrt[8]{49}=49^{1/8}=(7^2{)}^{1/8}=7^{2/8}=7^{1/4}$

So the expression becomes:

$\log \left(\frac{2^{4/3}\cdot 5^{3/4}}{7^{1/4}}\right)$

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✅ Step 2: Apply log rules

$\log \left(\frac{2^{4/3}\cdot 5^{3/4}}{7^{1/4}}\right)=\log \left(2^{4/3}\right)+\log \left(5^{3/4}\right)-\log \left(7^{1/4}\right)$ $=\frac{4}{3}\log 2+\frac{3}{4}\log 5-\frac{1}{4}\log 7$

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✅ Step 3: Express $\log 5$ in terms of $\log 2$ and $\log 3$

Since $5=\frac{10}{2}=\frac{2\cdot 5}{2}$, we use:

$\log 5=\log \left(10\right)-\log \left(2\right)=1-\log 2\text{(if base 10)}$

But if we want everything in terms of $\log 2$, $\log 3$, and $\log 7$, we can use:

$\log 5=\log \left(\frac{10}{2}\right)=\log 10-\log 2=1-\log 2$

So:

$\frac{3}{4}\log 5=\frac{3}{4}(1-\log 2)=\frac{3}{4}-\frac{3}{4}\log 2$

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✅ Final Answer:

$\log \left(\frac{\sqrt[3]{16}\cdot \sqrt[4]{125}}{\sqrt[8]{49}}\right)=\frac{4}{3}\log 2+\frac{3}{4}\log 5-\frac{1}{4}\log 7$

If you express $\log 5=\log 10-\log 2=1-\log 2$, then:

$=\frac{4}{3}\log 2+(\frac{3}{4}-\frac{3}{4}\log 2)-\frac{1}{4}\log 7=(\frac{4}{3}-\frac{3}{4})\log 2+\frac{3}{4}-\frac{1}{4}\log 7$ $=\left(\frac{16-9}{12}\right)\log 2+\frac{3}{4}-\frac{1}{4}\log 7=\frac{7}{12}\log 2+\frac{3}{4}-\frac{1}{4}\log 7$

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@Dr. Microbiota

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(d) Find the value of the following trigonometric ratios without using a calculator

(i)

$$\text{Sin}(300^\circ)$$

Solution:

$$\sin(300^\circ) = -\frac{\sqrt{3}}{2}$$
300° is in quadrant IV, reference angle = 360° - 300° = 60°.

$$\sin (300^{\circ })=-\sin (60^{\circ })$$

$$\sin(300^\circ) = -\sin(60^\circ) = -\frac{\sqrt{3}}{2}$$

(ii)

$$\text{Sec} \left( \frac{7\pi}{6} \right)$$

Solution:

$$\sec \left( \frac{7\pi}{6} \right) = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$$
$\frac{7\pi}{6}$ is in quadrant III, reference angle = $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$.

$$\cos \left(\frac{7\pi }{6}\right)=-\cos \left(\frac{\pi }{6}\right)$$

$$\cos \left( \frac{7\pi}{6} \right) = -\cos \left( \frac{\pi}{6} \right) = -\frac{\sqrt{3}}{2}$$

So:

$$\sec \left(\frac{7\pi }{6}\right)=\frac{1}{\cos }=\frac{1}{-\frac{\sqrt{3}}{2}}$$

$$=-\frac{2}{\sqrt{3}}$$

$$\sec \left( \frac{7\pi}{6} \right) = \frac{1}{\cos} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$$

@Dr. Microbiota

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END OF SOLUTIONS