INTRODUCTION TO MATHEMATICAL METHODS TEST TWO JAN INTAKE 2024

DATE : 25th APRIL, 2024

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QUESTION ONE

(a) Express the following as partial fractions

$$\frac{x-8}{(x+1)(x-2)}[3]$$

ii.

$$\frac{x-8}{(x+1)(x-2)} \quad [3]$$ $$\frac{x^3-3}{(x+1)(x-2)} \quad [5]$$

i) Solution for $\frac{x-8}{(x+1)(x-2)}$:

To decompose into partial fractions, set:

$$\frac{x-8}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$$

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

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

Expand and collect like terms:

$$x - 8 = (A + B)x + (-2A + B)$$

Equate coefficients:

• For $x$: $A + B = 1$
• Constant: $-2A + B = -8$
  Solve the system:
  Subtract the first equation from the second:

$$(-2A + B) - (A + B) = -8 - 1 \implies -3A = -9 \implies A = 3$$

Substitute $A = 3$ into $A + B = 1$:

$$3 + B = 1 \implies B = -2$$

Thus:

$$\frac{x-8}{(x+1)(x-2)} = \frac{3}{x+1} - \frac{2}{x-2}$$

ii). Solution for $\frac{x^3-3}{(x+1)(x-2)}$:
The numerator degree (3) is greater than the denominator degree (2), so perform polynomial division first.
Denominator: $(x+1)(x-2) = x^2 - x - 2$.
Divide $x^3 + 0x^2 + 0x - 3$ by $x^2 - x - 2$:

• Divide leading terms: $\frac{x^3}{x^2} = x$.
• Multiply: $x \cdot (x^2 - x - 2) = x^3 - x^2 - 2x$.
• Subtract: $(x^3 + 0x^2 + 0x - 3) - (x^3 - x^2 - 2x) = x^2 + 2x - 3$.
• Divide: $\frac{x^2}{x^2} = 1$.
• Multiply: $1 \cdot (x^2 - x - 2) = x^2 - x - 2$.
• Subtract: $(x^2 + 2x - 3) - (x^2 - x - 2) = 3x - 1$.
  So:

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

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

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

Now decompose $\frac{3x-1}{(x+1)(x-2)}$:
Set:

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

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

$$3x - 1 = C(x - 2) + D(x + 1)$$

Solve for $C$ and $D$ by substituting convenient values:

• Let $x = -1$: $3(-1) - 1 = C(-1 - 2) \implies -4 = -3C \implies C = \frac{4}{3}$.
• Let $x = 2$: $3(2) - 1 = D(2 + 1) \implies 5 = 3D \implies D = \frac{5}{3}$.
  Thus:

$$\frac{3x-1}{(x+1)(x-2)} = \frac{4/3}{x+1} + \frac{5/3}{x-2}$$

Combine:

$$\frac{x^3-3}{(x+1)(x-2)} = x + 1 + \frac{4}{3(x+1)} + \frac{5}{3(x-2)}$$

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

Solution:
Complete the square:

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

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

Add and subtract $1$ inside the parentheses:

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

So:

$$f(x)=-[(x+1{)}^2-1]-5$$

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

$$f(x) = -[(x+1)^2 - 1] - 5 = -(x+1)^2 + 1 - 5 = -(x+1)^2 - 4$$

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

• Y-intercept: Set $x = 0$: $$f(0) = -0 - 0 - 5 = -5 \quad \text{(Point: } (0, -5))$$
• X-intercepts: Set $f(x) = 0$: $$-x^2 - 2x - 5 = 0 \implies x^2 + 2x + 5 = 0$$ Discriminant: $d = 2^2 - 4(1)(5) = 4 - 20 = -16 < 0$, no real roots. So no x-intercepts.
• Turning point (vertex): From $f(x) = -(x+1)^2 - 4$, vertex is at $(-1, -4)$.
• Line of symmetry: $x = -1$.
• Shape: Leading coefficient is negative, so parabola opens downwards.

Graph Sketch Description:

• Vertex at $(-1, -4)$.
• Y-intercept at $(0, -5)$.
• No x-intercepts.
• Symmetry about $x = -1$.
• As $x \to \pm \infty$, $f(x) \to -\infty$.
• Passes through points like $(-2, -5)$, $(1, -8)$, etc. (using equation).
  Visual: Opens downward, vertex at $(-1, -4)$, y-axis crossed at $(0, -5)$,                                                           symmetric about $x = -1$, no x-axis crossings.

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(c) Given that the roots of the equation $2x^2 + 10x - 6 = 0$ are $\alpha$ and $\beta$.
Find an equation whose roots are $\alpha^2 + 2$ and $\beta^2 + 2$.
Hence determine the nature of its roots. $[4] + [5]$

Solution:
First, for $2x^2 + 10x - 6 = 0$, divide by 2: $x^2 + 5x - 3 = 0$.
Sum of roots: $\alpha + \beta = -\frac{b}{a} = -5$.
Product of roots: $\alpha\beta = \frac{c}{a} = -3$.

New roots: $p = \alpha^2 + 2$, $q = \beta^2 + 2$.
Sum $p + q = (\alpha^2 + 2) + (\beta^2 + 2) = \alpha^2 + \beta^2 + 4$.

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

$$=(-5{)}^2-2(-3)$$

$$=25+6$$

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-5)^2 - 2(-3) = 25 + 6 = 31$$

So:

$$p + q = 31 + 4 = 35$$

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

$=(\alpha \beta {)}^2+2(31)+4$

$=(-3{)}^2+62+4$

$=9+62+4$

$pq = (\alpha^2 + 2)(\beta^2 + 2) = \alpha^2\beta^2 + 2(\alpha^2 + \beta^2) + 4 = (\alpha\beta)^2 + 2(31) + 4 = (-3)^2 + 62 + 4 = 9 + 62 + 4 = 75$.

The quadratic equation with roots $p$ and $q$ is:

$$x^2 - (p+q)x + pq = 0 \implies x^2 - 35x + 75 = 0$$

Nature of roots: Discriminant $d = b^2 - 4ac = (-35)^2 - 4(1)(75) = 1225 - 300 = 925$.

• $d > 0$, so two distinct real roots.
• $d = 925$ is not a perfect square ($30^2 = 900$, $31^2 = 961$), so roots are irrational.

Thus, roots are real, distinct, and irrational.

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QUESTION TWO

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

Solution:
(i) Factor Theorem: If $f(c) = 0$ for a polynomial $f(x)$, then $(x - c)$ is a factor of $f(x)$.                                                                      Conversely, if $(x - c)$ is a factor, then $f(c) = 0$.
(ii) Polynomial of Degree 4: A polynomial where the highest power of $x$ is 4.
Example: $f(x) = x^4 + 2x^3 - x^2 + 4x - 3$.

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(b) Given that $f(x) = x^4 + 5x^3 - 3x^2 - 13x + 10$ is a polynomial of degree four.
(i) Factorise completely for which $f(x)$.
(ii) Find all solutions for which $f(x) = 0$.
(iii) Sketch the graph of $f(x)$.
$[5] + [5]$

Solution:
(i) Factorization:
Possible rational roots: $\pm 1, 2, 5, 10$.

• Test $x = 1$: $f(1) = 1 + 5 - 3 - 13 + 10 = 0$, so $(x - 1)$ is a factor.
  Synthetic division by $x - 1$ (root 1):

$$\begin{array}{r|rrrrr} 1 & 1 & 5 & -3 & -13 & 10 \\ & & 1 & 6 & 3 & -10 \\ \hline & 1 & 6 & 3 & -10 & 0 \\ \end{array}$$

Quotient: $x^3 + 6x^2 + 3x - 10$.

• Test $x = 1$ again on quotient: $1 + 6 + 3 - 10 = 0$, so $(x - 1)$ is a factor again.
  Synthetic division by $x - 1$ (root 1):

$$\begin{array}{r|rrrr} 1 & 1 & 6 & 3 & -10 \\ & & 1 & 7 & 10 \\ \hline & 1 & 7 & 10 & 0 \\ \end{array}$$

Quotient: $x^2 + 7x + 10$.
Factor: $x^2 + 7x + 10 = (x + 2)(x + 5)$.
Thus, complete factorization:

$$f(x) = (x - 1)^2 (x + 2)(x + 5)$$

(ii) Solutions to $f(x) = 0$:
Set each factor to zero:

• $(x - 1)^2 = 0 \implies x = 1$ (multiplicity 2)
• $x + 2 = 0 \implies x = -2$
• $x + 5 = 0 \implies x = -5$
  Solutions: $x = -5, -2, 1$ (with $x = 1$ having multiplicity 2).

(iii) Graph Sketch Description:

• Roots at $x = -5, -2, 1$.
• Y-intercept: $f(0) = 10$ (point $(0, 10)$).
• Behavior: Degree 4, positive leading coefficient, so as $x \to \pm \infty$, $f(x) \to \infty$.
• Multiplicity: At $x = 1$ (even multiplicity), graph touches x-axis and turns. At $x = -5$                                                            and $x = -2$ (odd multiplicity), graph crosses x-axis.
• Sign analysis:
  • For $x < -5$, e.g., $x = -6$, $f(-6) > 0$.
  • Between $-5$ and $-2$, e.g., $x = -3$, $f(-3) = (-4)^2(-1)(2) = 16 \cdot (-1) \cdot 2 = -32 < 0$.
  • Between $-2$ and $1$, e.g., $x = 0$, $f(0) = 10 > 0$.
  • For $x > 1$, e.g., $x = 2$, $f(2) = (1)^2(4)(7) > 0$.
• Turning points: Cubic-like behavior with turns near roots.
  Visual: Starts in quadrant II, crosses x-axis at $x = -5$ (downward), crosses again                                                                        at $x = -2$ (upward), touches at $x = 1$ (bounces up), ends in quadrant I. Y-intercept at $(0, 10)$.

                                           

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(c) Find the quotient and remainder of the polynomial $f(x) = 6x^3 - 5x^2 - 3x + 2$ when it                                                               is divided by $x + 2$ by using synthetic division.
$[4] + [4]$

Solution:
Divisor $x + 2 = 0 \implies x = -2$.
Synthetic division with root $-2$:

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

Interpret: Coefficients of quotient are $6, -17, 31$, so quotient is $6x^2 - 17x + 31$.
Remainder is $-60$.
Thus, quotient: $6x^2 - 17x + 31$, remainder: $-60$.

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QUESTION THREE

(a) Sketch the following logarithmic and exponential graphs on the same diagram.
$f(x) = \left( \frac{1}{3} \right)^x + 3 \quad \text{and} \quad f(x) = 3 + \log_3 x$
(Note: The second function should be labeled differently; assume $g(x) = \left( \frac{1}{3} \right)^x + 3$ and $h(x) = 3 + \log_3 x$ for clarity.)

Solution:
Key features for $g(x) = \left( \frac{1}{3} \right)^x + 3$:

• Exponential decay (base < 1), shifted up 3 units.
• As $x \to \infty$, $g(x) \to 3^+$ (horizontal asymptote: $y = 3$).
• As $x \to -\infty$, $g(x) \to \infty$.
• Y-intercept: $x = 0$, $g(0) = 1 + 3 = 4$ (point $(0, 4)$).
• Point: $x = 1$, $g(1) = \frac{1}{3} + 3 \approx 3.333$.

Key features for $h(x) = 3 + \log_3 x$:

• Logarithmic function, domain $x > 0$.
• Vertical asymptote: $x = 0$ (as $x \to 0^+$, $h(x) \to -\infty$).
• As $x \to \infty$, $h(x) \to \infty$.
• X-intercept: Set $h(x) = 0$: $3 + \log_3 x = 0 \implies \log_3 x = -3 \implies x = 3^{-3} = \frac{1}{27}$ (point $(\frac{1}{27}, 0)$).
• Point: $x = 1$, $h(1) = 3 + 0 = 3$.
• Point: $x = 3$, $h(3) = 3 + \log_3 3 = 3 + 1 = 4$.

Graph Sketch Description:

• Same diagram:
  • $g(x)$: Starts high left, passes through $(0,4)$, decays to $y=3$ as $x$ increases.
  • $h(x)$: Vertical asymptote at $x=0$, passes through $(\frac{1}{27}, 0)$, $(1,3)$, $(3,4)$.
• Intersections: Solve $g(x) = h(x)$: $\left( \frac{1}{3} \right)^x + 3 = 3 + \log_3 x \implies \left( \frac{1}{3} \right)^x = \log_3 x$.
  • At $x=1$, $g(1) \approx 3.333 > h(1) = 3$.
  • At $x=3$, $g(3) \approx 3.037 < h(3) = 4$.
  • As $x \to 0^+$, $g(x) \to \infty$, $h(x) \to -\infty$, so they intersect once in $(0,1)$.
  • They intersect again in $(1,3)$ (e.g., near $x \approx 1.5$).
• Sketch shows $g(x)$ decreasing from left, $h(x)$ increasing slowly from right of y-axis, intersecting twice.

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

$$d: \quad \log_2 (5x + 1) = 2 \log_4 (2x - 8)$$ $$y: \quad 2 \sin^2 x + \cos x - 1 = 0$$

Solution for $d$:
Simplify right side: $2 \log_4 (2x - 8)$.
Since $4 = 2^2$, $\log_4 z = \frac{\log_2 z}{\log_2 4} = \frac{\log_2 z}{2}$.
So:

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

Equation:

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

Arguments equal (same base, one-to-one):

$$5x + 1 = 2x - 8 \implies 3x = -9 \implies x = -3$$

Check domain:

• For $\log_2 (5x + 1)$: $5x + 1 > 0 \implies x > -\frac{1}{5}$.
• For $\log_2 (2x - 8)$: $2x - 8 > 0 \implies x > 4$.
  $x = -3$ satisfies neither ($5(-3)+1 = -14 < 0$, $2(-3)-8 = -14 < 0$), so not in domain.
  Thus, no solution.

Solution for $y$:

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

Use identity: $\sin^2 x = 1 - \cos^2 x$.
Substitute:

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

$$⟹2-2{\cos }^{2}x+\cos x-1=0$$

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

Multiply by $-1$:

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

Let $u = \cos x$:

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

Discriminant: $d = (-1)^2 - 4(2)(-1) = 1 + 8 = 9$.
Roots:

$$u = \frac{1 \pm \sqrt{9}}{4} = \frac{1 \pm 3}{4} \implies u = 1 \quad \text{or} \quad u = -\frac{1}{2}$$

So $\cos x = 1$ or $\cos x = -\frac{1}{2}$.

• $\cos x = 1 \implies x = 2k\pi, k \in \mathbb{Z}$.
• $\cos x = -\frac{1}{2} \implies x = \pm \frac{2\pi}{3} + 2k\pi, k \in \mathbb{Z}$.
  

Solutions: $x = 2k\pi, \quad x = \frac{2\pi}{3} + 2k\pi, \quad x = -\frac{2\pi}{3} + 2k\pi$ for integer $k$.

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(c) Prove the following trigonometric identities

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

Solution:
Left side:

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

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

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

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

Since $1 - \sin^2 x = \cos^2 x$.
Now, $\frac{2}{\cos^2 x} = 2 \sec^2 x$, as $\sec x = \frac{1}{\cos x}$.
Thus, left side equals right side.

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

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

Solution:

$$\sec \theta = \frac{1}{\cos \theta}$$

$\frac{7\pi}{6}$ radians = 210 degrees (third quadrant, reference angle $\frac{\pi}{6}$).
In third quadrant, cosine is negative:

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

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

Thus:

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

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

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

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(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

$$f(x) = 3 - 2 \cos 2x \quad \left[-\frac{\pi}{4}, 3\pi\right]$$

Solution:
Rewrite: $f(x) = -2 \cos(2x) + 3$.
Standard form: $A \cos(Bx + C) + D$.
Here, $A = -2$, $B = 2$, $C = 0$, $D = 3$.

• Amplitude: $|A| = |-2| = 2$.
• Period: $\frac{2\pi}{|B|} = \frac{2\pi}{2} = \pi$.
• Phase shift: $-\frac{C}{B} = 0$ (no shift).
• Vertical shift: $D = 3$.

Key points (one period $\pi$):

• $\cos(2x)$ cycles every $\pi$.
• When $\cos(2x) = 1$, $f(x) = 3 - 2(1) = 1$.
• When $\cos(2x) = -1$, $f(x) = 3 - 2(-1) = 5$.
• When $\cos(2x) = 0$, $f(x) = 3$.

Points in $[-\frac{\pi}{4}, 3\pi]$:

• $x = -\frac{\pi}{4}$: $2x = -\frac{\pi}{2}$, $\cos(-\frac{\pi}{2}) = 0$, $f = 3 - 2(0) = 3$.
• $x = 0$: $2x = 0$, $\cos(0) = 1$, $f = 3 - 2(1) = 1$.
• $x = \frac{\pi}{4}$: $2x = \frac{\pi}{2}$, $\cos(\frac{\pi}{2}) = 0$, $f = 3$.
• $x = \frac{\pi}{2}$: $2x = \pi$, $\cos(\pi) = -1$, $f = 3 - 2(-1) = 5$.
• $x = \frac{3\pi}{4}$: $2x = \frac{3\pi}{2}$, $\cos(\frac{3\pi}{2}) = 0$, $f = 3$.
• $x = \pi$: $2x = 2\pi$, $\cos(2\pi) = 1$, $f = 1$.
• $x = \frac{5\pi}{4}$: $2x = \frac{5\pi}{2} = 2\pi + \frac{\pi}{2}$, $\cos = 0$, $f = 3$.
• $x = \frac{3\pi}{2}$: $2x = 3\pi$, $\cos(3\pi) = -1$, $f = 5$.
• $x = \frac{7\pi}{4}$: $2x = \frac{7\pi}{2} = 3\pi + \frac{\pi}{2}$, $\cos = 0$, $f = 3$.
• $x = 2\pi$: $2x = 4\pi$, $\cos(4\pi) = 1$, $f = 1$.
• $x = \frac{9\pi}{4}$: $2x = \frac{9\pi}{2} = 4\pi + \frac{\pi}{2}$, $\cos = 0$, $f = 3$. (But $\frac{9\pi}{4} > 3\pi$, so stop at $3\pi$.)
• $x = 3\pi$: $2x = 6\pi$, $\cos(6\pi) = 1$, $f = 1$.

Graph Sketch Description:

• Range: $x$ from $-\frac{\pi}{4}$ to $3\pi$.
• Values oscillate between 1 and 5 with period $\pi$.
• Starts at $(-\frac{\pi}{4}, 3)$, decreases to $(0,1)$, increases to $(\frac{\pi}{2},5)$, decreases to $(\pi,1)$, and repeats.
• At $x=3\pi$, ends at $(3\pi,1)$.
• Shape: Cosine wave reflected over midline (due to negative A),                                                                                                       vertically stretched by 2, shifted up 3. Midline $y=3$.

📌 Graph Features Recap:

• Amplitude: 2
• Period: $\pi$
• Vertical Shift: Up 3 units
• Midline: $y=3$
• Phase Shift: None
• Oscillation Range: Between 1 and 5

@Dr. Microbiota

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