INTRODUCTION TO MATHEMATICAL METHODS QUIZ TWO JAN INTAKE 2024

DATE : 10th APRIL, 2024

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QUESTION ONE

a) The functions $f$ and $g$ have equations:
$f\left(x\right)=\mid x\mid ,x\in R$
$g\left(x\right)=\mid 2x+1\mid$

(ii) Sketch in the same diagram the graph of $f\left(x\right)$ and the graph of $g\left(x\right)$. The sketch must include the coordinates of any points where the graphs meet the coordinate axes. [4]

Answers:

• Graph of $f\left(x\right)=\mid x\mid$:

  • V-shaped graph with vertex at $(0,0)$.
  • Passes through points $(-2,2)$, $(-1,1)$, $(1,1)$, $(2,2)$.
  • Intercepts: Meets both axes at $(0,0)$.

• Graph of $g\left(x\right)=\mid 2x+1\mid$:

  • V-shaped graph with vertex at $(-\frac{1}{2},0)$.
  • Passes through points $(-1,1)$, $(0,1)$, $(-\frac{1}{2},0)$.
  • Intercepts:
    • Meets $x$-axis at $(-\frac{1}{2},0)$.
    • Meets $y$-axis at $(0,1)$.

Sketch Description (for standard graph paper):

1. Draw coordinate axes with $x$ from $-3$ to $3$ and $y$ from $0$ to $6$.
2. Plot $f\left(x\right)=\mid x\mid$:
   • Line from $(-3,3)$ to $(0,0)$ (slope $-1$).
   • Line from $(0,0)$ to $(3,3)$ (slope $1$).
3. Plot $g\left(x\right)=\mid 2x+1\mid$:
   • Line from $(-3,5)$ to $(-\frac{1}{2},0)$ (slope $2$).
   • Line from $(-\frac{1}{2},0)$ to $(3,7)$ (slope $2$).
4. Label intercepts:
   • $f\left(x\right)$: $(0,0)$.
   • $g\left(x\right)$: $(-\frac{1}{2},0)$ and $(0,1)$.

(iii) Find the $x$-coordinates of the points of intersection between the two graphs. [3]

Answer:
$x=-1,x=-\frac{1}{3}$

Explanation (Step-by-Step):
To find intersection points, solve $\mid x\mid =\mid 2x+1\mid$.

• Critical points (where expressions inside absolute values change sign):
  • For $\mid x\mid$: $x=0$.
  • For $\mid 2x+1\mid$: $2x+1=0⟹x=-\frac{1}{2}$.
• Divide real line into intervals based on critical points: $(-\infty ,-\frac{1}{2})$, $[-\frac{1}{2},0)$, $[0,\infty )$.

Case 1:

• Here, and , so:
  $\mid x\mid =-x,\mid 2x+1\mid =-(2x+1)$
  Equation: $-x=-(2x+1)$
  $-x=-2x-1⟹x=-1$
• Check: (valid).

Case 2:

• Here, and $2x+1\ge 0$, so:
  $\mid x\mid =-x,\mid 2x+1\mid =2x+1$
  Equation: $-x=2x+1$
  $-x-2x=1⟹-3x=1⟹x=-\frac{1}{3}$
• Check: (valid).

Case 3: $x\ge 0$

• Here, $x\ge 0$ and $2x+1>0$, so:
  $\mid x\mid =x,\mid 2x+1\mid =2x+1$
  Equation: $x=2x+1$
  $x-2x=1⟹-x=1⟹x=-1$
• Check: $x=-1\ge ̸0$ (invalid).

Conclusion: Intersections at $x=-1$ and $x=-\frac{1}{3}$.

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b) Solve the equation $\mid -4x-1\mid =-\mid 3x-27\mid$ [3]

Answer:
No solution.

Explanation (Step-by-Step):

• Left side: $\mid -4x-1\mid \ge 0$ (absolute value is always non-negative).
• Right side: $-\mid 3x-27\mid \le 0$ (negative of absolute value is always non-positive).
• Equality holds only if both sides are zero:
  $\mid -4x-1\mid =0\text{and}-\mid 3x-27\mid =0$
  • Solve $\mid -4x-1\mid =0$:
    $-4x-1=0⟹x=-\frac{1}{4}$
  • Solve $-\mid 3x-27\mid =0$:
    $\mid 3x-27\mid =0⟹3x-27=0⟹x=9$
• Contradiction: $x=-\frac{1}{4}$ and $x=9$ cannot both be true.
• Conclusion: No $x$ satisfies the equation.

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QUESTION TWO

a) The function $f\left(x\right)=x^2-2x+6$

(i) Complete the square. [3]

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

Explanation (Step-by-Step):

• Formula: For $a{x}^2+bx+c$, rewrite as $a{(x+\frac{b}{2a})}^2+(c-\frac{b^2}{4a})$.
• Here, $a=1$, $b=-2$, $c=6$:
  $f\left(x\right)=x^2-2x+6$
  $f\left(x\right)=(x^2-2x+{\left(\frac{-2}{2}\right)}^2)+6-{\left(\frac{-2}{2}\right)}^2$
  $f\left(x\right)=(x^2-2x+1)+6-1$
  $f\left(x\right)=(x-1{)}^2+5$

(ii) Hence sketch the graph of the function, showing clearly the intercepts, turning point, and line of symmetry. [4]

Answer:

• Turning point (vertex): $(1,5)$ (from $(x-1{)}^2+5$).
• Line of symmetry: $x=1$.
• $y$-intercept: At $x=0$:
  $f\left(0\right)=0^2-2\left(0\right)+6=6⟹(0,6)$
• $x$-intercepts: Set $f\left(x\right)=0$:
  $(x-1{)}^2+5=0⟹(x-1{)}^2=-5$
  No real solutions (since square is non-negative), so no $x$-intercepts.

Sketch Description (for standard graph paper):

1. Draw coordinate axes with $x$ from $-2$ to $4$ and $y$ from $0$ to $10$.
2. Plot vertex: $(1,5)$.
3. Plot $y$-intercept: $(0,6)$.
4. Symmetry: Line $x=1$ (vertical line through vertex).
5. Additional points:
   • $x=2$: $f\left(2\right)=(2{)}^2-2(2)+6=6⟹(2,6)$.
   • $x=3$: $f\left(3\right)=9-6+6=9⟹(3,9)$.
   • Symmetric point to $(0,6)$: $(2,6)$ (since line of symmetry is $x=1$).
6. Shape: Parabola opening upwards (minimum at vertex).

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b) The roots of the quadratic equation $2x^2+4x-3=0$ are $\alpha$ and $\beta$. Without solving the equation, find the exact values of ${\alpha }^3+{\beta }^3$. [3]

Answer:
${\alpha }^3+{\beta }^3=-17$

Explanation (Step-by-Step):

• Quadratic relations: For $a{x}^2+bx+c=0$,
  $\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$
  Here, $a=2$, $b=4$, $c=-3$:
  $\alpha +\beta =-\frac{4}{2}=-2$
  $\alpha \beta =\frac{-3}{2}=-\frac{3}{2}$

• Identity for ${\alpha }^3+{\beta }^3$:
  ${\alpha }^3+{\beta }^3=(\alpha +\beta )({\alpha }^2-\alpha \beta +{\beta }^2)$
  ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$

• Compute ${\alpha }^2+{\beta }^2$:
  ${\alpha }^2+{\beta }^2=(-2{)}^2-2(-\frac{3}{2})=4+3=7$

• Compute ${\alpha }^2-\alpha \beta +{\beta }^2$:
  ${\alpha }^2-\alpha \beta +{\beta }^2=({\alpha }^2+{\beta }^2)-\alpha \beta =7-(-\frac{3}{2})=7+\frac{3}{2}=\frac{17}{2}$

• Compute ${\alpha }^3+{\beta }^3$:
  ${\alpha }^3+{\beta }^3=(-2)\times \frac{17}{2}=-17$

Conclusion: ${\alpha }^3+{\beta }^3=-17$.

OTHER WAYS TO SOLVE THESE 

QUESTION ONE

a) Functions $f(x) = |x|$ and $g(x) = |2x + 1|$
(ii) Sketch the graphs of $f(x)$ and $g(x)$ in the same diagram, including intercepts with axes.

• Graph of $f(x) = |x|$:

  • V-shaped graph with vertex at (0, 0).
  • Intercepts:
    • X-intercept and y-intercept: (0, 0).
  • Passes through points: (-2, 2), (-1, 1), (1, 1), (2, 2).
  • For $x < 0$, $f(x) = -x$ (slope = -1).
  • For $x \geq 0$, $f(x) = x$ (slope = 1).

• Graph of $g(x) = |2x + 1|$:

  • V-shaped graph with vertex at (-0.5, 0).
  • Intercepts:
    • X-intercept: (-0.5, 0) (since $|2(-0.5) + 1| = 0$).
    • Y-intercept: (0, 1) (since $|2(0) + 1| = 1$).
  • Passes through points: (-1, 1), (-0.5, 0), (0, 1), (1, 3).
  • For $x < -0.5$, $g(x) = -2x - 1$ (slope = -2).
  • For $x \geq -0.5$, $g(x) = 2x + 1$ (slope = 2).

• Intersection Points (from part iii): (-1, 1) and (-$\frac{1}{3}$, $\frac{1}{3}$) ≈ (-0.333, 0.333).

Sketch Description:

• Draw coordinate axes.
• For $f(x)$: Plot vertex (0, 0), and points (-1, 1), (1, 1). Draw lines with slopes ±1.
• For $g(x)$: Plot vertex (-0.5, 0), y-intercept (0, 1), and points (-1, 1), (1, 3). Draw lines with slopes ±2.
• Label intercepts: (0, 0) for $f$, (-0.5, 0) and (0, 1) for $g$.
• Label intersection points: (-1, 1) and (-$\frac{1}{3}$, $\frac{1}{3}$).
• The graphs intersect at two points, and $g(x)$ is steeper than $f(x)$.

(iii) Find the x-coordinates of the points of intersection.
To find where $f(x) = g(x)$:
$|x| = |2x + 1|$
The critical points (where expressions inside absolute values change sign) are $x = -0.5$ and $x = 0$. Divide into intervals: $x < -0.5$, $-0.5 \leq x < 0$, and $x \geq 0$.

• Case 1: $x < -0.5$
  Here, $|x| = -x$ and $|2x + 1| = -(2x + 1) = -2x - 1$.
  Equation:
  $-x = -2x - 1$
  Solve:
  $-x + 2x = -1 \implies x = -1$
  Check: $x = -1 < -0.5$, valid.

• Case 2: $-0.5 \leq x < 0$
  Here, $|x| = -x$ and $|2x + 1| = 2x + 1$.
  Equation:
  $-x = 2x + 1$
  Solve:
  $-x - 2x = 1 \implies -3x = 1 \implies x = -\frac{1}{3}$
  Check: $-0.5 \leq -\frac{1}{3} \approx -0.333 < 0$, valid.

• Case 3: $x \geq 0$
  Here, $|x| = x$ and $|2x + 1| = 2x + 1$.
  Equation:
  $x = 2x + 1 \implies -x = 1 \implies x = -1$
  Check: $x = -1 \not\geq 0$, invalid.

Solution: The x-coordinates of intersection are $x = -1$ and $x = -\frac{1}{3}$.

b) Solve the equation $|-4x - 1| = -|3x - 27|$

• The left side, $|-4x - 1|$, is always non-negative ($\geq 0$).
• The right side, $-|3x - 27|$, is always non-positive ($\leq 0$) because it is the negative of an absolute value.
• For equality, both sides must be zero simultaneously:
  $|-4x - 1| = 0 \quad \text{and} \quad -|3x - 27| = 0$
• Solve $|-4x - 1| = 0$:
  $-4x - 1 = 0 \implies x = -\frac{1}{4}$
• Solve $-|3x - 27| = 0$:
  $|3x - 27| = 0 \implies 3x - 27 = 0 \implies x = 9$
• At $x = -\frac{1}{4}$:
  Left side: $|-4(-\frac{1}{4}) - 1| = |1 - 1| = 0$
  Right side: $-|3(-\frac{1}{4}) - 27| = -|-\frac{3}{4} - 27| = -|-\frac{111}{4}| = -\frac{111}{4} \neq 0$
• At $x = 9$:
  Left side: $|-4(9) - 1| = |-36 - 1| = 37 \neq 0$
  Right side: $-|3(9) - 27| = -|27 - 27| = 0$
• No $x$ satisfies both conditions simultaneously.

Solution: The equation has no solution. The solution set is empty.

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QUESTION TWO

a) Function $f(x) = x^2 - 2x + 6$
(i) Complete the square.

• Formula: For $ax^2 + bx + c$, rewrite as $a(x + \frac{b}{2a})^2 + \text{constant}$.
• Here, $a = 1$, $b = -2$, $c = 6$.
• Half of $b$ is $-1$, square is $1$.
• Add and subtract $1$:
  $x^2 - 2x + 6 = (x^2 - 2x + 1) + 6 - 1 = (x - 1)^2 + 5$
• Completed square form: $f(x) = (x - 1)^2 + 5$.

(ii) Sketch the graph, showing intercepts, turning point, and line of symmetry.

• Turning Point (Vertex): From $f(x) = (x - 1)^2 + 5$, vertex is at (1, 5).
• Line of Symmetry: Vertical line through vertex, $x = 1$.
• Y-intercept: Set $x = 0$:
  $f(0) = (0)^2 - 2(0) + 6 = 6 \quad \text{→ Point: (0, 6)}$
• X-intercepts: Set $y = 0$:
  $x^2 - 2x + 6 = 0$
  Discriminant: $(-2)^2 - 4(1)(6) = 4 - 24 = -20 < 0$ → No real roots, so no x-intercepts.
• Behavior: Since coefficient of $x^2$ is positive, parabola opens upwards.
• Additional Point: At $x = 2$, $f(2) = (2)^2 - 2(2) + 6 = 4 - 4 + 6 = 6$ → Point (2, 6). Symmetric to (0, 6) about $x = 1$.

Sketch Description:

• Plot vertex (1, 5).
• Plot y-intercept (0, 6) and symmetric point (2, 6).
• Draw parabola opening upwards, symmetric about $x = 1$, with minimum at (1, 5), and passing through (0, 6) and (2, 6).
• Label: vertex (1, 5), y-intercept (0, 6), and line of symmetry $x = 1$.

b) Roots of $2x^2 + 4x - 3 = 0$ are $\alpha$ and $\beta$. Find $\alpha^3 + \beta^3$ without solving.

• Key Formulas:
  • Sum of roots: $\alpha + \beta = -\frac{b}{a}$
  • Product of roots: $\alpha \beta = \frac{c}{a}$
  • $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)$
  • $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$
• Here, $a = 2$, $b = 4$, $c = -3$.
• Calculate:
  $\alpha + \beta = -\frac{b}{a} = -\frac{4}{2} = -2$
  $\alpha \beta = \frac{c}{a} = \frac{-3}{2} = -\frac{3}{2}$
• Find $\alpha^2 + \beta^2$:
  $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = (-2)^2 - 2(-\frac{3}{2}) = 4 - (-3) = 4 + 3 = 7$
• Find $\alpha^2 - \alpha \beta + \beta^2$:
  $\alpha^2 - \alpha \beta + \beta^2 = (\alpha^2 + \beta^2) - \alpha \beta = 7 - (-\frac{3}{2}) = 7 + \frac{3}{2} = \frac{14}{2} + \frac{3}{2} = \frac{17}{2}$
• Find $\alpha^3 + \beta^3$:
  $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2) = (-2) \left( \frac{17}{2} \right) = -\frac{34}{2} = -17$

Solution: $\alpha^3 + \beta^3 = -17$ (exact value).

@Dr. Microbiota

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End of Solutions