INTRODUCTION TO MATHEMATICAL METHODS DIFFERED QUIZ/TEST TWO JULY INTAKE 2024

DATE : WEDNESDAY 6th NOVEMBER, 2024

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QUESTION ONE

a. Two functions $f$ and $g$ are defined as $f(x) = \frac{x + 6}{2x - 3}$ and $g(x) = 5x - 8$.
i. Find the value of $g$ of $f(3)$ [2]
ii. Find $g$ of $f^{-1}(x)$ [3]

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Solution to a(i):
Step 1: Compute $f(3)$.

$$f(3)=\frac{3+6}{2(3)-3}$$

$$=\frac{9}{6-3}$$

$$=\frac{9}{3}$$

$$f(3) = \frac{3 + 6}{2(3) - 3} = \frac{9}{6 - 3} = \frac{9}{3} = 3.$$

Step 2: Compute $g(f(3)) = g(3)$.

$$g(3)=5(3)-8$$

$$=15-8$$

$$g(3) = 5(3) - 8 = 15 - 8 = 7.$$

Answer: $\boxed{7}$

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Solution to a(ii):
Step 1: Find $f^{-1}(x)$.
Let $y = f(x) = \frac{x + 6}{2x - 3}$. Solve for $x$:

$$y(2x-3)=x+6$$

$$⟹2xy-3y=x+6$$

$$⟹2xy-x=3y+6$$

$$⟹x(2y-1)=3y+6.$$

$$y(2x - 3) = x + 6 \implies 2xy - 3y = x + 6 \implies 2xy - x = 3y + 6 \implies x(2y - 1) = 3y + 6.$$ $$x=\frac{3y+6}{2y-1}$$

$$x = \frac{3y + 6}{2y - 1} \implies f^{-1}(x) = \frac{3x + 6}{2x - 1}.$$

Step 2: Compute $g(f^{-1}(x)) = g\left(\frac{3x + 6}{2x - 1}\right)$.

$$g\left(\frac{3x+6}{2x-1}\right)=5\left(\frac{3x+6}{2x-1}\right)-8$$

$$=\frac{5(3x+6)}{2x-1}-8$$

$$g\left(\frac{3x + 6}{2x - 1}\right) = 5\left(\frac{3x + 6}{2x - 1}\right) - 8 = \frac{5(3x + 6)}{2x - 1} - 8 = \frac{15x + 30}{2x - 1} - \frac{8(2x - 1)}{2x - 1}.$$ $$$$

$$=\frac{15x+30-16x+8}{2x-1}$$

$$= \frac{15x + 30 - 16x + 8}{2x - 1} = \frac{-x + 38}{2x - 1}.$$

Answer: $\boxed{\dfrac{-x + 38}{2x - 1}}$

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b. Let $f(x) = 2x^2 - 5x + 3$ be a quadratic function.
i. Find the turning point, x-intercepts, and y-intercepts for this function.
ii. Hence, sketch $f(x)$. [5]

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Solution to b(i):
Turning Point (Vertex):

$$x=-\frac{b}{2a}$$

$$=\frac{5}{2X2}$$

$$=\frac{5}{4}$$

$$=\mathrm{1.25.}$$

$$x = -\frac{b}{2a} = \frac{5}{2 \cdot 2} = \frac{5}{4} = 1.25.$$ $$f(1.25)=2(1.25{)}^2-5(1.25)+3$$

$$=2(1.5625)-6.25+3$$

$$=3.125-6.25+3$$

$$f(1.25) = 2(1.25)^2 - 5(1.25) + 3 = 2(1.5625) - 6.25 + 3 = 3.125 - 6.25 + 3 = -0.125.$$

Vertex: $\boxed{(1.25, -0.125)}$ or $\left(\dfrac{5}{4}, -\dfrac{1}{8}\right)$.

x-intercepts: Solve $2x^2 - 5x + 3 = 0$.

$$x=\frac{5\pm \sqrt{(-5{)}^2-4\cdot 2\cdot 3}}{4}$$

$$=\frac{5\pm \sqrt{25-24}}{4}$$

$$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 3}}{4} = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm 1}{4}.$$ $$$$

$$x = \frac{6}{4} = 1.5 \quad \text{or} \quad x = \frac{4}{4} = 1.$$

x-intercepts: $\boxed{(1, 0), (1.5, 0)}$ or $\left(1, 0\right), \left(\dfrac{3}{2}, 0\right)$.

y-intercept: $f(0) = 3$.
y-intercept: $\boxed{(0, 3)}$.

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Solution to b(ii): Sketch of $f(x)$

• Shape: Parabola opening upwards (since $a = 2 > 0$).
• Vertex: $\left(1.25, -0.125\right)$.
• x-intercepts: $(1, 0)$, $(1.5, 0)$.
• y-intercept: $(0, 3)$.

Graph:

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C. Show that $f\left(x\right)=7x^5+x^3-3x$ is an odd but not even function. [2]

Solution:
Odd Function Test: $f(-x) = -f(x)$.

$$f(-x)=7(-x{)}^5+(-x{)}^3-3(-x)$$

$$=-7x^5-x^3+3x$$

$$=-(7x^5+x^3-3x)$$

$$f(-x) = 7(-x)^5 + (-x)^3 - 3(-x) = -7x^5 - x^3 + 3x = -(7x^5 + x^3 - 3x) = -f(x).$$

Thus, $f$ is odd.

Even Function Test: $f(-x) = f(x)$.

$$f(-x) = -7x^5 - x^3 + 3x \neq f(x) \quad (\text{since } f(x) = 7x^5 + x^3 - 3x).$$

Thus, $f$ is not even.
Answer: $\boxed{\text{Odd but not even}}$

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d. Given two equations $xy = 12$ and $2x^2 + 3xy - y = 58$. Solve simultaneously, given $x > y$. [5]

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Solution:
Step 1: Substitute $y = \frac{12}{x}$ into the second equation:

$$2x^2+3x\left(\frac{12}{x}\right)-\frac{12}{x}=58$$

$$2x^2 + 3x\left(\frac{12}{x}\right) - \frac{12}{x} = 58 \implies 2x^2 + 36 - \frac{12}{x} = 58.$$

Step 2: Multiply through by $x$ to eliminate denominator:

$$2x^3+36x-12=58x$$

$$2x^3 + 36x - 12 = 58x \implies 2x^3 - 22x - 12 = 0.$$

Step 3: Divide by 2:

$$x^3 - 11x - 6 = 0.$$

Step 4: Factor using Rational Root Theorem. Possible roots: $\pm1, 2, 3, 6$.
Try $x = 3$: $3^3 - 11(3) - 6 = 27 - 33 - 6 = -12 \neq 0$.
Try $x = -2$: $(-2)^3 - 11(-2) - 6 = -8 + 22 - 6 = 8 \neq 0$.
Try $x = -3$: $(-3)^3 - 11(-3) - 6 = -27 + 33 - 6 = 0$.
Thus, $(x + 3)$ is a factor.

Step 5: Polynomial division:

$$x^3 - 11x - 6 \div (x + 3).$$

Quotient: $x^2 - 3x - 2$.

$$x^3-11x-\mathrm{6\; =\; 0}$$

$$x^3 - 11x - 6 = (x + 3)(x^2 - 3x - 2) = 0.$$

Step 6: Solve:
$x+3=0$

$x + 3 = 0 \implies x = -3$.

AND
$x^2-3x-2=0$

$⟹x=\frac{3\pm \sqrt{9+8}}{2}$

$x^2 - 3x - 2 = 0 \implies x = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}$.

Step 7: Find corresponding $y = \frac{12}{x}$:

• If $x = -3$, $y = \frac{12}{-3} = -4$.
• If $x = \frac{3 + \sqrt{17}}{2} \approx 3.56$, $y = \frac{12}{3.56} \approx 3.37$.
• If $x = \frac{3 - \sqrt{17}}{2} \approx -0.56$, $y = \frac{12}{-0.56} \approx -21.43$.

Step 8: Apply $x > y$:

• $(-3, -4)$: $-3 > -4$ ✓
• $(3.56, 3.37)$: $3.56 > 3.37$ ✓
• $(-0.56, -21.43)$: $-0.56 > -21.43$ ✓
  All pairs satisfy $x > y$.

Answer: $\boxed{(-3, -4)}$, $\boxed{\left( \dfrac{3 + \sqrt{17}}{2}, \dfrac{24}{3 + \sqrt{17}} \right)}$, $\boxed{\left( \dfrac{3 - \sqrt{17}}{2}, \dfrac{24}{3 - \sqrt{17}} \right)}$

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e. Find $k$ such that $x^2 - 3kx + (k - 1) = 0$ has equal roots. [4]

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Solution:
Condition for equal roots: Discriminant $D = 0$.

$$D=b^2-4ac$$

$$=(-3k{)}^2-4(1)(k-1)$$

$$D = b^2 - 4ac = (-3k)^2 - 4(1)(k - 1) = 9k^2 - 4k + 4.$$

Set $D = 0$:

$$9k^2 - 4k + 4 = 0.$$

Solve for $k$:

$$k=\frac{4\pm \sqrt{(-4{)}^2-4\cdot 9\cdot 4}}{18}$$

$$=\frac{4\pm \sqrt{16-144}}{18}$$

$$k = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 9 \cdot 4}}{18} = \frac{4 \pm \sqrt{16 - 144}}{18} = \frac{4 \pm \sqrt{-128}}{18}.$$

$\sqrt{-128} = i\sqrt{128}$ (complex). Recheck discriminant:

$$D=(-3k{)}^2-4(1)(k-1)$$

$$=9k^2-4(k-1)$$

$$=9k^2-4k+4.$$

$$D = (-3k)^2 - 4(1)(k - 1) = 9k^2 - 4(k - 1) = 9k^2 - 4k + 4.$$ $$D = 9k^2 - 4k + 4 \implies \text{Discriminant of this quadratic: } (-4)^2 - 4 \cdot 9 \cdot 4 = 16 - 144 = -128 < 0.$$

No real $k$ satisfies $D = 0$. Error in problem?
Assuming real $k$ expected:

$$D=b^2-4ac$$

$$=(-3k{)}^2-4(1)(k-1)$$

$$D = b^2 - 4ac = (-3k)^2 - 4(1)(k - 1) = 9k^2 - 4k + 4.$$

Minimum value of $D$: Vertex at $k = \frac{4}{18} = \frac{2}{9}$,

$$D\left(\frac{2}{9}\right)=9\left(\frac{4}{81}\right)-4\left(\frac{2}{9}\right)+4$$

$$=\frac{36}{81}-\frac{8}{9}+4$$

$$=\frac{4}{9}-\frac{8}{9}+4$$

$$D\left(\frac{2}{9}\right) = 9\left(\frac{4}{81}\right) - 4\left(\frac{2}{9}\right) + 4 = \frac{36}{81} - \frac{8}{9} + 4 = \frac{4}{9} - \frac{8}{9} + 4 = -\frac{4}{9} + 4 > 0.$$

Correction:

$$D=(-3k{)}^2-4(1)(k-1)$$

$$=9k^2-4(k-1)$$

$$D = (-3k)^2 - 4(1)(k - 1) = 9k^2 - 4(k - 1) = 9k^2 - 4k + 4.$$

Set to zero: $9k^2 - 4k + 4 = 0$.
Discriminant: $d = (-4)^2 - 4 \cdot 9 \cdot 4 = 16 - 144 = -128 < 0$.
No real solutions. Possible typo in problem?

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QUESTION TWO

a. Resolve $\frac{2x^3 + x^2 - 15x - 5}{(x + 3)(x - 2)}$ into partial fractions. [5]

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Solution:
Step 1: Degree check: Numerator degree (3) > Denominator degree (2). Perform polynomial division.
Divide $2x^3 + x^2 - 15x - 5$ by $(x+3)(x-2) = x^2 + x - 6$:

$$\begin{array}{r}2x-1\end{array}$$

$$\begin{array}{r}\mathrm{}\\ x^2+x-6\stackrel{}{)2x^3+x^2-15x-5}\end{array}$$

$$\begin{array}{r}\overline{\mathrm{}}\\ \underset{}{-(2x^3+2x^2-12x)}\end{array}$$

$$\begin{array}{r}\underset{‾}{}\\ -x^2-3x-5\end{array}$$

$$\begin{array}{r}\mathrm{}\\ \underset{}{-(-x^2-x+6)}\end{array}$$

$$\begin{array}{r}\underset{‾}{}\\ -2x-11\end{array}$$

$$\begin{array}{r} 2x - 1 \\ x^2 + x - 6 \overline{)2x^3 + x^2 - 15x - 5} \\ \underline{-(2x^3 + 2x^2 - 12x)} \\ -x^2 - 3x - 5 \\ \underline{-(-x^2 - x + 6)} \\ -2x - 11 \\ \end{array}$$

So,

$$\frac{2x^3 + x^2 - 15x - 5}{(x + 3)(x - 2)} = 2x - 1 + \frac{-2x - 11}{(x + 3)(x - 2)}.$$

Step 2: Partial fractions for remainder:

$$\frac{-2x - 11}{(x + 3)(x - 2)} = \frac{A}{x + 3} + \frac{B}{x - 2}.$$ $$$$

$$-2x - 11 = A(x - 2) + B(x + 3).$$

Step 3: Solve for $A$ and $B$:
Set $x = -3$:

$$-2(-3)-11=A(-3-2)$$

$$⟹6-11=A(-5)$$

$$⟹-5=-5A$$

$$-2(-3) - 11 = A(-3 - 2) \implies 6 - 11 = A(-5) \implies -5 = -5A \implies A = 1.$$

Set $x = 2$:

$$-2(2)-11=B(2+3)$$

$$⟹-4-11=5B$$

$$⟹-15=5B$$

$$-2(2) - 11 = B(2 + 3) \implies -4 - 11 = 5B \implies -15 = 5B \implies B = -3.$$

Step 4: Combine:

$$\frac{2x^3 + x^2 - 15x - 5}{(x + 3)(x - 2)} = 2x - 1 + \frac{1}{x + 3} - \frac{3}{x - 2}.$$

Answer: $\boxed{2x - 1 + \dfrac{1}{x + 3} - \dfrac{3}{x - 2}}$

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b. Divide $P(x) = x^4 + 2x^3 - 3x^2 + 4x - 4$ by $x + 2$. Write quotient $q(x)$ and remainder $r(x)$. [5]

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Solution:
Use synthetic division (root = $-2$):

$$\begin{array}{rrrrrr}-2& 1& 2& -3& 4& -4\\ & & -2& 0& 6& -20\\ & 1& 0& -3& 10& -24\end{array}$$

To divide
P(x) = x⁴ + 2x³ − 3x² + 4x − 4
by x + 2, we use synthetic division with −2 (the root of x + 2).

• Coefficients of P(x): 1 2 −3 4 −4
• Bring down 1.
• Multiply 1 × (−2) = −2, add to 2 → 0.
• Multiply 0 × (−2) = 0, add to −3 → −3.
• Multiply −3 × (−2) = 6, add to 4 → 10.
• Multiply 10 × (−2) = −20, add to −4 → −24.

From these steps, the quotient polynomial is
q(x) = x³ + 0x² − 3x + 10 ≡ x³ − 3x + 10

and the remainder is
r(x) = −24

$$\begin{array}{r|rrrrr} -2 & 1 & 2 & -3 & 4 & -4 \\ & & -2 & 0 & 6 & -20 \\ \hline & 1 & 0 & -3 & 10 & -24 \\ \end{array}$$

Interpretation:

• Quotient: $x^3 + 0x^2 - 3x + 10 = \boxed{x^3 - 3x + 10}$
• Remainder: $\boxed{-24}$

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c. Solve the equations:
i. $\log_9(x + 3) + \log_9(x - 3) = \log_3 x$ [4]
ii. $3^{2x} - 3^x = -2$ [3]

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Solution to c(i):
Step 1: Domain: $x > 3$ (since $x + 3 > 0$, $x - 3 > 0$, $x > 0$).
Step 2: Use $\log_b a = \frac{\log_k a}{\log_k b}$. Rewrite all as base 3:

$$\log_9(a) = \frac{\log_3 a}{\log_3 9} = \frac{\log_3 a}{2}.$$

Equation:

$$\frac{1}{2} \log_3(x + 3) + \frac{1}{2} \log_3(x - 3) = \log_3 x.$$

Step 3: Simplify:

$$\frac{1}{2}[{\log }_3(x+3)+{\log }_3(x-3)]={\log }_{3}x$$

$$\frac{1}{2} \left[ \log_3(x + 3) + \log_3(x - 3) \right] = \log_3 x \implies \log_3 \left( (x + 3)(x - 3) \right)^{\frac{1}{2}} = \log_3 x.$$ $${}_{}$$

$$\log_3 \sqrt{x^2 - 9} = \log_3 x.$$

Step 4: Equate arguments:

$$\sqrt{x^2-9}=x\text{(since }{\log }_{b}a={\log }_{b}c⟹a=c\text{)}\mathrm{.}$$

$$\sqrt{x^2 - 9} = x \quad \text{(since $\log_b a = \log_b c \implies a = c$)}.$$ $$x^2-9=x^2$$

$$x^2 - 9 = x^2 \implies -9 = 0 \quad \text{Contradiction!}$$

Re-examine: $\sqrt{x^2 - 9} = x$ only if $x \geq 0$, but squaring:

$$x^2-9=x^2$$

$$x^2 - 9 = x^2 \implies -9 = 0 \quad \text{impossible}.$$

Step 5: Alternative approach:

$${\log }_3\sqrt{x^2-9}={\log }_{3}x$$

$$\log_3 \sqrt{x^2 - 9} = \log_3 x \implies \sqrt{x^2 - 9} = x.$$

This holds only if $x^2 - 9 = x^2$ and $x \geq 0$, which is impossible. Check domain: $x > 3$.
Correct step after log equality:

$$\sqrt{x^2-9}=x$$

$$\sqrt{x^2 - 9} = x \implies x^2 - 9 = x^2 \quad \text{false}.$$

Error in step: $\log_3 \sqrt{x^2 - 9} = \log_3 x$ implies $\sqrt{x^2 - 9} = x$ only if the log is one-to-one, but $x$ must be positive, which is true.
Correct solution:
From:

$$\log_3 \sqrt{x^2 - 9} = \log_3 x \implies \sqrt{x^2 - 9} = x.$$

Square both sides:

$$x^2 - 9 = x^2 \implies -9 = 0 \quad \text{no solution}.$$

Possible extraneous solution?  Check original equation.

At $x = 3$ (boundary):
Left side: $\log_9(6) + \log_9(0)$ undefined.

Conclusion: No solution.

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Solution to c(ii):
Step 1: Let $u = 3^x$. Then $3^{2x} = u^2$.

$$u^2-u=-2$$

$$u^2 - u = -2 \implies u^2 - u + 2 = 0.$$

Step 2: Discriminant: $d=(-1{)}^2-4X1X2$

$=1-8$

$d = (-1)^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7 < 0$.
No real solutions.
Answer: $\boxed{\text{No real solutions}}$

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d. Sketch the graph of $f(x) = 4^x - 3$. Show all intercepts. [3]

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Solution:
Intercepts:

• y-intercept: $f(0) = 4^0 - 3 = 1 - 3 = -2$. Point: $(0, -2)$.
• x-intercept: $4^x - 3 = 0 \implies 4^x = 3 \implies x = \log_4 3 \approx 0.792$. Point: $(\log_4 3, 0)$.
  Behavior:
• As $x \to \infty$, $f(x) \to \infty$.
• As $x \to -\infty$, $4^x \to 0$, so $f(x) \to -3$ (horizontal asymptote: $y = -3$).
  Graph:

Note: Graph passes through $(0, -2)$, $(\log_4 3, 0)$, and increases rapidly for $x > 0$.

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e. Let $\alpha$ and $\beta$ be roots of $3x^2 - 9x + 1 = 0$. Find:
i. $\alpha^3 + \beta^3$
ii. The equation with roots $\frac{2\alpha}{\beta}$ and $\frac{2\beta}{\alpha}$

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Solution to e(i):
Step 1: Sum $\alpha + \beta = -\frac{b}{a} = \frac{9}{3} = 3$.
Product $\alpha\beta = \frac{c}{a} = \frac{1}{3}$.
Step 2: Use identity:

$${\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )$$

$$=3^3-3\cdot \frac{1}{3}\cdot 3$$

$$=27-3$$

$$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = 3^3 - 3 \cdot \frac{1}{3} \cdot 3 = 27 - 3 = 24.$$

Answer: $\boxed{24}$

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Solution to e(ii):
Step 1: Let roots be $r = \frac{2\alpha}{\beta}$, $s = \frac{2\beta}{\alpha}$.
Step 2: Sum:

$$r+s=2(\frac{\alpha }{\beta }+\frac{\beta }{\alpha })$$

$$=2\left(\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }\right)$$

$$r + s = 2\left( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \right) = 2\left( \frac{\alpha^2 + \beta^2}{\alpha\beta} \right).$$ $${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$$

$$=9-2X\frac{1}{3}$$

$$=9-\frac{2}{3}$$

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 9 - 2 \cdot \frac{1}{3} = 9 - \frac{2}{3} = \frac{25}{3}.$$ $$$$

$$r+s=2X\frac{\frac{25}{3}}{\frac{1}{3}}$$

$$=2X25$$

$$r + s = 2 \cdot \frac{\frac{25}{3}}{\frac{1}{3}} = 2 \cdot 25 = 50.$$

Step 3: Product:

$$rs = \left( \frac{2\alpha}{\beta} \right) \left( \frac{2\beta}{\alpha} \right) = 4.$$

Step 4: Quadratic equation: $t^2 - (r + s)t + rs = 0$:

$$t^2 - 50t + 4 = 0.$$

Answer: $\boxed{t^2 - 50t + 4 = 0}$

@Dr. Microbiota

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