INTRODUCTION TO MATHEMATICAL METHODS TEST TWO JULY INTAKE 2024

DATE : 30th OCTOBER, 2024

QUESTION ONE

A. The functions $f(x)$ and $g(x)$ are given by the following $f(x) = \frac{x-5}{x+10}$ and $g(x) = 2x + 4$
I. Calculate the value of $gof(2)$
Solution:

$$gof(2) = g(f(2))$$

First, compute $f(2)$:

$$f(2)=\frac{2-5}{2+10}$$

$$=\frac{-3}{12}$$

$$f(2) = \frac{2-5}{2+10} = \frac{-3}{12} = -\frac{1}{4}$$

Now, compute $g(f(2))$:

$$g(-\frac{1}{4})=2(-\frac{1}{4})+4$$

$$=-\frac{1}{2}+4$$

$$g\left(-\frac{1}{4}\right) = 2 \left(-\frac{1}{4}\right) + 4 = -\frac{1}{2} + 4 = \frac{7}{2}$$

Answer: $\frac{7}{2}$

II. Calculate the value of $fof(x)^{-1}$
Solution:
The notation $fof(x)^{-1}$ is interpreted as the inverse of $f \circ f$, denoted as $(f \circ f)^{-1}(x)$.
First, compute $f \circ f(x) = f(f(x))$:

$$f(f(x))=f\left(\frac{x-5}{x+10}\right)=\frac{\frac{x-5}{x+10}-5}{\frac{x-5}{x+10}+10}$$

$$=\frac{\frac{x-5-5(x+10)}{x+10}}{\frac{x-5+10(x+10)}{x+10}}$$

$$=\frac{\frac{x-5-5x-50}{x+10}}{\frac{x-5+10x+100}{x+10}}$$

$$=\frac{\frac{-4x-55}{x+10}}{\frac{11x+95}{x+10}}$$

$$f(f(x)) = f\left(\frac{x-5}{x+10}\right) = \frac{\frac{x-5}{x+10} - 5}{\frac{x-5}{x+10} + 10} = \frac{\frac{x-5 - 5(x+10)}{x+10}}{\frac{x-5 + 10(x+10)}{x+10}} = \frac{\frac{x-5-5x-50}{x+10}}{\frac{x-5+10x+100}{x+10}} = \frac{\frac{-4x-55}{x+10}}{\frac{11x+95}{x+10}} = \frac{-4x-55}{11x+95}$$

Now, find the inverse of $y = \frac{-4x-55}{11x+95}$:

$$y = \frac{-4x-55}{11x+95}$$

Solve for $x$:

$$y(11x+95)=-4x-55$$

$$y(11x + 95) = -4x - 55$$ $$11xy+95y=-4x-55$$

$$11xy + 95y = -4x - 55$$ $$11xy+4x=-55-95y$$

$$11xy + 4x = -55 - 95y$$ $$x(11y+4)=-55-95y$$

$$x(11y + 4) = -55 - 95y$$ $$x = \frac{-55 - 95y}{11y + 4}$$

Thus, the inverse function is:

$$(f \circ f)^{-1}(x) = \frac{-55 - 95x}{11x + 4}$$

Answer: $\frac{-55 - 95x}{11x + 4}$

B. Show whether the function $f(x) = x^4 + x^3$ is even, odd or neither.
Solution:
A function is even if $f(-x) = f(x)$, and odd if $f(-x) = -f(x)$.
Compute $f(-x)$:

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

Compare to $f(x)$ and $-f(x)$:

$$f(x) = x^4 + x^3, \quad -f(x) = -x^4 - x^3$$

Since $f(-x) = x^4 - x^3 \neq f(x)$ and $f(-x) = x^4 - x^3 \neq -f(x)$, the function is neither even nor odd.
Answer: Neither

C. Given the function $f(x) = -4 - \sqrt{2x - 3}$

I. Find the $x$-intercept and $y$-intercept if they exist.
Solution:

• $y$-intercept: Set $x = 0$:

$$f(0)=-4-\sqrt{2(0)-3}$$

$$f(0) = -4 - \sqrt{2(0) - 3} = -4 - \sqrt{-3}$$

The expression $\sqrt{-3}$ is not real. Thus, no $y$-intercept.

• $x$-intercept: Set $f(x) = 0$:

$$0=-4-\sqrt{2x-3}$$

$$0 = -4 - \sqrt{2x - 3} \implies \sqrt{2x - 3} = -4$$

A square root is non-negative, so $\sqrt{2x - 3} \geq 0$. The equation $\sqrt{2x - 3} = -4$                                                                                           has no real solution. Thus, no $x$-intercept.
Answer: No $x$-intercept or $y$-intercept

II. Sketch the graph of $f(x)$
Solution:

• Domain: $2x - 3 \geq 0 \implies x \geq \frac{3}{2}$.
• Behavior:
  • At $x = \frac{3}{2}$, $f\left(\frac{3}{2}\right) = -4 - \sqrt{0} = -4$.
  • As $x$ increases, $\sqrt{2x - 3}$ increases, so $-\sqrt{2x - 3}$ decreases, and $f(x)$ decreases.
  • As $x \to \infty$, $f(x) \to -\infty$.
• Vertex: The function can be rewritten as:

$$y+4=-\sqrt{2x-3}$$

$$⟹(y+4{)}^2=2x-3$$

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

This is a parabola opening right with vertex at $\left(\frac{3}{2}, -4\right)$.
Graph Sketch:

• Starts at $\left(\frac{3}{2}, -4\right)$.
• Decreases as $x$ increases.
• Passes through points:
  • $\left(\frac{3}{2}, -4\right)$
  • $(2, -4 - \sqrt{1}) = (2, -5)$
  • $(6, -4 - \sqrt{9}) = (6, -7)$

D. Solve $2x^2 - 12x + 13 = 0$ using completing the square method.
Solution:

$$2x^2 - 12x + 13 = 0$$

Divide by 2:

$$x^2 - 6x + \frac{13}{2} = 0$$

Isolate the constant term:

$$x^2 - 6x = -\frac{13}{2}$$

Complete the square:

$$x^2-6x+9=-\frac{13}{2}+9$$

$$⟹(x-3{)}^2=-\frac{13}{2}+\frac{18}{2}$$

$$\frac{\mathrm{}}{}$$$\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ⟹(x-3{)}^2=\frac{5}{2}$

$$Solve for $x$:

$$x-3=\pm \sqrt{\frac{5}{2}}$$

$$=\pm \frac{\sqrt{10}}{2}$$

$$x - 3 = \pm \sqrt{\frac{5}{2}} = \pm \frac{\sqrt{10}}{2}$$ $$x=3\pm \frac{\sqrt{10}}{2}$$

$$x = 3 \pm \frac{\sqrt{10}}{2} = \frac{6 \pm \sqrt{10}}{2}$$

Answer: $x = \frac{6 \pm \sqrt{10}}{2}$

E. Given the simultaneous equations $2x + y = 1$ and $x^2 - 4ky + 5k = 0$ where $k$ is non-zero constant.
I. Show that $x^2 + 8kx + k = 0$
Solution:
From $2x + y = 1$, solve for $y$:

$$y = 1 - 2x$$

Substitute into the second equation:

$$x^2-4k(1-2x)+5k=0$$

$$x^2 - 4k(1 - 2x) + 5k = 0$$ $$x^2-4k+8kx+5k=0$$

$$x^2 - 4k + 8kx + 5k = 0$$ $$x^2 + 8kx + k = 0$$

Answer: Shown as required.

II. Given that $x^2 + 8kx + k = 0$ has equal roots, find the value of $k$.
Solution:
For equal roots, discriminant $D = 0$:

$$D=(8k{)}^2-4(1)(k)$$

$$D = (8k)^2 - 4(1)(k) = 64k^2 - 4k$$

Set $D = 0$:

$$64k^2-4k=0$$

$$64k^2 - 4k = 0 \implies 4k(16k - 1) = 0$$ $$k = 0 \quad \text{or} \quad k = \frac{1}{16}$$

Since $k$ is non-zero, $k = \frac{1}{16}$.
Answer: $k = \frac{1}{16}$

F. If the profit $p$ in the manufacture and sale of $x$ units of a product is given by $p(x) = 4x + 3x - 2x^2$
Note: Simplify to $p(x) = 7x - 2x^2$.
I. Find the value of $x$ that yields the maximum profit.
Solution:
The function $p(x) = -2x^2 + 7x$ is a downward-opening quadratic. Maximum at vertex:

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

$$=-\frac{7}{2(-2)}$$

$$x = -\frac{b}{2a} = -\frac{7}{2(-2)} = \frac{7}{4}$$

Answer: $x = \frac{7}{4}$

II. Find the maximum profit if each item is sold at K500.
Solution:
The selling price (K500) is irrelevant as the profit function is given. Substitute $x = \frac{7}{4}$:

$$p\left(\frac{7}{4}\right)=7\left(\frac{7}{4}\right)-2{\left(\frac{7}{4}\right)}^2$$

$$=\frac{49}{4}-2X\frac{49}{16}$$

$$=\frac{49}{4}-\frac{49}{8}$$

$$=\frac{98}{8}-\frac{49}{8}$$

$$p\left(\frac{7}{4}\right) = 7\left(\frac{7}{4}\right) - 2\left(\frac{7}{4}\right)^2 = \frac{49}{4} - 2 \cdot \frac{49}{16} = \frac{49}{4} - \frac{49}{8} = \frac{98}{8} - \frac{49}{8} = \frac{49}{8}$$

Answer: Maximum profit = $\frac{49}{8}$

III. Sketch the graph of the function $p(x)$
Solution:

• Shape: Parabola opening downward.
• Vertex: $\left( \frac{7}{4}, \frac{49}{8} \right) = (1.75, 6.125)$.
• $x$-intercepts: Set $p(x) = 0$:

$$7x-2x^2=0$$

$$⟹x(7-2x)=0$$

$$7x - 2x^2 = 0 \implies x(7 - 2x) = 0 \implies x = 0 \quad \text{or} \quad x = \frac{7}{2} = 3.5$$

• Points:
  • $(0, 0)$
  • $(3.5, 0)$
  • Vertex $(1.75, 6.125)$
    Graph Sketch:

@Dr. Microbiota

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