Level 5 — MasteryFunctions

Functions

⏱ 75 minutes50 marksprintable — key stays hidden on paper

Chapter: Functions Level: 5 — Mastery (cross-domain: math + physics + coding, build/prove) Time limit: 75 minutes Total marks: 50

Question 1 — Build & Prove: Inverses, Composition, Parity (18 marks)

Consider the function $f(x) = \frac{3x - 2}{x + 1}, \quad x \neq -1.$

(a) State the domain and range of $f$ . Justify the range by expressing $y = f(x)$ and solving for $x$ . (4)

(b) Prove algebraically that $f$ is one-to-one (injective), and hence explain why $f^{-1}$ exists. Reference the horizontal line test in your reasoning. (4)

(c) Find $f^{-1}(x)$ explicitly, stating its domain. (4)

(d) Let $g(x) = f(f(x))$ . Compute $g(x)$ as a single simplified rational expression and state any excluded values of $x$ . (4)

(e) Define $h(x) = f(x) - f(-x)$ on the common domain. Determine algebraically whether $h$ is even, odd, or neither. (2)

Question 2 — Physics Model: Piecewise Motion & Transformations (18 marks)

A drone's height above ground (in metres) as a function of time $t$ (seconds) is modelled by the piecewise function

H(t) = \begin{cases} 4t & 0 \le t < 3 \\ 12 & 3 \le t < 7 \\ 12 - 2(t-7)^2 & 7 \le t \le 9. \end{cases}

(a) Sketch $H(t)$ over $[0,9]$ , clearly labelling the key features at $t = 0, 3, 7, 9$ . State the height at each. (5)

(b) Verify that $H$ is continuous at $t=3$ and at $t=7$ by checking the piece values. Identify the interval(s) on which $H$ is increasing, constant, and decreasing. (5)

(c) The drone's average vertical velocity on $[a,b]$ is $\dfrac{H(b)-H(a)}{b-a}$ . Compute it on $[0,3]$ and on $[7,9]$ , and interpret the sign of each physically. (4)

(d) A second drone follows the model $K(t) = H(t-1) + 5$ . Describe in words the transformation from $H$ to $K$ , and state the new domain of $K$ and $K(4)$ . (4)

Question 3 — Coding & Reasoning: Function Composition Machine (14 marks)

A student writes pseudocode for numeric functions:

function f(x):  return 2*x + 1
function g(x):  return x*x
function apply(F, G, x):  return F(G(x))

(a) Using the definitions above, compute apply(f, g, 3) and apply(g, f, 3) by hand, showing the composition being evaluated. Explain why the two results differ (comment on non-commutativity). (4)

(b) Write the closed-form algebraic expressions for $(f\circ g)(x)$ and $(g\circ f)(x)$ , then solve the equation $(f\circ g)(x) = (g\circ f)(x)$ . (5)

(c) The student wants to add a function h so that apply(f, h, x) returns the identity, i.e. $f(h(x)) = x$ for all $x$ . Derive the required rule for h(x) (i.e. find $f^{-1}$ ), and verify by substitution that $f(h(2)) = 2$ . (5)

Answer keyMark scheme & solutions

Question 1

(a) Domain: all real $x$ except $x=-1$ , i.e. $\{x\in\mathbb{R}: x\neq -1\}$ . (1) Set $y = \frac{3x-2}{x+1}$ . Solve: $y(x+1)=3x-2 \Rightarrow yx + y = 3x - 2 \Rightarrow x(y-3) = -2 - y \Rightarrow x = \frac{-2-y}{y-3} = \frac{y+2}{3-y}$ . (2) This has a solution for every $y\neq 3$ . Hence range $=\{y\in\mathbb{R}: y\neq 3\}$ . (1)

(b) Suppose $f(a)=f(b)$ : $\frac{3a-2}{a+1}=\frac{3b-2}{b+1}$ . Cross-multiply: $(3a-2)(b+1)=(3b-2)(a+1)$ . $3ab+3a-2b-2 = 3ab+3b-2a-2 \Rightarrow 3a-2b = 3b-2a \Rightarrow 5a = 5b \Rightarrow a=b$ . (3) Since $f(a)=f(b)\Rightarrow a=b$ , $f$ is injective; every horizontal line meets the graph at most once (horizontal line test passes), so $f^{-1}$ exists. (1)

(c) From (a), swapping roles: $f^{-1}(x) = \dfrac{x+2}{3-x}$ , domain $\{x\in\mathbb{R}: x\neq 3\}$ (the range of $f$ ). (4)

(d) $g(x)=f(f(x))=\dfrac{3f(x)-2}{f(x)+1}$ . With $f(x)=\frac{3x-2}{x+1}$ : Numerator: $3\cdot\frac{3x-2}{x+1}-2 = \frac{9x-6-2(x+1)}{x+1}=\frac{7x-8}{x+1}$ . Denominator: $\frac{3x-2}{x+1}+1 = \frac{3x-2+x+1}{x+1}=\frac{4x-1}{x+1}$ . (2) So $g(x)=\dfrac{7x-8}{4x-1}$ . Excluded values: $x\neq -1$ (original domain) and $x\neq \tfrac14$ (where $4x-1=0$ , i.e. $f(x)=-1$ ). (2)

(e) $h(x)=f(x)-f(-x)$ . Then $h(-x)=f(-x)-f(x)=-h(x)$ . Hence $h$ is odd (as a difference of $f(x)$ and its reflected version, the antisymmetric combination). (2)

Question 2

(a) Values: $H(0)=0$ ; $H(3)=12$ (from $4t$ : $4\times3=12$ ); $H(7)=12$ ; $H(9)=12-2(2)^2=12-8=4$ . Graph: straight line $0\to12$ on $[0,3)$ , horizontal at $12$ on $[3,7)$ , downward parabola from $12$ to $4$ on $[7,9]$ . (5)

(b) At $t=3$ : left piece $4(3)=12$ , right piece $=12$ → continuous. At $t=7$ : middle piece $=12$ , right piece $12-2(0)^2=12$ → continuous. (3) Increasing on $[0,3]$ ; constant on $[3,7]$ ; decreasing on $[7,9]$ . (2)

(c) On $[0,3]$ : $\frac{H(3)-H(0)}{3-0}=\frac{12-0}{3}=4$ m/s (positive → rising). (2) On $[7,9]$ : $\frac{H(9)-H(7)}{9-7}=\frac{4-12}{2}=-4$ m/s (negative → descending). (2)

(d) $K(t)=H(t-1)+5$ : horizontal shift right by 1 unit and vertical shift up by 5 units. (2) New domain: $[1,10]$ (original $[0,9]$ shifted right by 1). (1) $K(4)=H(3)+5=12+5=17$ m. (1)

Question 3

(a) apply(f,g,3): $g(3)=9$ , then $f(9)=2(9)+1=19$ . (1) apply(g,f,3): $f(3)=7$ , then $g(7)=49$ . (1) Results $19 \neq 49$ . (1) Composition is not commutative: applying $g$ then $f$ generally differs from $f$ then $g$ . (1)

(b) $(f\circ g)(x)=f(x^2)=2x^2+1$ . (1) $(g\circ f)(x)=g(2x+1)=(2x+1)^2=4x^2+4x+1$ . (1) Equation: $2x^2+1=4x^2+4x+1 \Rightarrow 0=2x^2+4x \Rightarrow 2x(x+2)=0$ . (2) So $x=0$ or $x=-2$ . (1)

(c) Need $f(h(x))=x$ : $2h(x)+1=x \Rightarrow h(x)=\dfrac{x-1}{2}=f^{-1}(x)$ . (3) Check: $h(2)=\frac{1}{2}$ , $f(\tfrac12)=2\cdot\tfrac12+1=2$ . ✓ (2)

[
  {"claim":"g(x)=f(f(x)) simplifies to (7x-8)/(4x-1)", "code":"x=symbols('x'); f=(3*x-2)/(x+1); g=f.subs(x,f); result = simplify(g-(7*x-8)/(4*x-1))==0"},
  {"claim":"f inverse is (x+2)/(3-x)", "code":"x=symbols('x'); f=(3*x-2)/(x+1); inv=(x+2)/(3-x); result = simplify(f.subs(x,inv)-x)==0"},
  {"claim":"H average velocity on [7,9] is -4", "code":"result = ((12-2*(9-7)**2)-12)/(9-7)==-4"},
  {"claim":"(f o g)=(g o f) solutions are 0 and -2", "code":"x=symbols('x'); sol=solve(Eq(2*x**2+1,(2*x+1)**2),x); result = set(sol)=={0,-2}"}
]