Functions

Question 1 [8]

(a) Denominator cannot be zero: $x - 1 \neq 0 \Rightarrow x \neq 1$. [1] Domain: $\{x \in \mathbb{R} : x \neq 1\}$. The restriction arises because division by zero is undefined. [1]

(b) Set $y = \dfrac{2x+3}{x-1}$. Multiply: $y(x-1) = 2x+3$ [1] $yx - y = 2x + 3 \Rightarrow yx - 2x = y + 3 \Rightarrow x(y-2) = y+3$ [1] $x = \dfrac{y+3}{y-2}$. This requires $y \neq 2$. [1] So range $= \{y \in \mathbb{R} : y \neq 2\}$; $y=2$ is impossible because it would require division by zero when solving for $x$ (it is the horizontal asymptote). [1]

(c) $f$ is one-to-one: each horizontal line $y=k$ ($k\neq2$) meets the curve exactly once, since solving gives a unique $x$. Hence it passes the horizontal line test. [2]

Question 2 [10]

(a) $f(x) = x^2 - 4x + 1 = (x^2 - 4x + 4) - 4 + 1 = (x-2)^2 - 3$. [2] Vertex: $(2, -3)$. [1] Since $(x-2)^2 \ge 0$, minimum value is $-3$; range $= [-3, \infty)$. [1]

(b) Reflection in $x$-axis: $-f(x)$. Then shift up 3: $g(x) = -f(x) + 3$. [2] Order matters: reflect first, then translate. [1] $g(x) = -(x^2 - 4x + 1) + 3 = -x^2 + 4x - 1 + 3 = -x^2 + 4x + 2$. [1]

(c) Reflecting vertex $(2,-3)$ gives $(2,3)$; shift up 3 gives $(2, 6)$. Vertex $(2,6)$. [1] Opens downward, so range $= (-\infty, 6]$. [1] (Check: $g(x)=-x^2+4x+2=-(x-2)^2+6$, vertex $(2,6)$. ✓)

Question 3 [9]

(a) $(f\circ g)(x) = f(g(x)) = 3(x^2+1) - 2 = 3x^2 + 3 - 2 = 3x^2 + 1$. [2] $(g\circ f)(x) = g(f(x)) = (3x-2)^2 + 1 = 9x^2 - 12x + 4 + 1 = 9x^2 - 12x + 5$. [2]

(b) $3x^2 + 1 = 9x^2 - 12x + 5$ [1] $0 = 6x^2 - 12x + 4 \Rightarrow 0 = 3x^2 - 6x + 2$. [1] $x = \dfrac{6 \pm \sqrt{36 - 24}}{6} = \dfrac{6 \pm 2\sqrt{3}}{6} = 1 \pm \dfrac{\sqrt{3}}{3}$. [1]

(c) Since $(f\circ g)(x) \neq (g\circ f)(x)$ for all $x$ (they are equal only at the two solutions above), composition is not commutative — the order of applying functions changes the result. [2]

Question 4 [9]

(a) Radicand must be $\ge 0$: $x - 2 \ge 0 \Rightarrow x \ge 2$. [1] At $x=2$, $f=1$; as $x\to\infty$, $f\to\infty$; $\sqrt{x-2}\ge 0$ so $f\ge 1$. [1] Range $= [1, \infty)$. [1]

(b) Let $y = \sqrt{x-2} + 1$. Swap: $x = \sqrt{y-2}+1$... using swap-and-solve on original: $y - 1 = \sqrt{x-2}$ [1] $(y-1)^2 = x - 2$ [1] $x = (y-1)^2 + 2$, so $f^{-1}(x) = (x-1)^2 + 2$. [1] Domain of $f^{-1}$ = range of $f$ = $[1, \infty)$. [1]

(c) Take $x = 3$ (valid, $\ge 1$): $f^{-1}(3) = (3-1)^2 + 2 = 6$. Then $f(6) = \sqrt{4}+1 = 3$. ✓ [2]

Question 5 [8]

(a) $f(-x) = (-x)^4 - 3(-x)^2 = x^4 - 3x^2 = f(x)$ → even. [2]

(b) $g(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x) = -g(x)$ → odd. [2]

(c) $h(-x) = (-x)^2 + (-x) = x^2 - x \neq h(x)$ and $\neq -h(x)$ → neither. [2]

(d) Even functions are symmetric about the $y$-axis (mirror reflection). Odd functions have $180°$ rotational symmetry about the origin. [2]

Question 6 [6]

(a) $f(-3) = -(-3) = 3$; $f(0) = 0^2 = 0$; $f(2) = 2^2 = 4$; $f(5) = 4$. [2] (½ each)

(b) Decreasing on $(-\infty, 0)$ (since $-x$ falls); increasing on $(0, 2)$ (since $x^2$ rises); constant on $(2,\infty)$. [2]

(c) Yes — every vertical line meets the graph at most once (each $x$ has one output, pieces don't overlap in $x$). This confirms it is a genuine function. [2]

[
  {"claim":"Q2a vertex form: x^2-4x+1 = (x-2)^2-3","code":"x=symbols('x'); result = simplify((x-2)**2 - 3 - (x**2-4*x+1))==0"},
  {"claim":"Q2b g(x) = -x^2+4x+2 = -(x-2)^2+6","code":"x=symbols('x'); result = simplify(-(x-2)**2+6 - (-x**2+4*x+2))==0"},
  {"claim":"Q3b solutions of 3x^2+1=9x^2-12x+5 are 1 +/- sqrt(3)/3","code":"x=symbols('x'); sol=solve(Eq(3*x**2+1,9*x**2-12*x+5),x); result = set(sol)==set([1+sqrt(3)/3,1-sqrt(3)/3])"},
  {"claim":"Q4 inverse: f(f^{-1}(x))=x where f=sqrt(x-2)+1, finv=(x-1)^2+2","code":"x=symbols('x'); f=lambda t: sqrt(t-2)+1; finv=lambda t:(t-1)**2+2; result = simplify(f(finv(x)) - x)==0 if True else False"},
  {"claim":"Q4c f(6)=3","code":"result = sqrt(6-2)+1 == 3"}
]