A Level H2 Mathematics Practice Paper 5

TuitionGoWhere Practice Paper - Maths H2 A-Level (Version 5) Answer Key

Total Marks: 80

Section A: Functions and Inverse

Q1 [3 marks]

• $f(x) = 2x - 5$. Let $y = 2x - 5 \Rightarrow x = \frac{y+5}{2}$. So $f^{-1}(x) = \frac{x+5}{2}$.
• Domain of $f^{-1}$: since $f$ is defined on $\mathbb{R}$, range of $f$ is $\mathbb{R}$, thus domain of $f^{-1}$ is $\mathbb{R}$.
  Marks: 2 for inverse, 1 for domain.

Q2 [4 marks]

• $g(x) = x^2 + 1$, $x \ge 0$. It is one-to-one on this domain (strictly increasing), so $g^{-1}$ exists.
• $y = x^2 + 1 \Rightarrow x = \sqrt{y-1}$ (since $x \ge 0$). Thus $g^{-1}(x) = \sqrt{x-1}$.
• Domain of $g^{-1}$: range of $g$ is $[1, \infty)$, so domain is $x \ge 1$.
  Marks: 1 explanation, 2 inverse, 1 domain.

Q3 [4 marks]

• $y = \frac{3x}{x-2} \Rightarrow y(x-2) = 3x \Rightarrow yx - 2y = 3x \Rightarrow x(y-3) = 2y \Rightarrow x = \frac{2y}{y-3}$.
• $h^{-1}(x) = \frac{2x}{x-3}$, $x \ne 3$.
• Range of $h^{-1}$ = domain of $h$ = $\mathbb{R} \setminus \{2\}$.
  Marks: 2 algebra, 1 expression, 1 range.

Q4 [4 marks]

• $p(x) = e^x - 1$ is strictly increasing on $\mathbb{R}$, hence one-to-one; $p^{-1}$ exists.
• $y = e^x - 1 \Rightarrow e^x = y+1 \Rightarrow x = \ln(y+1)$. So $p^{-1}(x) = \ln(x+1)$.
• Domain of $p^{-1}$: range of $p$ is $(-1, \infty)$, so $x > -1$.
  Marks: 1 exists, 2 inverse, 1 domain.

Q5 [5 marks]

• $y = \ln(x+3) \Rightarrow x+3 = e^y \Rightarrow x = e^y - 3$. So $q^{-1}(x) = e^x - 3$.
• Domain of $q^{-1}$: range of $q$ is $\mathbb{R}$, so domain is $\mathbb{R}$.
• Range of $q^{-1}$: domain of $q$ is $x > -3$, so range is $y > -3$.
• Sketch: $y = \ln(x+3)$ passes through $(0,\ln3)$ with vertical asymptote $x=-3$; inverse is $y=e^x-3$ with horizontal asymptote $y=-3$, reflection in $y=x$.
  Marks: 1 inverse, 1 domain, 1 range, 2 sketch.

Section B: Composite Functions

Q6 [4 marks]

• $g(x)=x^2$, $x\ge0$ → range $[0,\infty)$. Domain of $f$ is $\mathbb{R}$, so range of $g \subseteq$ domain of $f$; $fg$ exists.
• $fg(x) = f(g(x)) = x^2 + 1$. Range: $[1,\infty)$.
  Marks: 1 existence, 2 expr, 1 range.

Q7 [4 marks]

• $u(x)=\sqrt{x}$, $x\ge0$ → range $[0,\infty)$. $v$ domain $x\ge4$. Range of $u$ not subset of domain of $v$ (e.g. $u(0)=0<4$), but $vu$ defined only where $u(x)\ge4 \Rightarrow x\ge16$. So $vu$ exists with domain $x\ge16$.
• $vu(x) = v(u(x)) = \sqrt{x} - 4$, domain $x\ge16$.
  Marks: 2 existence+domain, 2 expr.

Q8 [4 marks]

• $f(x)=1/x$, $x>0$ → range $(0,\infty)$. $g$ domain $x>0$, so subset ok; $gf$ exists.
• $gf(x) = g(f(x)) = 2(1/x)+3 = 2/x + 3$. Domain: $x>0$.
  Marks: 1 existence, 2 expr, 1 domain.

Q9 [4 marks]

• $k(x)=\sqrt{x}$, domain $x\ge0$, range $[0,\infty)$. $h$ domain $\mathbb{R}$, so $hk$ exists: $hk(x)=(\sqrt{x})^2-1 = x-1$.
• $kh(x)=\sqrt{x^2-1}$ requires $x^2-1\ge0$, but $h$ range includes negatives (e.g. $h(0)=-1$), so $kh$ not defined for all real $x$; composite $kh$ does not exist as a function $\mathbb{R}\to\mathbb{R}$.
  Marks: 2 each.

Q10 [4 marks]

• $fg(x)= f(g(x)) = 3(x/2)-2 = 1.5x - 2$, domain $\mathbb{R}$, range $\mathbb{R}$.
• $gf(x)= g(f(x)) = (3x-2)/2 = 1.5x - 1$, domain $\mathbb{R}$, range $\mathbb{R}$.
  Marks: 1 each expr, 1 each domain/range.

Section C: Graphs and Transformations

Q11 [3 marks]

• Vertex when $2x-1=0 \Rightarrow x=0.5$, $y=0$. $y$-intercept $x=0 \Rightarrow y=1$.
• V-shape with vertex $(0.5,0)$, intercept $(0,1)$.
  Marks: 1 vertex, 1 intercept, 1 sketch.

Q12 [4 marks]

• Since $f(x)>0$ and $\to0$, $1/f(x) \to \infty$ as $f\to0$; at $x=0$, $1/f(0)=0.5$. Horizontal asymptote $y=0$ becomes vertical? Actually $f$ has $y=0$ asymptote meaning $f\to0$ as $x\to\infty$, so $1/f\to\infty$; reciprocal has vertical asymptote where $f=0$ (none). Key: passes $(0,0.5)$, approaches $y=0$ from above as $x\to-\infty$ if $f$ large.
  Marks: 2 sketch, 2 features.

Q13 [4 marks]

• $y=f(x-3)+2$ is translation right 3, up 2. Original vertex $(0,0)$ → $(3,2)$.
  Marks: 2 description, 2 sketch.

Q14 [4 marks]

• Vertical asymptote: denominator zero $\Rightarrow x=1$. Horizontal: $\lim_{x\to\infty} \frac{2x+1}{x-1} = 2 \Rightarrow y=2$.
• Sketch hyperbola branches.
  Marks: 1 each asymptote, 2 sketch.

Q15 [5 marks]

• $x=2t \Rightarrow t = x/2$. $y = (x/2)^2 + 1 = x^2/4 + 1$. Cartesian: $y = \frac{x^2}{4} + 1$.
• Since $t\in\mathbb{R}$, $x\in\mathbb{R}$, $y \ge 1$. Range $y \ge 1$.
  Marks: 3 elimination, 2 range.

Section D: Equations and Inequalities

Q16 [3 marks]

• $|x-4|<3 \Leftrightarrow -3 < x-4 < 3 \Leftrightarrow 1 < x < 7$.
  Marks: 3 for solution.

Q17 [4 marks]

• Critical points: $x=-2$, $x=1$. Test intervals: $x<-2$: (+)/(-) neg; $-2<x<1$: (+)/(-) neg? Actually $(x+2)$ neg, $(x-1)$ neg → pos. $x>1$: pos/pos pos. So $>0$ for $x<-2$ or $x>1$.
  Marks: 2 critical, 2 intervals.

Q18 [4 marks]

• $|2x+1|>5 \Leftrightarrow 2x+1 < -5$ or $2x+1 > 5 \Rightarrow x < -3$ or $x > 2$.
  Marks: 4.

Q19 [4 marks]

• $x^2-3x-4 = (x-4)(x+1) < 0 \Rightarrow -1 < x < 4$.
  Marks: 2 factor, 2 solution.

Q20 [5 marks]

• $m(x)= \frac{(x-2)(x+2)}{x+1} \le 0$. Critical: $-2, -1, 2$. Sign chart: $(-∞,-2]$ neg/neg? Actually test: $x<-2$: num pos, den neg → neg; $-2<x<-1$: num neg, den neg → pos; $-1<x<2$: num neg, den pos → neg; $x>2$: pos/pos pos. Include roots $-2,2$; exclude $-1$. Solution: $x \le -2$ or $-1 < x \le 2$.
• Number line with closed at $-2,2$, open at $-1$, shaded left of $-2$ and between $-1$ and $2$.
  Marks: 2 algebra, 2 intervals, 1 diagram.