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A Level H2 Mathematics Practice Paper 1

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Questions

TuitionGoWhere Exam Practice (AI) - Maths H2 A-Level

Subject: Mathematics (H2)
Level: A-Level
Paper: Practice Paper 1 (Version 1 of 5) - Algebra & Functions
Duration: 1 hour 30 minutes
Total Marks: 60
Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Answer all questions.
Write your answers in the spaces provided.
Unless otherwise stated, give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees.
You are expected to use an approved graphing calculator.
Unsupported answers from a graphing calculator are allowed unless the question specifically requires otherwise.
Clear presentation in working is essential.

Section A: Functions and Inverses [20 Marks]

1 The function $f$ is defined by $f(x) = \frac{2x+1}{x-3}$ , for $x \in \mathbb{R}, x \neq 3$ .

(a) Find an expression for $f^{-1}(x)$ and state its domain.
[3]

(b) Solve the inequality $|f^{-1}(x)| < 2$ .
[4]

2 The functions $g$ and $h$ are defined by: $g(x) = \sqrt{x+2}, \quad x \ge -2$ $h(x) = x^2 - 4, \quad x \in \mathbb{R}$

(a) Explain why the composite function $hg$ exists, but the composite function $gh$ does not exist.
[2]

(b) Restrict the domain of $h$ to $x \ge k$ such that the composite function $gh$ exists. State the smallest possible value of $k$ .
[2]

3 The function $\phi$ is defined by $\phi(x) = e^{2x} - 4e^x + 3$ , for $x \in \mathbb{R}$ .

(a) Show that $\phi(x)$ can be written in the form $(e^x - a)(e^x - b)$ , stating the values of constants $a$ and $b$ .
[2]

(b) Hence, find the exact range of $\phi$ .
[3]

4 Let $f(x) = \ln(x^2 - 4)$ for $x > 2$ and $g(x) = e^x + 2$ for $x \in \mathbb{R}$ .

(a) Find $fg(x)$ in its simplest form.
[2]

(b) State the domain and range of $fg$ .
[2]

Section B: Graphs and Transformations [20 Marks]

5 The diagram below shows the graph of $y = f(x)$ for $-3 \le x \le 5$ . The graph has a maximum point at $A(1, 4)$ , a minimum point at $B(4, -2)$ , and crosses the x-axis at $(-1, 0)$ and $(3, 0)$ . The y-intercept is $(0, 3)$ .

(Note: In a real exam, a sketch would be provided here. Assume standard smooth curve behavior between points.)

On separate diagrams, sketch the graph of:

(a) $y = |f(x)|$ , stating the coordinates of the turning points and intercepts.
[3]

(b) $y = f(|x|)$ , stating the coordinates of the turning points and intercepts.
[3]

6 The curve $C$ has parametric equations: $x = t^2 - 1$ $y = t(t^2 - 4)$ for $t \in \mathbb{R}$ .

(a) Find the cartesian equation of $C$ in the form $y^2 = P(x)$ , where $P(x)$ is a polynomial in $x$ .
[3]

(b) Find the coordinates of the points where $C$ crosses the x-axis.
[2]

(c) Determine the set of values of $x$ for which the curve is not defined in the real plane if we consider the restriction that $y$ must be real. (Hint: Consider the sign of $y^2$ ).
[2]

7 Sketch the graph of $y = \frac{2x^2 - 8}{x^2 - 1}$ . In your sketch, you must show:

The equations of any asymptotes.
The coordinates of any axial intercepts.
The coordinates of any stationary points.

[4]

Section C: Equations, Inequalities and Modulus [20 Marks]

8 Solve the inequality: $\frac{x+1}{2x-3} \le 1$ [4]

9 Given that $x$ is real, solve the equation: $|2x - 1| = |x + 3| + 2$ [4]

10 The variables $x$ and $y$ are related by the equation $y = Ax^B$ , where $A$ and $B$ are constants.

(a) State the gradient and vertical intercept of the straight line graph obtained by plotting $\ln y$ against $\ln x$ .
[2]

(b) The following data is recorded:

$\ln x$	1.0	2.0	3.0	4.0	5.0
$\ln y$	2.5	4.1	5.6	7.2	8.8

Using the method of least squares or a graphing calculator, estimate the values of $A$ and $B$ .
[3]

(c) Use your model to estimate the value of $y$ when $x = 10$ . Comment on the reliability of this estimate if the original data for $x$ was in the range $[2, 150]$ .
[2]

11 The function $f$ is defined by $f(x) = |x^2 - 6x + 5|$ .

(a) Sketch the graph of $y = f(x)$ .
[2]

(b) Hence, determine the number of real roots for the equation $f(x) = k$ for the following cases: (i) $k < 0$ (ii) $k = 0$ (iii) $0 < k < 4$ (iv) $k = 4$ (v) $k > 4$

[5]

End of Paper

Answers

TuitionGoWhere Exam Practice (AI) - Maths H2 A-Level

Answer Key & Marking Scheme

Paper: Practice Paper 1 (Version 1 of 5) - Algebra & Functions

Section A: Functions and Inverses

1 (a) Let $y = \frac{2x+1}{x-3}$ . Interchange $x$ and $y$ : $x = \frac{2y+1}{y-3}$ $x(y-3) = 2y+1$ $xy - 3x = 2y + 1$ $xy - 2y = 3x + 1$ $y(x-2) = 3x + 1$ $y = \frac{3x+1}{x-2}$ $\therefore f^{-1}(x) = \frac{3x+1}{x-2}$ Domain of $f^{-1}$ is Range of $f$ . As $x \to \infty, f(x) \to 2$ . $f(x) \neq 2$ . Domain: $x \in \mathbb{R}, x \neq 2$ . [3] (1 for expression, 1 for method, 1 for domain)

(b) $|\frac{3x+1}{x-2}| < 2$ $\Rightarrow -2 < \frac{3x+1}{x-2} < 2$

Case 1: $\frac{3x+1}{x-2} < 2$ $\frac{3x+1 - 2(x-2)}{x-2} < 0$ $\frac{x+5}{x-2} < 0$ Critical values: $x = -5, 2$ . Solution: $-5 < x < 2$ .

Case 2: $\frac{3x+1}{x-2} > -2$ $\frac{3x+1 + 2(x-2)}{x-2} > 0$ $\frac{5x-3}{x-2} > 0$ Critical values: $x = 3/5, 2$ . Solution: $x < 3/5$ or $x > 2$ .

Intersection of Case 1 and Case 2: $(-5 < x < 2) \cap (x < 0.6 \text{ or } x > 2)$ $\Rightarrow -5 < x < 0.6$ (or $3/5$ ) [4] (1 for splitting inequality, 1 for solving first part, 1 for solving second part, 1 for final intersection)

2 (a) Range of $g$ : Since $x \ge -2$ , $\sqrt{x+2} \ge 0$ . $R_g = [0, \infty)$ . Domain of $h$ : $\mathbb{R}$ . Since $R_g \subseteq D_h$ , $hg$ exists.

Range of $h$ : $x^2 - 4 \ge -4$ . $R_h = [-4, \infty)$ . Domain of $g$ : $[-2, \infty)$ . Since $R_h \not\subseteq D_g$ (e.g., $-4 \notin D_g$ ), $gh$ does not exist. [2] (1 for correct reasoning for $hg$ , 1 for correct reasoning for $gh$ )

(b) For $gh$ to exist, we need $R_h \subseteq D_g$ . $h(x) = x^2 - 4$ . If we restrict domain to $x \ge k$ , we need the range of this restricted $h$ to be within $[-2, \infty)$ . Minimum of $h(x)$ on $x \ge 0$ is $-4$ (at $x=0$ ). This is not $\ge -2$ . We need $x^2 - 4 \ge -2 \Rightarrow x^2 \ge 2 \Rightarrow x \ge \sqrt{2}$ (since $x>0$ ). Smallest $k = \sqrt{2}$ . [2] (1 for condition, 1 for value)

(c) $gh(x) = g(h(x)) = \sqrt{x^2 - 4 + 2} = \sqrt{x^2 - 2}$ . Domain: $x \ge \sqrt{2}$ . At $x = \sqrt{2}$ , $gh(\sqrt{2}) = 0$ . As $x \to \infty$ , $gh(x) \to \infty$ . Range: $[0, \infty)$ . [3] (1 for expression, 1 for domain consideration, 1 for range)

3 (a) $\phi(x) = (e^x)^2 - 4(e^x) + 3$ . Let $u = e^x$ . $u^2 - 4u + 3 = (u-3)(u-1)$ . $\phi(x) = (e^x - 3)(e^x - 1)$ . $a=3, b=1$ (or vice versa). [2]

(b) Let $u = e^x$ . Since $x \in \mathbb{R}$ , $u > 0$ . Consider $y = u^2 - 4u + 3$ for $u > 0$ . Vertex at $u = -(-4)/2 = 2$ . Min value $y = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1$ . Since $u=2$ is in domain $u>0$ , the minimum is attained. As $u \to 0^+$ , $y \to 3$ . As $u \to \infty$ , $y \to \infty$ . Range is $[-1, \infty)$ . [3] (1 for substitution/vertex, 1 for min value, 1 for correct range notation)

4 (a) $fg(x) = f(g(x)) = f(e^x + 2)$ . $f(u) = \ln(u^2 - 4)$ . $fg(x) = \ln((e^x+2)^2 - 4) = \ln(e^{2x} + 4e^x + 4 - 4) = \ln(e^{2x} + 4e^x)$ . Simplify: $\ln(e^x(e^x + 4)) = \ln(e^x) + \ln(e^x + 4) = x + \ln(e^x + 4)$ . [2]

(b) Domain of $fg$ : Domain of $g$ is $\mathbb{R}$ . We need $g(x)$ in Domain of $f$ . Domain of $f$ : $x^2 - 4 > 0 \Rightarrow |x| > 2$ . So we need $|e^x + 2| > 2$ . Since $e^x + 2 > 2$ for all real $x$ , this is always true. Domain: $x \in \mathbb{R}$ . Range: As $x \to -\infty$ , $e^x \to 0$ , $fg(x) \to \ln(4)$ . As $x \to \infty$ , $fg(x) \to \infty$ . Function is strictly increasing. Range: $(\ln 4, \infty)$ . [2] (1 for domain, 1 for range)

Section B: Graphs and Transformations

5 (a) $y = |f(x)|$ : Reflect negative part of graph in x-axis. $B(4, -2)$ becomes $(4, 2)$ . Intercepts $(-1,0)$ and $(3,0)$ remain. $y$ -int $(0,3)$ remains. Max $A(1,4)$ remains. New local min at $(4,2)$ ? No, it's a "bounce" or sharp point if it crossed axis, but here B was a min. The graph goes from 0 at x=3 down to -2 at x=4 then up. So $|f|$ goes from 0 at x=3 up to 2 at x=4. Coordinates: $(-1,0), (0,3), (1,4), (3,0), (4,2)$ . [3]

(b) $y = f(|x|)$ : Even function. Keep $x \ge 0$ part, reflect in y-axis. Right side points: $(0,3), (1,4), (3,0), (4,-2)$ . Reflected points: $(-1,4), (-3,0), (-4,-2)$ . Coordinates: $(-4,-2), (-3,0), (-1,4), (0,3), (1,4), (3,0), (4,-2)$ . [3]

6 (a) $x = t^2 - 1 \Rightarrow t^2 = x+1$ . $y = t(t^2 - 4)$ . Square both sides: $y^2 = t^2 (t^2 - 4)^2$ . Substitute $t^2 = x+1$ : $y^2 = (x+1) ( (x+1) - 4 )^2$ $y^2 = (x+1)(x-3)^2$ . [3]

(b) Crosses x-axis when $y=0$ . $(x+1)(x-3)^2 = 0$ . $x = -1$ or $x = 3$ . If $x = -1, t^2 = 0 \Rightarrow t=0 \Rightarrow y=0$ . Point $(-1,0)$ . If $x = 3, t^2 = 4 \Rightarrow t=\pm 2$ . $t=2 \Rightarrow y = 2(4-4)=0$ . Point $(3,0)$ . $t=-2 \Rightarrow y = -2(4-4)=0$ . Point $(3,0)$ . Points: $(-1, 0)$ and $(3, 0)$ . [2]

(c) For $y$ to be real, $y^2 \ge 0$ . $(x+1)(x-3)^2 \ge 0$ . Since $(x-3)^2 \ge 0$ always, we need $x+1 \ge 0 \Rightarrow x \ge -1$ . The curve is not defined for $x < -1$ . [2]

7 $y = \frac{2(x^2-4)}{x^2-1} = \frac{2(x-2)(x+2)}{(x-1)(x+1)}$ . Asymptotes: Vertical: $x = 1, x = -1$ . Horizontal: $y = 2$ (coeff of $x^2$ / coeff of $x^2$ ). Intercepts: $x=0 \Rightarrow y = \frac{-8}{-1} = 8$ . $(0,8)$ . $y=0 \Rightarrow x = 2, -2$ . $(2,0), (-2,0)$ . Stationary Points: Quotient rule or rewrite: $y = \frac{2x^2-8}{x^2-1}$ . $y' = \frac{(x^2-1)(4x) - (2x^2-8)(2x)}{(x^2-1)^2} = \frac{4x^3-4x - 4x^3+16x}{(x^2-1)^2} = \frac{12x}{(x^2-1)^2}$ . $y'=0 \Rightarrow x=0$ . At $x=0, y=8$ . Point $(0,8)$ . Check nature: For $x<0$ (near 0), $y'<0$ . For $x>0$ (near 0), $y'>0$ . Min? Wait, check asymptotes. Between $x=-1$ and $x=1$ : At $x=0, y=8$ . As $x \to 1^-, y \to -\infty$ ? Let $x=0.9$ . Num $\approx -7.4$ . Den $\approx -0.19$ . $y \approx 39$ . Positive. Let $x=0.5$ . Num $-7.5$ . Den $-0.75$ . $y=10$ . Actually, let's check limits. $x \to 1^-$ : Num $\to -6$ . Den $\to 0^-$ . $y \to +\infty$ . $x \to -1^+$ : Num $\to -6$ . Den $\to 0^-$ . $y \to +\infty$ . So $(0,8)$ is a local Minimum in the central region. Sketch: Left branch ( $x<-1$ ): Comes from $y=2$ , goes through $(-2,0)$ , down to $-\infty$ at $x=-1$ . Middle branch ( $-1<x<1$ ): Down from $+\infty$ at $x=-1$ , min at $(0,8)$ , up to $+\infty$ at $x=1$ . Correction: Re-eval $x=0.5$ . $y = \frac{2(0.25)-8}{0.25-1} = \frac{-7.5}{-0.75} = 10$ . Re-eval $x=0$ . $y=8$ . Re-eval $x=-0.5$ . $y=10$ . So $(0,8)$ is a local Minimum. Right branch ( $x>1$ ): From $+\infty$ at $x=1$ , down through $(2,0)$ , approaches $y=2$ from below? Let $x=3$ . $y = \frac{10}{8} = 1.25 < 2$ . So it crosses axis at 2 and approaches 2 from below. [4] (1 for asymptotes, 1 for intercepts, 1 for stationary point, 1 for correct shape)

Section C: Equations, Inequalities and Modulus

8 $\frac{x+1}{2x-3} - 1 \le 0$ $\frac{x+1 - (2x-3)}{2x-3} \le 0$ $\frac{-x+4}{2x-3} \le 0$ Multiply by -1 (flip inequality): $\frac{x-4}{2x-3} \ge 0$ Critical values: $x=4, x=1.5$ . Test intervals: $x > 4$ : $(+) / (+) > 0$ . (Valid) $1.5 < x < 4$ : $(-) / (+) < 0$ . (Invalid) $x < 1.5$ : $(-) / (-) > 0$ . (Valid) Solution: $x < 1.5$ or $x \ge 4$ . [4]

9 $|2x - 1| = |x + 3| + 2$ Critical points: $x=1/2, x=-3$ .

Case 1: $x < -3$ $-(2x-1) = -(x+3) + 2$ $-2x+1 = -x-3+2$ $-2x+1 = -x-1$ $-x = -2 \Rightarrow x=2$ . (Reject, $2 \not< -3$ )

Case 2: $-3 \le x < 0.5$ $-(2x-1) = (x+3) + 2$ $-2x+1 = x+5$ $-3x = 4 \Rightarrow x = -4/3$ . Check: $-3 \le -1.33 < 0.5$ . (Accept)

Case 3: $x \ge 0.5$ $2x-1 = x+3+2$ $2x-1 = x+5$ $x = 6$ . Check: $6 \ge 0.5$ . (Accept)

Solutions: $x = -4/3, 6$ . [4]

10 (a) $\ln y = \ln(Ax^B) = \ln A + B \ln x$ . Plot $\ln y$ (Y-axis) vs $\ln x$ (X-axis). Gradient: $B$ . Vertical Intercept: $\ln A$ . [2]

(b) Using GC (Linear Regression $Y = A' + BX$ where $Y=\ln y, X=\ln x$ ): Data: X: 1, 2, 3, 4, 5 Y: 2.5, 4.1, 5.6, 7.2, 8.8 Gradient $B \approx 1.57$ (Calc: $S_{xy}/S_{xx}$ ). Intercept $\ln A \approx 0.93$ . $B = 1.57$ (3 s.f.) $A = e^{0.93} \approx 2.53$ (3 s.f.) [3]

(c) $x=10 \Rightarrow \ln x \approx 2.3$ . $\ln y = 1.57(2.3) + 0.93 \approx 4.54$ . $y = e^{4.54} \approx 93.7$ . Reliability: $x=10$ corresponds to $\ln x \approx 2.3$ . Original data $\ln x$ range: $[1, 5]$ . $2.3$ is within the data range (Interpolation). Therefore, the estimate is reliable. [2]

11 (a) $y = x^2 - 6x + 5 = (x-1)(x-5)$ . Parabola opening up, roots at 1, 5. Vertex at $x=3, y = 9-18+5 = -4$ . $|f(x)|$ reflects the part below x-axis (between 1 and 5) upwards. Vertex becomes $(3, 4)$ . Roots $(1,0), (5,0)$ become sharp points (cusps). Y-int: $x=0, y=5$ . [2]

(b) Line $y=k$ intersecting graph. (i) $k < 0$ : 0 roots (graph is non-negative). (ii) $k = 0$ : 2 roots ( $x=1, 5$ ). (iii) $0 < k < 4$ : 4 roots (line cuts the "W" shape 4 times). (iv) $k = 4$ : 3 roots (touches peak at $x=3$ , cuts outer arms). (v) $k > 4$ : 2 roots (cuts outer arms only). [5] (1 for each case)

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