Graphs of functions, modulus and proportion - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 18 exam-style questions on Graphs of functions, modulus and proportion for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

18 questions
3–9 marks each
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18 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

4 marks

The curve \(C\) has equation
\[y = \dfrac{4}{x + 3} - 2, \quad x ≠ -3\]
Find the equations of the two asymptotes of \(C\), and the coordinates of the points where \(C\) crosses the coordinate axes.

Model answer

Vertical asymptote where the denominator is zero: \(x = -3\)
As \(x \to \pm\infty\), \(\dfrac{4}{x + 3} \to 0\), so the horizontal asymptote is \(y = -2\)
\(x = 0\): \(y = \dfrac{4}{3} - 2 = -\dfrac{2}{3}\), giving \(\left(0, -\dfrac{2}{3}\right)\)
\(y = 0\): \(\dfrac{4}{x + 3} = 2 \Rightarrow x + 3 = 2 \Rightarrow x = -1\), giving \((-1, 0)\)

Explanation

The curve is the graph of \(y = \dfrac{4}{x}\) translated 3 units left and 2 units down, so its asymptotes move from \(x = 0\) and \(y = 0\) to \(x = -3\) and \(y = -2\). The intercepts come from putting \(x = 0\) and \(y = 0\) in turn. Common errors are giving \(x = 3\) for the vertical asymptote and writing the asymptotes as numbers rather than equations.

Mark scheme

1 mark
B1 \(x = -3\)
Correct answers with no working score full marks. Each asymptote must be given as an equation: \(-3\) alone scores B0. \(x = 3\) scores B0.
1 mark
B1 \(y = -2\)
1 mark
B1 \(\left(0, -\dfrac{2}{3}\right)\)
Accept \(x = 0, y = -\dfrac{2}{3}\). Do not accept \(-0.667\) unless \(-\dfrac{2}{3}\) is also seen.
1 mark
B1 \((-1, 0)\)
Accept \(x = -1, y = 0\).

Question 2

5 marks

Use algebra to solve the inequality
\[|3x - 4| > x + 2\]
writing the solution in set notation.

Model answer

Critical values:
\(3x - 4 = x + 2 \Rightarrow 2x = 6 \Rightarrow x = 3\)
\(-(3x - 4) = x + 2 \Rightarrow 4 - 3x = x + 2 \Rightarrow x = \dfrac{1}{2}\)
The graph of \(y = |3x - 4|\) is a V shape with vertex \(\left(\dfrac{4}{3}, 0\right)\); it is above the line \(y = x + 2\) outside the critical values (e.g. at \(x = 0\): \(4 > 2\)).
\(\{x : x < \dfrac{1}{2}\} \cup \{x : x > 3\}\)

Explanation

The critical values are where the V-shaped graph of \(y = |3x - 4|\) meets the line \(y = x + 2\): one on each branch, at \(x = \dfrac{1}{2}\) and \(x = 3\). The V is above the line outside these values, which can be checked by testing a value such as \(x = 0\), and the two separate regions are joined with \(\cup\). Common errors are finding only one critical value, choosing the region between them and using \(\cap\).

Mark scheme

1 mark
M1 Solves \(3x - 4 = x + 2\) (or the corresponding inequality) to find a critical value. An answer with no working scores 0 marks
An answer of \(\{x : x < \dfrac{1}{2}\} \cup \{x : x > 3\}\) with no working scores 0 marks: the question says 'using algebra'. Squaring both sides, \((3x - 4)^2 > (x + 2)^2\), giving \(8x^2 - 28x + 12 > 0\) and so \((2x - 1)(x - 3) > 0\), is valid here and scores the marks in the same order (M1 A1 for one critical value, M1 A1 for the other, A1 for the set).
1 mark
A1 \(x = 3\)
1 mark
M1 Solves \(-(3x - 4) = x + 2\) (or the corresponding inequality) to find the other critical value
1 mark
A1 \(x = \dfrac{1}{2}\)
1 mark
A1 \(\{x : x < \dfrac{1}{2}\} \cup \{x : x > 3\}\)
Requires the outside region, set brackets and \(\cup\) (typed U is fine). Do not accept \(\cap\), 'and', \(\{x : \dfrac{1}{2} < x < 3\}\), non-strict inequalities, or an answer not in set notation. Interval notation, e.g. \((-\infty, \tfrac{1}{2}) \cup (3, \infty)\), is not the set notation asked for: A0.

Question 3

6 marks

The curves \(C_1\) and \(C_2\) have equations
\[C_1: y = \dfrac{2}{x}, \ x ≠ 0 \qquad C_2: y = 3 - x^2\]
Show that the \(x\)-coordinates of the points where \(C_1\) and \(C_2\) meet satisfy the equation
\[x^3 - 3x + 2 = 0\]
Hence find the coordinates of the points where \(C_1\) and \(C_2\) meet, and explain what the repeated root of this equation tells you about \(C_1\) and \(C_2\).

Model answer

\(\dfrac{2}{x} = 3 - x^2\)
\(2 = 3x - x^3\)
\(x^3 - 3x + 2 = 0\), as required.
\(x = 1\): \(1 - 3 + 2 = 0\), so \((x - 1)\) is a factor.
\(x^3 - 3x + 2 = (x - 1)(x^2 + x - 2) = (x - 1)(x - 1)(x + 2) = (x - 1)^2(x + 2)\)
\(x = 1\): \(y = 2\); \(x = -2\): \(y = \dfrac{2}{-2} = -1\)
The curves meet at \((1, 2)\) and \((-2, -1)\).
\(x = 1\) is a repeated root, so at \((1, 2)\) the curves touch: they have a common tangent there and do not cross.

Explanation

Equating the two expressions for \(y\) and multiplying by \(x\) gives the cubic, and the factor theorem with \(x = 1\) leads to \((x - 1)^2(x + 2) = 0\). A repeated root means the curves meet at that point without crossing, just as a tangent touches a curve, while the single root \(x = -2\) is a point where they cross. Common errors are not using the factor theorem to find a first root and forgetting the \(y\)-coordinates.

Mark scheme

1 mark
M1 Sets \(\dfrac{2}{x} = 3 - x^2\) and multiplies through by \(x\). An answer with no working scores 0 marks
An answer with no working scores 0 marks: the first part is a 'show that', and stating the points \((1, 2)\) and \((-2, -1)\) without the factorisation does not use 'hence'.
1 mark
A1* \(2 = 3x - x^3\), leading to \(x^3 - 3x + 2 = 0\)
cso; the '= 0' must be seen.
1 mark
M1 Finds a root by the factor theorem (e.g. \(1 - 3 + 2 = 0\)) and divides to obtain a quadratic factor: \((x - 1)(x^2 + x - 2)\)
1 mark
A1 \((x - 1)^2(x + 2) = 0\), so \(x = 1\) (repeated) and \(x = -2\)
1 mark
A1 \((1, 2)\) and \((-2, -1)\)
Both points, with both coordinates.
1 mark
B1 At \((1, 2)\), where the root is repeated, the curves touch: they meet there without crossing and have the same tangent (gradient \(-2\))
Accept 'the curves are tangent to each other at \((1, 2)\)' or 'the curves touch at \((1, 2)\)'. Do not accept 'the curves meet twice at \((1, 2)\)'.

Question 4

8 marks

The curve \(C\) has equation
\[y = (2x - 1)^2(x + 2)\]
Find the coordinates of the points where \(C\) meets the coordinate axes, and state, with a reason, whether \(C\) crosses or touches the \(x\)-axis at each point where it meets the \(x\)-axis. Describe the behaviour of \(y\) as \(x \to \infty\) and as \(x \to -\infty\).

Using algebra, find the exact solutions of the equation
\[(2x - 1)^2(x + 2) = 2\]

Model answer

\(y = 0\): \(x = \dfrac{1}{2}\) or \(x = -2\), giving \(\left(\dfrac{1}{2}, 0\right)\) and \((-2, 0)\)
\(x = 0\): \(y = (-1)^2(2) = 2\), giving \((0, 2)\)
\((2x - 1)^2\) is a repeated factor, so \(C\) touches the \(x\)-axis at \(\left(\dfrac{1}{2}, 0\right)\); \((x + 2)\) is a single factor, so \(C\) crosses the \(x\)-axis at \((-2, 0)\).
The \(x^3\) coefficient is 4, which is positive, so \(y \to \infty\) as \(x \to \infty\) and \(y \to -\infty\) as \(x \to -\infty\).
\((4x^2 - 4x + 1)(x + 2) = 4x^3 + 4x^2 - 7x + 2\)
\(4x^3 + 4x^2 - 7x + 2 = 2 \Rightarrow 4x^3 + 4x^2 - 7x = 0 \Rightarrow x(4x^2 + 4x - 7) = 0\)
\(x = 0\) or \(x = \dfrac{-4 \pm \sqrt{16 + 112}}{8} = \dfrac{-4 \pm 8\sqrt{2}}{8} = \dfrac{-1 \pm 2\sqrt{2}}{2}\)
Solutions: \(x = 0\), \(x = \dfrac{-1 + 2\sqrt{2}}{2}\), \(x = \dfrac{-1 - 2\sqrt{2}}{2}\)

Explanation

A squared factor gives a point where the curve touches the \(x\)-axis and a single factor a point where it crosses, and the positive \(x^3\) coefficient fixes the end behaviour. The line \(y = 2\) passes through the \(y\)-intercept \((0, 2)\), so \(x = 0\) is one solution, and the other two come from the quadratic factor by the formula. Common errors are dividing by \(x\) and losing \(x = 0\), and expanding \((2x - 1)^2\) as \(4x^2 + 1\).

Mark scheme

1 mark
B1 \((-2, 0)\) and \(\left(\dfrac{1}{2}, 0\right)\); for the final part an answer with no working scores 0 marks (see comments)
An answer of \(x = 0\) and \(x = \dfrac{-1 \pm 2\sqrt{2}}{2}\) with no working scores 0 marks for the final part, as the question says 'using algebra'; the B marks for the features can still be awarded without working.
1 mark
B1 \((0, 2)\)
1 mark
B1 Touches the \(x\)-axis at \(\left(\dfrac{1}{2}, 0\right)\), because \((2x - 1)\) is a repeated (squared) factor, and crosses at \((-2, 0)\), because \((x + 2)\) is a single factor
Both statements needed, with a reason for the touching point.
1 mark
B1 \(y \to \infty\) as \(x \to \infty\) and \(y \to -\infty\) as \(x \to -\infty\)
Both needed.
1 mark
M1 Expands \((2x - 1)^2(x + 2)\) and forms a cubic equation \(= 0\): \(4x^3 + 4x^2 - 7x + 2 - 2 = 0\)
Expanding \((2x - 1)^2\) as \(4x^2 + 1\) scores M0.
1 mark
A1 \(4x^3 + 4x^2 - 7x = 0\), so \(x(4x^2 + 4x - 7) = 0\)
1 mark
M1 (dM1) Solves \(4x^2 + 4x - 7 = 0\) using the formula or completing the square
Depends on the previous M mark. Roots from a calculator with no working score M0.
1 mark
A1 \(x = 0\), \(x = \dfrac{-1 + 2\sqrt{2}}{2}\) and \(x = \dfrac{-1 - 2\sqrt{2}}{2}\)
All three, exact. Accept \(-\dfrac{1}{2} \pm \sqrt{2}\) or \(\dfrac{-4 \pm \sqrt{128}}{8}\). Dividing by \(x\) and losing \(x = 0\) scores A0; decimals alone (0.914, \(-1.91\)) score A0.

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