Quadratic functions - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 18 exam-style questions on Quadratic functions 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

3 marks

The curve with equation \(y = x^2 + px + q\), where \(p\) and \(q\) are constants, has a turning point at \((3, -7)\).

Find \(p\) and \(q\).

Model answer

The turning point is \((3, -7)\), so \(y = (x - 3)^2 - 7\).
\(y = x^2 - 6x + 9 - 7 = x^2 - 6x + 2\)
So \(p = -6\) and \(q = 2\).

Explanation

A curve \(y = x^2 + px + q\) with turning point \((3, -7)\) can be written as \(y = (x - 3)^2 - 7\), because the completed-square form shows the turning point directly. Expanding gives \(p = -6\) and \(q = 2\). A common sign error is to write \((x + 3)^2\) for a turning point at \(x = 3\).

Mark scheme

1 mark
M1 Uses the turning point, e.g. \(y = (x - 3)^2 - 7\) and expands; or \(-\dfrac{p}{2} = 3\) and substitutes \((3, -7)\); or a correct final answer of \(p = -6\), \(q = 2\) with no working
An answer of \(p = -6\) and \(q = 2\) with no working scores 3 marks. Alternative: \(\dfrac{dy}{dx} = 2x + p = 0\) at \(x = 3\), then substitutes \((3, -7)\). \(y = (x + 3)^2 - 7\) is M0.
1 mark
A1 \(p = -6\)
1 mark
A1 \(q = 2\)
From \(9 - 18 + q = -7\) or the expansion \(x^2 - 6x + 9 - 7\).

Question 2

4 marks

Solutions relying entirely on calculator technology are not acceptable.

Find the exact solution of the equation
\[4\mathrm{e}^x - 6\mathrm{e}^{-x} = 5\]

Model answer

Multiply by \(\mathrm{e}^x\): \(4\mathrm{e}^{2x} - 6 = 5\mathrm{e}^x\)
\(4\left(\mathrm{e}^x\right)^2 - 5\mathrm{e}^x - 6 = 0\)
\(\left(4\mathrm{e}^x + 3\right)\left(\mathrm{e}^x - 2\right) = 0\)
\(\mathrm{e}^x = 2\) or \(\mathrm{e}^x = -\dfrac{3}{4}\)
\(\mathrm{e}^x > 0\) for all real \(x\), so \(\mathrm{e}^x = -\dfrac{3}{4}\) has no solution.
\(\mathrm{e}^x = 2\), so \(x = \ln 2\)

Explanation

Multiplying by \(\mathrm{e}^x\) uses \(\mathrm{e}^x \times \mathrm{e}^{-x} = 1\) and gives a quadratic in \(\mathrm{e}^x\), which factorises as \(\left(4\mathrm{e}^x + 3\right)\left(\mathrm{e}^x - 2\right) = 0\). Since \(\mathrm{e}^x\) is always positive, only \(\mathrm{e}^x = 2\) gives a solution, \(x = \ln 2\). Marks are lost for keeping \(\ln\left(-\dfrac{3}{4}\right)\), for giving a decimal instead of the exact answer and for writing \(\mathrm{e}^x \times \mathrm{e}^{-x}\) as \(\mathrm{e}^{-x^2}\).

Mark scheme

1 mark
M1 Multiplies through by \(\mathrm{e}^x\) to form a quadratic in \(\mathrm{e}^x\): \(4\mathrm{e}^{2x} - 5\mathrm{e}^x - 6 = 0\) (allow sign slips). An answer with no working scores 0 marks.
An answer of \(x = \ln 2\) with no working scores 0 marks: solutions relying entirely on calculator technology are not acceptable. A decimal answer (0.6931...) with no working scores 0 marks.
1 mark
A1 \(4\left(\mathrm{e}^x\right)^2 - 5\mathrm{e}^x - 6 = 0\) oe (e.g. \(4y^2 - 5y - 6 = 0\) with \(y = \mathrm{e}^x\))
1 mark
M1 (dM1) Solves their quadratic with working: \(\left(4\mathrm{e}^x + 3\right)\left(\mathrm{e}^x - 2\right) = 0\), so \(\mathrm{e}^x = 2\) or \(\mathrm{e}^x = -\dfrac{3}{4}\)
Depends on the first M mark. The formula or completing the square is equally acceptable if the working is seen.
1 mark
A1 \(x = \ln 2\) only, with \(\mathrm{e}^x = -\dfrac{3}{4}\) rejected (\(\mathrm{e}^x > 0\) for all \(x\))
cso. The rejection must be seen, or \(\ln\left(-\dfrac{3}{4}\right)\) must not be offered as a solution. A decimal only (0.693) scores A0.

Question 3

5 marks

The equation
\[x^2 - 2kx + 2k^2 - 2k - 3 = 0\]
where \(k\) is a constant, has real roots.

Find the set of possible values of \(k\). Give your answer in set notation.

Model answer

Real roots, so \(b^2 - 4ac \geqslant 0\):
\((-2k)^2 - 4(1)(2k^2 - 2k - 3) \geqslant 0\)
\(4k^2 - 8k^2 + 8k + 12 \geqslant 0\)
\(-4k^2 + 8k + 12 \geqslant 0\)
\(k^2 - 2k - 3 \leqslant 0\)
\((k - 3)(k + 1) \leqslant 0\)
Critical values \(k = -1\) and \(k = 3\). The quadratic in \(k\) is negative or zero between them, so \(-1 \leqslant k \leqslant 3\).
\(\{k : -1 \leqslant k \leqslant 3\}\)

Explanation

Real roots need \(b^2 - 4ac \geqslant 0\), which simplifies to \(k^2 - 2k - 3 \leqslant 0\), so \(k\) lies between the critical values \(-1\) and 3. The end points are included because "real roots" includes the case of a repeated root, when the discriminant is zero. Common errors are sign slips when expanding \(4(2k^2 - 2k - 3)\), dividing by \(-4\) without reversing the inequality, and using strict inequalities.

Mark scheme

1 mark
M1 Uses \(b^2 - 4ac\) with the values in the correct places and a comparison with 0: \((-2k)^2 - 4(1)(2k^2 - 2k - 3) \geqslant 0\) (condone \(>\) or \(=\) for this mark). An answer of \(\{k : -1 \leqslant k \leqslant 3\}\) with no working scores 2 marks (special case).
An answer of \(\{k : -1 \leqslant k \leqslant 3\}\) with no working scores 2 marks (special case: the final M1 and A1), because the discriminant inequality must be seen for the first three marks; \(-1 \leqslant k \leqslant 3\) not in set notation with no working scores 1 mark.
1 mark
A1 \(k^2 - 2k - 3 \leqslant 0\) oe (e.g. \(-4k^2 + 8k + 12 \geqslant 0\))
A common error is \(4k^2 - 8k^2 - 8k - 12\) (a sign slip in \(-4(2k^2 - 2k - 3)\)): A0, but the later M marks are available.
1 mark
M1 (dM1) Solves their quadratic in \(k\) to find the critical values (\(k = -1\) and \(k = 3\))
Depends on the first M mark. The critical values may come from a calculator.
1 mark
M1 (ddM1) Chooses the region between their critical values
Depends on both previous M marks, except in the answer-only special case in the first mark's comments.
1 mark
A1 \(\{k : -1 \leqslant k \leqslant 3\}\)
Real roots include equal roots, so the end points are included: \(\{k : -1 < k < 3\}\) scores A0. Accept \(\{k : k \geqslant -1\} \cap \{k : k \leqslant 3\}\). An answer not in set notation, in interval notation such as \([-1, 3]\), or with \(\cup\), scores A0.

Question 4

7 marks

The fuel consumption, \(F\) litres per 100 km, of a car travelling at a constant speed of \(v\) km/h is modelled by
\[F = 0.002v^2 - 0.24v + 12, \quad 40 \leqslant v \leqslant 120\]
Write \(F\) in the form \(a(v - b)^2 + c\), where \(a\), \(b\) and \(c\) are constants.

Hence state, according to the model, the speed at which the fuel consumption is least, and the least fuel consumption.

Using algebra, find the set of speeds for which, according to the model, the fuel consumption is at most 6 litres per 100 km. Give the limits of your answer to 3 significant figures where necessary.

Model answer

\(F = 0.002\left(v^2 - 120v\right) + 12\)
\(= 0.002\left[(v - 60)^2 - 3600\right] + 12\)
\(= 0.002(v - 60)^2 - 7.2 + 12\)
\(= 0.002(v - 60)^2 + 4.8\)
So \(a = 0.002\), \(b = 60\), \(c = 4.8\).
\(0.002(v - 60)^2 \geqslant 0\), so the least fuel consumption is 4.8 litres per 100 km, at a speed of 60 km/h.
\(0.002(v - 60)^2 + 4.8 \leqslant 6\)
\((v - 60)^2 \leqslant 600\)
\(-\sqrt{600} \leqslant v - 60 \leqslant \sqrt{600}\)
\(60 - 10\sqrt{6} \leqslant v \leqslant 60 + 10\sqrt{6}\), i.e. \(35.505\ldots \leqslant v \leqslant 84.494\ldots\)
The model only applies for \(40 \leqslant v \leqslant 120\), so the fuel consumption is at most 6 litres per 100 km when \(40 \leqslant v \leqslant 84.5\) (3 s.f.).

Explanation

Take out 0.002 and complete the square to get \(F = 0.002(v - 60)^2 + 4.8\), so the least consumption is 4.8 litres per 100 km at 60 km/h. Setting \(F \leqslant 6\) gives \((v - 60)^2 \leqslant 600\), so \(v\) lies between \(60 - 10\sqrt{6}\) and \(60 + 10\sqrt{6}\); the lower critical value 35.5 is below the 40 km/h at which the model starts, so the set is \(40 \leqslant v \leqslant 84.5\). Common errors are not multiplying the \(-3600\) by 0.002 and ignoring the domain of the model.

Mark scheme

1 mark
M1 Takes out 0.002 and attempts to complete the square, e.g. \(0.002\left[(v - 60)^2 - 3600\right] + 12\), or compares coefficients of \(a(v^2 - 2bv + b^2) + c\); or a correct final answer of \(0.002(v - 60)^2 + 4.8\) with no working
An answer of \(0.002(v - 60)^2 + 4.8\), with the least consumption 4.8 litres per 100 km at 60 km/h, with no working scores 4 marks; the set of speeds with no working scores 0 marks, as the question demands algebra. Comparing coefficients: \(a = 0.002\), \(2ab = 0.24\), \(ab^2 + c = 12\).
1 mark
A1 \(a = 0.002\) and \(b = 60\)
Accept \(a = \dfrac{1}{500}\).
1 mark
A1 \(c = 4.8\), so \(F = 0.002(v - 60)^2 + 4.8\)
A common error is \(c = 12 - 3600 = -3588\) (the \(-3600\) not multiplied by 0.002): A0.
1 mark
B1ft Least fuel consumption 4.8 litres per 100 km, at a speed of 60 km/h
Follow through their \(b\) and \(c\). Both values needed, as for the coordinates of the minimum point. These values are exact. Units not required.
1 mark
M1 Sets \(0.002(v - 60)^2 + 4.8 \leqslant 6\) (or \(v^2 - 120v + 3000 \leqslant 0\)) and finds the critical values algebraically, e.g. \((v - 60)^2 = 600\) or by the quadratic formula
Critical values written down from a calculator score M0. Using < instead of \(\leqslant\) is condoned for this mark.
1 mark
A1 Critical values \(v = 60 \pm 10\sqrt{6}\), i.e. awrt 35.5 and awrt 84.5 (in [35.45, 35.55) and [84.45, 84.55))
Accept \(60 \pm \sqrt{600}\). The exact values are 35.505... and 84.494...
1 mark
A1 \(40 \leqslant v \leqslant 84.5\), using the domain of the model for the lower limit
\(35.5 \leqslant v \leqslant 84.5\) scores A0: the model only applies for \(v \geqslant 40\). Accept \(\{v : 40 \leqslant v \leqslant 84.5\}\), \(40 \leqslant v \leqslant 60 + 10\sqrt{6}\), or the set in words. The upper limit may be given as awrt 84.5, i.e. in [84.45, 84.55), or more accurately. Strict inequalities at 84.5 are condoned.

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