Quadratic functions - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 16 exam-style questions on Quadratic functions for Edexcel A-Level Maths (Paper 2). 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.

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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

Using algebra, solve the equation
\[x^2 - 3|x| - 10 = 0\]

Model answer

Since \(x^2 = |x|^2\), the equation is \(|x|^2 - 3|x| - 10 = 0\).
\((|x| - 5)(|x| + 2) = 0\)
\(|x| = 5\) or \(|x| = -2\)
\(|x| \geqslant 0\), so \(|x| = -2\) is impossible.
\(|x| = 5\), so \(x = 5\) or \(x = -5\).

Explanation

Because \(x^2 = |x|^2\), the equation is a quadratic in \(|x|\), which factorises as \((|x| - 5)(|x| + 2) = 0\). A modulus cannot be negative, so only \(|x| = 5\) gives solutions, and it gives two: \(x = 5\) and \(x = -5\). Common errors are keeping \(|x| = -2\) and giving only \(x = 5\).

Mark scheme

1 mark
M1 Uses \(x^2 = |x|^2\) to form a quadratic in \(|x|\) and solves it: \((|x| - 5)(|x| + 2) = 0\). An answer with no working scores 0 marks.
An answer of \(x = 5\) and \(x = -5\) with no working scores 0 marks: the question says "using algebra". Alternative: considers the cases \(x \geqslant 0\) (\(x^2 - 3x - 10 = 0\), so \(x = 5\), rejecting \(x = -2\)) and \(x < 0\) (\(x^2 + 3x - 10 = 0\), so \(x = -5\), rejecting \(x = 2\)): M1 for solving one case, A1 for each correct value with the other root rejected.
1 mark
A1 \(|x| = 5\), with \(|x| = -2\) rejected because \(|x| \geqslant 0\)
1 mark
A1 \(x = 5\) and \(x = -5\)
Depends on the M mark. Both needed, and no others: including \(x = \pm 2\) scores A0.

Question 2

5 marks

The equation
\[(k + 1)x^2 + 2kx + k - 2 = 0\]
where \(k\) is a constant, has two distinct real roots.

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

Model answer

For two distinct real roots, \(b^2 - 4ac > 0\):
\((2k)^2 - 4(k + 1)(k - 2) > 0\)
\(4k^2 - 4\left(k^2 - k - 2\right) > 0\)
\(4k + 8 > 0\), so \(k > -2\)
Also the equation must be a quadratic, so \(k + 1 ≠ 0\): when \(k = -1\) it becomes \(-2x - 3 = 0\), which has only one root.
So \(\{k : -2 < k < -1\} \cup \{k : k > -1\}\)

Explanation

The \(k^2\) terms cancel, so the discriminant \(4k + 8\) is linear in \(k\) and the condition is simply \(k > -2\). The value \(k = -1\) must then be removed, because it makes the coefficient of \(x^2\) zero and leaves a linear equation with only one root. Expansion errors in \(4(k + 1)(k - 2)\) and forgetting to exclude \(k = -1\) are the common faults.

Mark scheme

1 mark
M1 Uses \(b^2 - 4ac\) with the values in the correct places and a comparison with 0: \((2k)^2 - 4(k + 1)(k - 2) > 0\) (condone \(\geqslant\) or \(=\) for this mark); an answer of \(\{k : -2 < k < -1\} \cup \{k : k > -1\}\) with no working scores 2 marks
An answer of \(\{k : -2 < k < -1\} \cup \{k : k > -1\}\) with no working scores 2 marks (special case: the B1 and the final A1), because the discriminant must be seen for the other marks.
1 mark
A1 \(4k^2 - 4\left(k^2 - k - 2\right) = 4k + 8\)
A common error is \(-4(k + 1)(k - 2) = -4k^2 - 4k + 8\): A0.
1 mark
A1 \(4k + 8 > 0\), so \(k > -2\)
Depends on the M mark.
1 mark
B1 States that \(k ≠ -1\), with a reason: when \(k = -1\) the coefficient of \(x^2\) is 0 and the equation, \(-2x - 3 = 0\), is linear with only one root
A reason is required.
1 mark
A1 \(\{k : -2 < k < -1\} \cup \{k : k > -1\}\)
cso. Depends on the M mark and the B mark (apart from the answer-only special case in the first mark's comments). Accept \(\{k : k > -2, k ≠ -1\}\). Do not accept interval notation such as \((-2, -1) \cup (-1, \infty)\), \(\cap\), or \(\{k : k > -2\}\) (the value \(-1\) not excluded).

Question 3

6 marks

In this question you must show all stages of your working.

\[\mathrm{f}(x) = \dfrac{x^2 + 3}{x + 1}, \quad x \in \mathbb{R}, \ x ≠ -1\]
By considering the discriminant of a suitable quadratic equation in \(x\), find the range of \(\mathrm{f}\).

Model answer

Let \(y = \dfrac{x^2 + 3}{x + 1}\). Then \(y(x + 1) = x^2 + 3\), so \(x^2 - yx + 3 - y = 0\).
For a value \(y\) to be in the range, this quadratic must have a real root \(x\), so \(b^2 - 4ac \geqslant 0\):
\((-y)^2 - 4(1)(3 - y) \geqslant 0\)
\(y^2 + 4y - 12 \geqslant 0\)
\((y + 6)(y - 2) \geqslant 0\), so \(y \leqslant -6\) or \(y \geqslant 2\).
(\(x = -1\) would give \(1 + y + 3 - y = 4 ≠ 0\), so it is never a root.)
The range of \(\mathrm{f}\) is \(\mathrm{f}(x) \leqslant -6\) or \(\mathrm{f}(x) \geqslant 2\).

Explanation

A value \(y\) is in the range when \(\dfrac{x^2 + 3}{x + 1} = y\) has a real solution, and rearranging gives the quadratic \(x^2 - yx + 3 - y = 0\), which needs a non-negative discriminant. That condition, \((y + 6)(y - 2) \geqslant 0\), gives the outside region \(y \leqslant -6\) or \(y \geqslant 2\), and the end values are included because they give repeated roots. Common errors are choosing the inside region, using strict inequalities and writing the range in terms of \(x\).

Mark scheme

1 mark
M1 Sets \(y = \dfrac{x^2 + 3}{x + 1}\) and rearranges to a three-term quadratic in \(x\): \(x^2 - yx + 3 - y = 0\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question says that you must show all stages of your working and specifies the method. Finding the stationary points by calculus or from a graph instead of using the discriminant scores 0 marks. Any letter may be used for \(y\).
1 mark
A1 \(x^2 - yx + (3 - y) = 0\) oe
1 mark
M1 Uses the condition for real \(x\): \(b^2 - 4ac \geqslant 0\), i.e. \((-y)^2 - 4(1)(3 - y) \geqslant 0\)
Condone \(>\) for this mark.
1 mark
A1 \(y^2 + 4y - 12 \geqslant 0\)
1 mark
M1 (dM1) Solves to find the critical values \(y = -6\) and \(y = 2\) and chooses the outside region
Depends on the previous M mark. \((y + 6)(y - 2) \geqslant 0\).
1 mark
A1 \(\mathrm{f}(x) \leqslant -6\) or \(\mathrm{f}(x) \geqslant 2\)
cso. Accept \(y\) for \(\mathrm{f}(x)\), and \(\{y : y \leqslant -6\} \cup \{y : y \geqslant 2\}\). Do not accept \(x \leqslant -6\) or \(x \geqslant 2\), strict inequalities, "and", or \(-6 \leqslant \mathrm{f}(x) \leqslant 2\). (The value \(x = -1\) never satisfies the quadratic, since it gives \(4 = 0\), so no values need to be removed.)

Question 4

8 marks

The curve \(C\) has equation \(y = 3x^2 - 5x + 4\) and the line \(l\) has equation \(y = x + 13\).

Using algebra, find the coordinates of the two points where \(l\) meets \(C\).

For values of \(x\) between the \(x\)-coordinates of these two points, \(l\) lies above \(C\). At each such value of \(x\), the vertical distance from \(C\) up to \(l\) is \(d\).

Show that \(d = 9 + 6x - 3x^2\) and, by completing the square, find the greatest value of \(d\) and the value of \(x\) at which it occurs.

The line \(l\) is translated parallel to the \(y\)-axis until it just touches \(C\). Find an equation of the translated line and the coordinates of the point where it touches \(C\).

Model answer

\(3x^2 - 5x + 4 = x + 13\)
\(3x^2 - 6x - 9 = 0\), so \(x^2 - 2x - 3 = 0\)
\((x - 3)(x + 1) = 0\), so \(x = -1\) or \(x = 3\)
\(x = -1\): \(y = -1 + 13 = 12\); \(x = 3\): \(y = 3 + 13 = 16\)
The points are \((-1, 12)\) and \((3, 16)\).
\(d = (x + 13) - \left(3x^2 - 5x + 4\right) = x + 13 - 3x^2 + 5x - 4 = 9 + 6x - 3x^2\)
\(d = -3\left(x^2 - 2x\right) + 9 = -3\left[(x - 1)^2 - 1\right] + 9 = 12 - 3(x - 1)^2\)
\((x - 1)^2 \geqslant 0\), so the greatest value of \(d\) is 12, when \(x = 1\).
Moving \(l\) down by this greatest distance, 12, makes it just touch \(C\): \(y = x + 13 - 12\), i.e. \(y = x + 1\).
It touches \(C\) where \(x = 1\): \(y = 1 + 1 = 2\), so at \((1, 2)\).
(Check: \(3x^2 - 5x + 4 = x + 1\) gives \(3x^2 - 6x + 3 = 3(x - 1)^2 = 0\), a repeated root at \(x = 1\).)

Explanation

Equating the two expressions for \(y\) gives \(x^2 - 2x - 3 = 0\), so \(l\) meets \(C\) at \((-1, 12)\) and \((3, 16)\). Completing the square on the vertical gap gives \(d = 12 - 3(x - 1)^2\), so the greatest gap is 12, at \(x = 1\), and moving \(l\) down by exactly 12 makes it touch \(C\) there: \(y = x + 1\), touching at \((1, 2)\) (setting the discriminant of \(3x^2 - 6x + 4 - c = 0\) to zero gives the same line). Common errors are subtracting the line from the curve, which reverses the sign of \(d\), and not multiplying the \(-1\) by \(-3\) when completing the square.

Mark scheme

1 mark
M1 Sets \(3x^2 - 5x + 4 = x + 13\) and solves the resulting three-term quadratic \(3x^2 - 6x - 9 = 0\) (or \(x^2 - 2x - 3 = 0\)) by factorising, the formula or completing the square. An answer with no working scores 0 marks for this part
An answer of \((-1, 12)\) and \((3, 16)\) with no working scores 0 marks, because the question says "using algebra"; likewise a greatest value of 12 at \(x = 1\) without completing the square scores 0 marks. For the last part, \(y = x + 1\) and \((1, 2)\) with no working score 2 marks (the two B marks). Points written down from a graphical calculator with no algebra score M0.
1 mark
A1 \(x = -1\) and \(x = 3\)
Both needed: \((x - 3)(x + 1) = 0\).
1 mark
A1 \((-1, 12)\) and \((3, 16)\)
Depends on the M mark. Both points needed, as coordinates or as correctly paired values \(x = -1, y = 12\) and \(x = 3, y = 16\). The \(y\)-values may come from either equation.
1 mark
B1* \(d = (x + 13) - \left(3x^2 - 5x + 4\right) = 9 + 6x - 3x^2\)
cso. The subtraction with the bracket (or a correct expansion of it) must be seen; restating the given result scores B0. \(d = 3x^2 - 5x + 4 - (x + 13)\) is the negative of \(d\) and scores B0.
1 mark
M1 Completes the square on \(9 + 6x - 3x^2\): \(-3\left[(x - 1)^2 - 1\right] + 9\), or a form \(p - 3(x - 1)^2\) seen, or compares coefficients with \(p - q(x - r)^2\)
Differentiating instead is not completing the square: M0 A0. A common error is \(10 - 3(x - 1)^2\) (the \(-1\) not multiplied by \(-3\)): M1 A0.
1 mark
A1 \(d = 12 - 3(x - 1)^2\), so the greatest value of \(d\) is 12, when \(x = 1\)
Both the greatest value 12 and \(x = 1\) are needed. Accept \(-3(x - 1)^2 + 12\).
1 mark
B1ft \(y = x + 1\)
The line moves down by the greatest value of \(d\): follow through \(y = x + 13 - \) their greatest \(d\). Alternative method: sets \(3x^2 - 5x + 4 = x + c\) and uses \(b^2 - 4ac = 0\): \(36 - 12(4 - c) = 0\), so \(c = 1\). Accept any equivalent form, e.g. \(x - y + 1 = 0\).
1 mark
B1 \((1, 2)\)
The line touches \(C\) where \(d\) is greatest, \(x = 1\); \(y = 3 - 5 + 4 = 2\) (or \(1 + 1 = 2\)). Accept \(x = 1, y = 2\).

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