Proof - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Proof 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.

18 questions
2–9 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 equation
\[x^2 + mx + n = 0\]
where \(m\) and \(n\) are real constants, has two distinct real roots.

Prove that the equation
\[x^2 + (m + 2)x + m + n + 1 = 0\]
also has two distinct real roots.

Model answer

The first equation has two distinct real roots, so its discriminant is positive: \(m^2 - 4n > 0\).
For the second equation, \(b^2 - 4ac = (m + 2)^2 - 4(1)(m + n + 1)\)
\(= m^2 + 4m + 4 - 4m - 4n - 4\)
\(= m^2 - 4n\)
\(> 0\)
So the second equation also has two distinct real roots.

Explanation

Two distinct real roots means a positive discriminant, so compare the two discriminants: that of the second equation simplifies to exactly \(m^2 - 4n\), the discriminant of the first, which is given to be positive. (The second quadratic is \((x + 1)^2 + m(x + 1) + n\), so its roots are the roots of the first equation minus 1.) Marks are lost for sign slips in \(-4(m + n + 1)\) and for not linking \(m^2 - 4n > 0\) back to the given information.

Mark scheme

1 mark
M1 Finds the discriminant of the second equation with the coefficients in the correct places: \((m + 2)^2 - 4(1)(m + n + 1)\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a proof. Testing particular values of \(m\) and \(n\) is not a proof and scores 0 marks. Alternative method: writes \(x^2 + (m + 2)x + m + n + 1 = (x + 1)^2 + m(x + 1) + n\) (M1), deduces that its roots are \(\alpha - 1\) and \(\beta - 1\), where \(\alpha\) and \(\beta\) are the two distinct real roots of the first equation (A1), and concludes (A1*).
1 mark
A1 Simplifies to \(m^2 - 4n\)
From \(m^2 + 4m + 4 - 4m - 4n - 4\). A common error is \(-4(m + n + 1) = -4m - 4n + 4\) (a sign slip): A0.
1 mark
A1* States that \(m^2 - 4n > 0\) because the first equation has two distinct real roots, so the discriminant of the second equation is positive, and concludes that the second equation has two distinct real roots
cso. Depends on the M mark. Both the reason (\(m^2 - 4n\) is the discriminant of the first equation, which is positive) and the conclusion are needed.

Question 2

5 marks

Show that \(x^4 + 2x^2 - 8x + 9\) can be written in the form
\[(x^2 - 1)^2 + k(x - 1)^2 + 4\]
where \(k\) is a constant to be found.

Hence prove that \(x^4 + 2x^2 - 8x + 9 \geqslant 4\) for all real values of \(x\), and find the value of \(x\) for which \(x^4 + 2x^2 - 8x + 9 = 4\).

Model answer

\((x^2 - 1)^2 + k(x - 1)^2 + 4 = x^4 - 2x^2 + 1 + kx^2 - 2kx + k + 4 = x^4 + (k - 2)x^2 - 2kx + (k + 5)\)
Comparing with \(x^4 + 2x^2 - 8x + 9\): \(k - 2 = 2\), \(-2k = -8\) and \(k + 5 = 9\), so \(k = 4\).
So \(x^4 + 2x^2 - 8x + 9 = (x^2 - 1)^2 + 4(x - 1)^2 + 4\).
For all real \(x\), \((x^2 - 1)^2 \geqslant 0\) and \(4(x - 1)^2 \geqslant 0\), so \(x^4 + 2x^2 - 8x + 9 \geqslant 0 + 0 + 4 = 4\).
Equality needs \(x^2 - 1 = 0\) and \(x - 1 = 0\), so \(x = 1\).

Explanation

Expanding the given form and comparing coefficients gives \(k = 4\), and all three comparisons must agree. Once the quartic is written as two squares plus 4, it can never be less than 4, because neither square can be negative. Equality needs both squares to be zero at the same time, which happens only at \(x = 1\); giving \(x = \pm 1\) is a common error, since \(x = -1\) makes the second square 16.

Mark scheme

1 mark
M1 Expands \((x^2 - 1)^2 + k(x - 1)^2 + 4\) to \(x^4 - 2x^2 + 1 + kx^2 - 2kx + k + 4\) and compares coefficients with \(x^4 + 2x^2 - 8x + 9\). An answer with no working scores 0 marks for the show-that and the proof; an answer of \(x = 1\) with no working scores 1 mark.
An answer with no working scores 0 marks for the show-that and the proof (\(k = 4\) stated alone scores 0 marks); an answer of \(x = 1\) with no working scores 1 mark (the B1). Allow one slip in the expansion for M1.
1 mark
A1 \(k = 4\), with the comparisons consistent (\(x^2\): \(k - 2 = 2\); \(x\): \(-2k = -8\); constant: \(k + 5 = 9\))
At least two of the comparisons must be seen. Alternatively, \((x^2 - 1)^2 + 4(x - 1)^2 + 4\) fully expanded to \(x^4 + 2x^2 - 8x + 9\).
1 mark
M1 Uses \((x^2 - 1)^2 \geqslant 0\) and \(4(x - 1)^2 \geqslant 0\) for all real \(x\)
Both terms must be considered. Finding the minimum of the quartic by calculus does not use "hence": M0 for this mark and the next.
1 mark
A1* So \(x^4 + 2x^2 - 8x + 9 = (x^2 - 1)^2 + 4(x - 1)^2 + 4 \geqslant 4\) for all real \(x\), with a conclusion
cso. "Squares are always positive" is not enough: accept "never negative" or "\(\geqslant 0\)".
1 mark
B1 \(x = 1\) (equality needs \(x^2 - 1 = 0\) and \(x - 1 = 0\) at the same time)
Do not accept \(x = \pm 1\): \(x = -1\) gives \(0 + 16 + 4 = 20\).

Question 3

6 marks

In a sale, a shop reduces the price of a coat by \(y\%\), where \(0 < y < 100\). After the sale, the shop increases the reduced price by \(z\%\), so that the coat returns to its original price.

Show that
\[z = \dfrac{100y}{100 - y}\]
Prove that \(z > y\).

Find the set of values of \(y\) for which \(z \geqslant 2y\), giving your answer in set notation.

Model answer

Let the original price be £\(P\). After the reduction the price is \(P\left(1 - \dfrac{y}{100}\right)\), and after the increase it is \(P\left(1 - \dfrac{y}{100}\right)\left(1 + \dfrac{z}{100}\right)\).
This equals \(P\), so \(\left(1 - \dfrac{y}{100}\right)\left(1 + \dfrac{z}{100}\right) = 1\).
\(1 + \dfrac{z}{100} = \dfrac{1}{1 - \dfrac{y}{100}} = \dfrac{100}{100 - y}\)
\(\dfrac{z}{100} = \dfrac{100}{100 - y} - 1 = \dfrac{y}{100 - y}\), so \(z = \dfrac{100y}{100 - y}\)
\(z - y = \dfrac{100y}{100 - y} - y = \dfrac{100y - 100y + y^2}{100 - y} = \dfrac{y^2}{100 - y}\)
\(y^2 > 0\) because \(y > 0\), and \(100 - y > 0\) because \(y < 100\), so \(z - y > 0\). Hence \(z > y\).
\(z \geqslant 2y\) means \(\dfrac{100y}{100 - y} \geqslant 2y\). Since \(y > 0\) and \(100 - y > 0\), multiplying both sides by the positive number \(\dfrac{100 - y}{y}\) gives \(100 \geqslant 2(100 - y)\), so \(2y \geqslant 100\) and \(y \geqslant 50\).
So the set of values is \(\{y : 50 \leqslant y < 100\}\).

Explanation

Treating each percentage change as a multiplier gives \(\left(1 - \dfrac{y}{100}\right)\left(1 + \dfrac{z}{100}\right) = 1\), which rearranges to \(z = \dfrac{100y}{100 - y}\). The difference \(z - y\) simplifies to \(\dfrac{y^2}{100 - y}\), which is positive because \(0 < y < 100\), so the increase needed to undo a reduction is always a larger percentage than the reduction, and it is at least double the reduction exactly when \(y \geqslant 50\). Marks are lost for multiplying an inequality by an expression without saying that it is positive, and for leaving out the condition \(y < 100\) in the final set.

Mark scheme

1 mark
M1 Forms a correct equation linking \(y\) and \(z\), e.g. \(P\left(1 - \dfrac{y}{100}\right)\left(1 + \dfrac{z}{100}\right) = P\), where £\(P\) is the original price, or \((100 - y)(100 + z) = 10\,000\). An answer with no working scores 0 marks for the show-that and the proof; an answer of \(\{y : 50 \leqslant y < 100\}\) with no working scores 1 mark.
An answer with no working scores 0 marks for the show-that and the proof; an answer of \(\{y : 50 \leqslant y < 100\}\) with no working scores 1 mark (the final A1). Numerical examples (e.g. \(y = 20\), \(z = 25\)) are not a proof. Adding or subtracting the percentages (e.g. \(z = y\)) is M0.
1 mark
A1* Rearranges with at least one intermediate step to \(z = \dfrac{100y}{100 - y}\), e.g. \(1 + \dfrac{z}{100} = \dfrac{100}{100 - y}\), so \(\dfrac{z}{100} = \dfrac{100}{100 - y} - 1 = \dfrac{y}{100 - y}\)
cso. The given answer must not be reached in one step from the equation.
1 mark
M1 Considers \(z - y\) over a common denominator: \(\dfrac{100y}{100 - y} - y = \dfrac{100y - y(100 - y)}{100 - y}\)
Alternative: writes \(z = y \times \dfrac{100}{100 - y}\) and uses \(0 < 100 - y < 100\), so \(\dfrac{100}{100 - y} > 1\), and \(y > 0\), to deduce \(z > y\): M1 A1*.
1 mark
A1* \(z - y = \dfrac{y^2}{100 - y}\), which is positive because \(y^2 > 0\) and \(100 - y > 0\), so \(z > y\)
cso. Both reasons (\(y^2 > 0\), or \(y > 0\), and \(100 - y > 0\)) and a conclusion are needed.
1 mark
M1 Solves \(\dfrac{100y}{100 - y} \geqslant 2y\) by multiplying by a positive quantity, with a reason that it is positive, e.g. multiplies by \(\dfrac{100 - y}{y}\) (positive as \(0 < y < 100\)) to get \(100 \geqslant 2(100 - y)\), or multiplies by \(100 - y\) to get \(100y \geqslant 2y(100 - y)\), i.e. \(2y(y - 50) \geqslant 0\)
Multiplying by \(100 - y\) or \(y\) without saying that it is positive is M0.
1 mark
A1 \(\{y : 50 \leqslant y < 100\}\)
Depends on the previous M mark, except in the answer-only special case in the first mark's comments. Accept \(\{y \in \mathbb{R} : 50 \leqslant y < 100\}\). Do not accept \(50 \leqslant y < 100\) without set notation, interval notation such as \([50, 100)\), \(\{y : y \geqslant 50\}\) (the condition \(y < 100\) must be included), or a strict inequality at 50 (when \(y = 50\), \(z = 100 = 2y\)).

Question 4

7 marks

A three-digit number is called balanced if its middle digit is the mean of its first digit and its last digit. For example, 246, 753 and 555 are balanced numbers.

Prove that every balanced number is a multiple of 3.

Prove that a balanced number is a multiple of 9 if and only if its middle digit is a multiple of 3.

Find all the balanced numbers that are multiples of 10.

Model answer

Let the balanced number be \(N = 100a + 10b + c\), where \(a\), \(b\) and \(c\) are its digits and \(a \geqslant 1\).
The middle digit is the mean of the other two, so \(b = \dfrac{a + c}{2}\), i.e. \(c = 2b - a\).
\(N = 100a + 10b + 2b - a = 99a + 12b = 3(33a + 4b)\)
\(33a + 4b\) is an integer, so every balanced number is a multiple of 3.
\(N = 99a + 12b = 9(11a + b) + 3b\)
If \(b\) is a multiple of 3, say \(b = 3m\), then \(N = 9(11a + b) + 9m = 9(11a + b + m)\), a multiple of 9.
If \(N\) is a multiple of 9, then \(3b = N - 9(11a + b)\) is a multiple of 9, so \(b\) is a multiple of 3.
So a balanced number is a multiple of 9 if and only if its middle digit is a multiple of 3.
A multiple of 10 has last digit \(c = 0\), so \(a = 2b\).
\(1 \leqslant a \leqslant 9\), so \(b = 1, 2, 3\) or 4, giving \(a = 2, 4, 6\) or 8.
The balanced multiples of 10 are 210, 420, 630 and 840.

Explanation

Writing the number as \(100a + 10b + c\) and using \(c = 2b - a\) gives \(N = 99a + 12b = 3(33a + 4b)\), which proves the first result, and \(N = 9(11a + b) + 3b\) shows that \(N\) is a multiple of 9 exactly when \(3b\) is, that is when \(b\) is a multiple of 3; both directions must be argued. For a multiple of 10 the last digit is 0, so the first digit is twice the middle one, giving 210, 420, 630 and 840. Common errors are proving only one direction of the "if and only if" and checking examples instead of using algebra.

Mark scheme

1 mark
B1 Writes the number as \(N = 100a + 10b + c\), where \(a\), \(b\) and \(c\) are its digits (\(a \geqslant 1\)), with the condition \(a + c = 2b\) (or \(b = \dfrac{a + c}{2}\)). An answer with no working scores 0 marks for the proofs; the numbers 210, 420, 630 and 840 with no working score 1 mark
An answer with no working scores 0 marks for the two proofs; the list 210, 420, 630, 840 with no working scores 1 mark (the final A1, as a special case), because a method is needed to show that there are no others. Checking examples (e.g. \(246 = 3 \times 82\)) is not a proof.
1 mark
M1 Uses the condition to write \(N\) in terms of two of the digits, e.g. \(c = 2b - a\) gives \(N = 100a + 10b + 2b - a = 99a + 12b\) (or \(10b = 5a + 5c\) gives \(N = 105a + 6c\))
Alternative: the digit sum is \(a + b + c = 3b\), and \(N = 99a + 9b + (a + b + c)\) (or the digit-sum test for divisibility by 3 is quoted): M1 for this, A1* for the conclusion.
1 mark
A1* \(N = 3(33a + 4b)\) (or \(3(35a + 2c)\)), where \(33a + 4b\) is an integer, so every balanced number is a multiple of 3
cso. A conclusion is needed.
1 mark
M1 Writes \(N\) as a multiple of 9 plus a remainder term, e.g. \(N = 99a + 12b = 9(11a + b) + 3b\)
Alternative: uses the digit-sum test for divisibility by 9 with the digit sum \(3b\): M1 for \(a + b + c = 3b\) and the test, A1* for both directions (\(3b\) is a multiple of 9 exactly when \(b\) is a multiple of 3).
1 mark
A1* Both directions: if \(b = 3m\), then \(N = 9(11a + b + m)\), a multiple of 9; if \(N\) is a multiple of 9, then \(3b = N - 9(11a + b)\) is a multiple of 9, so \(b\) is a multiple of 3
cso. Both directions of the "if and only if" are needed (a single chain of equivalences, \(9 \mid N \Leftrightarrow 9 \mid 3b \Leftrightarrow 3 \mid b\), is acceptable).
1 mark
M1 For a multiple of 10 the last digit is 0, so \(a = 2b\) with \(1 \leqslant a \leqslant 9\), giving \(b = 1, 2, 3, 4\)
Must use \(c = 0\) and the condition. \(b = 0\) gives \(a = 0\), which is not a three-digit number.
1 mark
A1 210, 420, 630 and 840, and no others
Depends on the previous M mark, except in the answer-only special case in the first mark's comments. All four and no extras. Including 000 or 0 scores A0.

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