Pearson Edexcel International Advanced Level Mathematics: Proof, factor theorem and circles (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover P2 sections 1.1 to 1.3 (proof), 2.1 (factor and remainder theorems) and 3.1 (the circle) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. Learn the methods first from the study guide and the revision notes. See also the course hub, the printable checklist and the free diagnostics. For the whole unit, see the Pure Mathematics 2 practice set.

Questions

1. Prove that the sum of the squares of any two consecutive integers is always odd. [3]

2. Prove that the cube of any integer is either a multiple of 9, or one more than a multiple of 9, or one less than a multiple of 9. [5]

3. Show by counter example that each statement is false.

(a) “4n² + 1 is prime for all positive integers n.” [2] (b) “For real numbers a and b, if a² > b² then a > b.” [2]

4. The polynomial f(x) = 2x³ + px² + qx − 8 has a factor (2x + 1). When f(x) is divided by (x + 1) the remainder is 5.

(a) Determine p and q. [4] (b) Hence solve f(x) = 0. [3]

5. f(x) = 6x³ − 11x² − 12x + 5.

(a) Show that (2x − 5) is a factor of f(x). [2] (b) Hence factorise f(x) completely. [3]

6. A circle has equation x² + y² + 6x − 4y − 23 = 0.

(a) State the coordinates of the centre and the value of the radius. [3] (b) Show that the point (3, 2) lies on the circle. [2]

7. The line segment PQ, where P is (−2, 3) and Q is (4, −1), is a diameter of a circle.

(a) Find an equation of the circle. [4] (b) The point R has x-coordinate 3 and positive y-coordinate and lies on the circle. Find the y-coordinate of R and show that angle PRQ = 90°. [3]

8. The circle C has equation (x − 2)² + (y + 3)² = 34.

(a) Show that T(7, −6) lies on C. [1] (b) Find an equation of the tangent to C at T, in the form ax + by = c where a, b and c are integers. [3] (c) Find the coordinates of the point where this tangent crosses the x-axis. [2]

9. The line y = x + 2 meets the circle x² + y² − 8x + 2y − 8 = 0 at the points A and B.

(a) Work out the coordinates of A and B. [4] (b) The centre of the circle is C. Show that the perpendicular from C to the line AB meets AB at the midpoint of AB. [3]

10. f(x) = x³ − 6x² + 3x + 10.

(a) Show that (x + 1) is a factor of f(x). [2] (b) Factorise f(x) completely. [3] (c) The curve y = f(x) crosses the x-axis at three points. The circle S has as a diameter the line joining the outer two of these points. Find an equation of S. [2] (d) Show that P(3, 2√2) lies on S and that the angle at P between the lines joining P to the ends of the diameter is a right angle. [2]

Answers

1. Let the integers be n and n + 1, with n an integer. [1] Sum of squares = n² + (n + 1)² = 2n² + 2n + 1 [1] = 2(n² + n) + 1, which is twice a whole number plus 1, so it is odd. [1] [3] Examiner insight: A proof built from a few numerical cases earns little; the general argument with a letter, and a closing sentence that links 2(n² + n) + 1 back to the claim, is what the question asks for.

2. Every integer has the form 3m, 3m + 1 or 3m + 2, where m is an integer, so these cases cover all integers. [1] Case 3m: (3m)³ = 27m³ = 9(3m³), a multiple of 9. [1] Case 3m + 1: (3m + 1)³ = 27m³ + 27m² + 9m + 1 = 9(3m³ + 3m² + m) + 1, one more than a multiple of 9. [1] Case 3m + 2: (3m + 2)³ = 27m³ + 54m² + 36m + 8 = 9(3m³ + 6m² + 4m + 1) − 1, one less than a multiple of 9. [1] Every case gives one of the three forms, so the statement holds for all integers. [1] [5] Examiner insight: In exhaustion, say why the list of cases is complete and write a conclusion; checking only some of the cases leaves the proof unfinished.

3. (a) Take n = 4: 4(16) + 1 = 65. [1] 65 = 5 × 13 is not prime, so the statement is false. [1] (b) Take a = −5 and b = 2: a² = 25 and b² = 4, so a² > b². [1] But a = −5 is less than b = 2, so a > b fails. [1] [4] Examiner insight: A counter example must satisfy the condition of the statement and then break the conclusion; show both calculations rather than quoting the numbers.

4. (a) Factor theorem: f(−1/2) = 0, so −1/4 + p/4 − q/2 − 8 = 0, which gives p − 2q = 33. [1] Remainder theorem: f(−1) = 5, so −2 + p − q − 8 = 5, which gives p − q = 15. [1] Subtract the second equation from the first: −q = 18, so q = −18. [1] p = −3, q = −18 [1] (b) Division by (2x + 1) gives f(x) = (2x + 1)(x² − 2x − 8). [1] x² − 2x − 8 = (x − 4)(x + 2). [1] x = −1/2, x = 4, x = −2 [1] [7] Examiner insight: Write the substitution value for each condition before you simplify, so the equations are clear even if an arithmetic slip follows; then check your p and q in both conditions.

5. (a) f(5/2) = 6(125/8) − 11(25/4) − 12(5/2) + 5 [1] = 375/4 − 275/4 − 30 + 5 = 25 − 25 = 0, so (2x − 5) is a factor. [1] (b) Dividing gives f(x) = (2x − 5)(3x² + 2x − 1). [1] 3x² + 2x − 1 = (3x − 1)(x + 1). [1] f(x) = (2x − 5)(3x − 1)(x + 1) [1] [5] Examiner insight: For “show that”, write out the substitution and the zero result, then state the conclusion; “completely” means the quadratic factor must also be factorised.

6. (a) Complete the square: (x + 3)² − 9 + (y − 2)² − 4 − 23 = 0. [1] (x + 3)² + (y − 2)² = 36 [1] Centre (−3, 2), radius 6 [1] (b) (3 + 3)² + (2 − 2)² = 36 + 0 [1] = 36, equal to r², so (3, 2) lies on the circle. [1] [5] Examiner insight: The centre has the opposite signs to the brackets, and the radius is the square root of the right-hand side; both slips are common on expanded equations.

7. (a) Centre = midpoint of PQ = ((−2 + 4)/2, (3 − 1)/2) = (1, 1). [1] PQ² = 6² + 4² = 52. [1] r² = (1/4)(52) = 13. [1] (x − 1)² + (y − 1)² = 13 [1] (b) (3 − 1)² + (y − 1)² = 13, so (y − 1)² = 9, giving y = 4 or y = −2. Since y is positive, y = 4. [1] Gradient PR = (4 − 3)/(3 + 2) = 1/5 and gradient RQ = (−1 − 4)/(4 − 3) = −5. [1] The product is −1, so PR is perpendicular to RQ and angle PRQ = 90°. [1] [7] Examiner insight: Use the positive condition to reject the second root explicitly, and show the gradient product rather than quoting the semicircle property alone, since the question says “show”.

8. (a) (7 − 2)² + (−6 + 3)² = 25 + 9 = 34, so T lies on C. [1] (b) Centre (2, −3). Gradient of radius = (−6 + 3)/(7 − 2) = −3/5. [1] Gradient of tangent = 5/3. [1] y + 6 = (5/3)(x − 7), so 3y + 18 = 5x − 35, giving 5x − 3y = 53. [1] (c) Put y = 0: 5x = 53. [1] (53/5, 0) [1] [6] Examiner insight: Flip and negate the radius gradient to get the tangent gradient; check your final line by substituting T, and give integer coefficients because the form is specified.

9. (a) Substitute y = x + 2: x² + (x + 2)² − 8x + 2(x + 2) − 8 = 0. [1] 2x² − 2x = 0, so 2x(x − 1) = 0. [1] x = 0 or x = 1. [1] A(0, 2) and B(1, 3) (using y = x + 2) [1] (b) Complete the square: (x − 4)² + (y + 1)² = 25, so C is (4, −1). [1] Midpoint M of AB = (1/2, 5/2). Gradient CM = (5/2 + 1)/(1/2 − 4) = −1. [1] Gradient AB = 1, and 1 × (−1) = −1, so CM is perpendicular to AB. The perpendicular from C meets AB at M. [1] [7] Examiner insight: Substitute the line into the circle, not the other way round, and find the y-values from the line; then state the conclusion in words, because the question asks you to show a property.

10. (a) f(−1) = −1 − 6 − 3 + 10 = 0. [1] So (x + 1) is a factor. [1] (b) Division gives f(x) = (x + 1)(x² − 7x + 10). [1] x² − 7x + 10 = (x − 2)(x − 5). [1] f(x) = (x + 1)(x − 2)(x − 5) [1] (c) The outer crossing points are (−1, 0) and (5, 0), so the centre is (2, 0) and the radius is 3. [1] (x − 2)² + y² = 9 [1] (d) (3 − 2)² + (2√2)² = 1 + 8 = 9, so P lies on S. [1] Gradient from (−1, 0) to P = 2√2/4 = √2/2; gradient from P to (5, 0) = −2√2/2 = −√2. The product is −1, so the angle at P is 90°. [1] [9] Examiner insight: Name the three crossing points from your factorisation before building the circle, and use the angle-in-a-semicircle property in your explanation as well as the gradient check.

Where marks are usually lost

• Checking one or two numbers in a proof instead of using general letters.
• Missing a final sentence that links the algebra to the statement being proved.
• Listing exhaustion cases that do not cover every integer.
• Offering a counter example that does not meet the condition in the statement.
• Substituting x = b/a instead of x = −b/a for the divisor (ax + b).
• Stopping at a quadratic factor when the question says “completely”.
• Reading the centre with the wrong signs, or giving r² as the radius.
• Using the radius gradient for the tangent instead of its negative reciprocal.
• Stating a circle property without showing the numerical check when the question says “show that”.

Next steps

• Re-read the revision notes for the methods behind the questions you missed.
• Work back through the study guide.
• Use the Edexcel A Level Mathematics hub and the checklist to plan which unit to revise next.
• Take the free diagnostics when you want a quick check on another topic.
• Book a free trial class: /trial/.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics: Specification, Issue 3, April 2019, first teaching September 2018, published by Pearson Education Limited. These questions cover P2 sections 1.1, 1.2, 1.3, 2.1 and 3.1.