9709 Pure Mathematics 1 Oct/Nov 2023 Paper 11

26 questions and parts, 75 marks in total. Marks come from the published mark scheme.

1. Question 1 (a)

   2 marks · 2/75

   This tests the use of the binomial theorem to obtain specified terms of an expansion.

   Examiner report: Most candidates used the binomial method successfully. A frequent mistake was mishandling the squared factor when forming the third term.

   Question paper p.3 · Mark scheme p.6

2. Question 1 (b)

   2 marks · 4/75

   This tests selecting and combining only the products that contribute to a particular coefficient in a product of polynomials.

   Examiner report: The report found that stronger responses used the three relevant products rather than expanding every term; a complete expansion was viewed as unnecessary and time-consuming.

   Question paper p.3 · Mark scheme p.6

3. Question 2

   4 marks · 8/75

   This tests comparing a line and a curve by forming a quadratic intersection equation and analysing its discriminant.

   Examiner report: Sign errors occurred while forming the quadratic or its discriminant. Some candidates incorrectly treated a positive discriminant as zero, so the link between discriminant sign and the number of intersections was not always understood.

   Question paper p.4 · Mark scheme p.6

4. Question 3

   3 marks · 11/75

   This tests related rates involving the volume of a cube and the chain rule.

   Examiner report: This was difficult for many candidates, with numerous attempts avoiding differentiation altogether. Successful work used the volume derivative followed by the chain rule.

   Question paper p.5 · Mark scheme p.7

5. Question 4 (a)

   2 marks · 13/75

   This tests applying a reflection and then a translation to the graph of a quadratic, while respecting the order of transformations.

   Examiner report: Sign mistakes were common, although stronger candidates obtained both transformed equations correctly.

   Question paper p.6 · Mark scheme p.7

6. Question 4 (b)

   2 marks · 15/75

   This tests applying the same two transformations in the reverse order and recognising that transformation order matters.

   Examiner report: Many candidates lost the first mark through a sign or order error but were more often able to gain the mark for presenting a quadratic form.

   Question paper p.6 · Mark scheme p.7

7. Question 4 (c)

   2 marks · 17/75

   This tests comparing two transformed graphs and describing the single transformation relating them.

   Examiner report: Candidates who had reliable results in the first two parts generally identified this translation. Follow-through credit was available when the earlier equations led to a vertical shift.

   Question paper p.6 · Mark scheme p.7

8. Question 5 (a)

   3 marks · 20/75

   This tests transforming a trigonometric equation containing tangent and reciprocal sine into a quadratic in cosine.

   Examiner report: A sound route was to replace tangent before multiplying through by sine. Algebraic manipulation errors prevented many candidates from completing the reduction.

   Question paper p.7 · Mark scheme p.8

9. Question 5 (b)

   3 marks · 23/75

   This tests solving a trigonometric equation through a quadratic in cosine and listing every solution in a full degree interval.

   Examiner report: Full credit depended on showing how the quadratic was solved and then finding both angles. Incomplete methods and omitted solutions were common; checking candidates' values in the original equation was also advised.

   Question paper p.7 · Mark scheme p.8

10. Question 6 (a)

    1 mark · 24/75

    Question paper p.8 · Mark scheme p.8

11. Question 6 (b)

    1 mark · 25/75

    Question paper p.8 · Mark scheme p.8

12. Question 6 (c)

    5 marks · 30/75

    This tests finding the exact area of a compound circular shape using sectors, triangles, and circular segments.

    Examiner report: This was one of the more demanding parts because candidates had to devise the area decomposition themselves. Partial attempts often found a sector, triangle, or segment, while a complete strategy was uncommon.

    Question paper p.9 · Mark scheme p.9

13. Question 7

    7 marks · 37/75

    This tests solving for the terms of a geometric progression from a finite sum and an infinite sum, with a sign restriction on the ratio.

    Examiner report: The initial sum equations were often correct. Methods that left the first-two-term expression unsimplified tended to create a more complicated equation and more errors, whereas stronger candidates used the negative solution as instructed.

    Question paper p.11 · Mark scheme p.12

14. Question 8 (a)

    4 marks · 41/75

    This tests using a substitution to solve the intersections of two polynomial curves and then recovering both coordinates.

    Examiner report: Some candidates forgot to take square roots after solving for the squared variable, while others treated the substituted variable as x. Correctly recovering both coordinates was essential.

    Question paper p.12 · Mark scheme p.15

15. Question 8 (b)

    5 marks · 46/75

    This tests finding the exact area between two curves by integrating their vertical difference over the intersection interval.

    Examiner report: Most candidates who attempted this part understood that integration and both limits were required. Calculator-only answers without displayed limit substitution did not receive the associated method credit.

    Question paper p.13 · Mark scheme p.16

16. Question 9 (a)

    2 marks · 48/75

    Question paper p.14 · Mark scheme p.17

17. Question 9 (b)

    3 marks · 51/75

    This tests finding the domain of a composite function by requiring the output of one function to lie within the domain of the next.

    Examiner report: Successful solutions worked directly from the range restriction on f. A number of candidates instead tried to form the composite function, which did not address the domain condition.

    Question paper p.14 · Mark scheme p.18

18. Question 9 (c)

    1 mark · 52/75

    This tests forming and simplifying a composite function by substituting one quadratic function into a linear function.

    Examiner report: Most candidates gained the available mark, and equivalent simplified forms were accepted.

    Question paper p.15 · Mark scheme p.18

19. Question 9 (d)

    3 marks · 55/75

    This tests finding an inverse function for a restricted quadratic by completing the square and selecting the branch determined by the domain.

    Examiner report: Completing the square first made this part accessible. The main difficulty was choosing the branch consistent with the original domain; attempts starting from the uncompleted quadratic generally made no progress.

    Question paper p.15 · Mark scheme p.18

20. Question 10 (a)

    3 marks · 58/75

    This tests integrating a second derivative and using a stationary-point condition to determine the integration constant.

    Examiner report: The reliable method was to integrate before applying the stationary condition. A common error was substituting the point's y-coordinate into an equation involving dy/dx.

    Question paper p.16 · Mark scheme p.19

21. Question 10 (b)

    3 marks · 61/75

    This tests integrating a first derivative and using a known point on the curve to determine the second constant.

    Examiner report: Most candidates who reached this part integrated their derivative and used the supplied point appropriately.

    Question paper p.16 · Mark scheme p.19

22. Question 10 (c)

    3 marks · 64/75

    This tests locating the remaining stationary point and classifying it using the second derivative.

    Examiner report: Candidates generally found the stationary x-values by setting the gradient to zero. Those who classified the point usually used the second derivative, which was the appropriate test.

    Question paper p.17 · Mark scheme p.20

23. Question 10 (d)

    2 marks · 66/75

    This tests finding a tangent line from the curve's y-axis intercept and the derivative at that point.

    Examiner report: Successful responses supplied both ingredients: the gradient from the derivative and the intercept from the curve equation. Omitting either prevented completion.

    Question paper p.17 · Mark scheme p.20

24. Question 11 (a)

    3 marks · 69/75

    This tests finding the chord joining the contact points of parallel tangents to a circle.

    Examiner report: Stronger candidates used the centre and the perpendicular-radius property to form the line. Some identified the centre but did not continue to the equation.

    Question paper p.18 · Mark scheme p.20

25. Question 11 (b)

    4 marks · 73/75

    This tests finding a circle-line intersection and expressing the selected point in surd form.

    Examiner report: Candidates who had a correct line in the previous part usually substituted successfully. The quadratic had to be solved and the correct intersection selected for A.

    Question paper p.19 · Mark scheme p.21

26. Question 11 (c)

    2 marks · 75/75

    This tests finding the tangent at a known point on a circle when its gradient is specified.

    Examiner report: Candidates with the coordinates of A often earned at least the method mark by using the tangent equation through that point.

    Question paper p.19 · Mark scheme p.21