9709 Pure Mathematics 3 May/June 2022 Paper 31

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

1. Question 1

   4 marks · 4/75

   Solving an exponential equation by applying logarithms and rearranging the resulting linear equation.

   Examiner report: A frequent error was treating powers as though they could be subtracted directly. Many candidates formed the logarithmic equation successfully but then failed to isolate the unknown; a smaller group used an equivalent index-based reduction effectively.

   Question paper p.3 · Mark scheme p.6

2. Question 2 (a)

   4 marks · 8/75

   Using the binomial expansion for a negative integer power and simplifying terms through the fourth power of the variable.

   Examiner report: Errors commonly arose from extracting the factor of 2 incorrectly or expanding the wrong expression. Even when the general expansion was right, arithmetic mistakes often spoiled the simplified coefficients.

   Question paper p.4 · Mark scheme p.7

3. Question 2 (b)

   1 mark · 9/75

   Question paper p.4 · Mark scheme p.7

4. Question 3

   6 marks · 15/75

   Question paper p.5 · Mark scheme p.8

5. Question 4

   7 marks · 22/75

   Solving a first-order separable differential equation using an initial condition.

   Examiner report: Most successful attempts began with separation and integrated the logarithmic side correctly. Some confused the x-integral with an inverse-tangent integral, while others stopped at a logarithmic form or introduced an incorrect constant when exponentiating.

   Question paper p.6 · Mark scheme p.9

6. Question 5 (a)

   5 marks · 27/75

   Using the factor theorem on a polynomial and its derivative to determine unknown coefficients.

   Examiner report: Long division was used unnecessarily by many candidates and often did not produce a usable remainder. The more direct factor-theorem approach would have supplied two simple equations.

   Question paper p.8 · Mark scheme p.10

7. Question 5 (b)

   3 marks · 30/75

   Factorising a cubic when a factor is repeated.

   Examiner report: Only a small proportion recognised immediately that the known factor must occur twice. Some used numerical roots but then failed to restore the leading coefficient in the final factorisation.

   Question paper p.9 · Mark scheme p.11

8. Question 6 (a)

   4 marks · 34/75

   Carrying out a trigonometric substitution in a definite integral, including the differential and changed limits.

   Examiner report: Very few candidates differentiated the substitution correctly. Others replaced x but not dx, or did not use the identity linking tangent and secant; candidates who completed the substitution usually handled the new limits well.

   Question paper p.10 · Mark scheme p.12

9. Question 6 (b)

   4 marks · 38/75

   Question paper p.11 · Mark scheme p.12

10. Question 7 (a)

    3 marks · 41/75

    Converting a quotient of complex numbers into Cartesian form.

    Examiner report: Rationalising the denominator was the usual method, but arithmetic slips in the numerator were common. A few candidates instead compared real and imaginary parts after multiplying out, which was also effective.

    Question paper p.12 · Mark scheme p.13

11. Question 7 (b)

    2 marks · 43/75

    Question paper p.13 · Mark scheme p.13

12. Question 7 (c)

    3 marks · 46/75

    Finding both square roots of a complex number in exponential form.

    Examiner report: Candidates who had a correct polar form generally made reasonable progress. Those who did not obtain that form in the previous part rarely found a viable alternative route.

    Question paper p.13 · Mark scheme p.13

13. Question 8 (a)

    4 marks · 50/75

    Implicitly differentiating an algebraic equation involving x, y and a product term.

    Examiner report: Many candidates knew the general implicit-differentiation process and obtained the derivative. Errors mainly came from differentiating the y-dependent terms incorrectly.

    Question paper p.14 · Mark scheme p.14

14. Question 8 (b)

    5 marks · 55/75

    Finding the angle between two tangents using their gradients and the tangent formula for two lines.

    Examiner report: This part caused considerable difficulty. Some candidates substituted only one coordinate and did not locate the actual points; even those who found both gradients often lacked a method for combining them to obtain the angle.

    Question paper p.15 · Mark scheme p.14

15. Question 9 (a)

    3 marks · 58/75

    Using cuboid geometry, midpoints and a section ratio to express three-dimensional vectors in component form.

    Examiner report: Several candidates produced correct vectors, with most failures caused by reading the diagram incorrectly or making simple component arithmetic errors.

    Question paper p.16 · Mark scheme p.15

16. Question 9 (b)

    2 marks · 60/75

    Forming a vector equation for a line through two points in three dimensions.

    Examiner report: Most candidates knew the required line structure and used their earlier vectors appropriately. A recurring presentation error was omitting the equality sign in the vector equation.

    Question paper p.16 · Mark scheme p.15

17. Question 9 (c)

    4 marks · 64/75

    Finding the shortest distance from the origin to a line in three-dimensional vector form.

    Examiner report: This was a more demanding vector calculation. Stronger responses located the perpendicular foot first; partial attempts often identified the need for perpendicularity but did not complete the parameter or distance calculation.

    Question paper p.17 · Mark scheme p.15

18. Question 10 (a)

    4 marks · 68/75

    Differentiating a product containing a square root of a trigonometric function and using the stationary-point condition.

    Examiner report: Most candidates recognised the product rule. The main technical weakness was mishandling the derivative of the square root of sine, so the final relation was not reached.

    Question paper p.18 · Mark scheme p.16

19. Question 10 (b)

    2 marks · 70/75

    Numerically verifying that a root lies within a stated interval by comparing the relevant function values.

    Examiner report: A number of complete arguments were seen, but incorrect calculator angle settings caused many of the errors.

    Question paper p.19 · Mark scheme p.16

20. Question 10 (c)

    2 marks · 72/75

    Showing that a fixed-point iteration has the same limiting value as the root of a trigonometric equation.

    Examiner report: Most candidates did not understand that the iteration had to be examined at its limiting value. A few wrote a suitable fixed-point equation, but only rarely was the trigonometric rearrangement justified fully.

    Question paper p.19 · Mark scheme p.16

21. Question 10 (d)

    3 marks · 75/75

    Using fixed-point iteration to approximate a root to a specified accuracy.

    Examiner report: Candidates who iterated with the correct angle setting generally obtained the required approximation. Inadequate working or too few displayed iterations made some otherwise plausible rounded results unsupported.

    Question paper p.19 · Mark scheme p.17