Mathematics 9709/22 May/June 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
1 Attempt to express equation in terms of sec θ only M1 or equivalent in terms of cos θ only, using a correct
identity, allow if ‘5’ omitted.
Obtain 6sec2 θ − 17secθ − 14 ( = 0 ) A1 or 14 cos 2 θ + 17 cos θ − 6 = 0 .
Attempt solution of 3-term quadratic equation to find one value of θ , from M1
cos θ = ...
Obtain 73.4 A1 or greater accuracy.
Obtain 286.6 A1 or greater accuracy; and no others between 0 and 360.
5
ln y
2.9, 14.1
0.4, 3.6 x
O
The variables x and y satisfy the equation y = Ae A−Bx , where A and B are constants. The graph of ln y against x is a straight line passing through the points 0.4, 3.6 and 2.9, 14.1, as shown in the diagram.
Find the values of A and B correct to 3 significant figures. [5]
Mark scheme
2 State or imply equation ln y = ln A + ( A − B ) x B1 Allow inclusion of ln e.
Equate A − B to gradient of line M1
Obtain A − B = 4.2 A1
Substitute appropriate values to find value of ln A M1
Obtain ln A = 1.92 and hence A = 6.82 and B = 2.62 A1 or greater accuracy.
Alternative Method 1
State or imply equation ln y = ln A + ( A − B ) x B1 Allow inclusion of ln e.
ln y − 3.6 x − 0.4 M1 condone use of y in place of ln y .
Use of coordinates to obtain equation of line =
14.1 − 3.6 2.9 − 0.4
Obtain gradient equal to 4.2 A1
Substitute appropriate values to find value of ln A M1
Obtain ln A = 1.92 and hence A = 6.82 and B = 2.62 A1 or greater accuracy.
Alternative Method 2
State or imply equation ln y = ln A + ( A − B ) x B1 Allow inclusion of ln e.
3.6 = ln A + 0.4 ( A − B ) M1 For one correct equation.
14.1 = ln A + 2.9 ( A − B ) A1 For both correct equations.
Obtain A = 6.82 and B = 2.62 M1 A1 For attempt at solution by elimination of ln A to obtain
both values. SC B1 for y = 4.2 x + 1.92 .
5
y
2
x
O 6
6
The diagram shows part of the curve y = . The shaded region is bounded by the curve and the
2x + 3 lines x = 6 and y = 2.
Find the exact area of the shaded region, giving your answer in the form a − ln b, where a and b are integers. [5]
Mark scheme
3 Integrate to obtain the form k ln(2 x + 3) *M1 Condone lack or misuse of brackets if recovered later.
Obtain correct 3ln(2 x + 3) A1 Allow unsimplified.
Apply limits 0 and 6 correctly to obtain k ln15 − k ln 3 *DM1 Allow unsimplified.
Apply relevant logarithm properties correctly to obtain form ln b DM1
Obtain 12 − ln125 A1
5
(a) y
x
O
− 1x
The diagram shows the graph of y = 3 − e 2 .
− 1x
On the diagram, sketch the graph of y = 5x − 4, and show that the equation 3 − e 2 = 5x − 4 has exactly two real roots. [2]
− 1x
It is given that the two roots of 3 − e 2 = 5x − 4 are denoted by ! and ", where ! < ".
(b) Show by calculation that ! lies between 0.36 and 0.37. [2]
−1x
!
(c) Use the iterative formula xn+1 = 15 7 − e 2 n to find " correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]
Mark scheme
4(a) Draw (more or less) correct sketch with vertex on positive x-axis *B1 crossing y-axis above given graph, may be implied by
extrapolation.
Indicate in some way the two roots DB1
2
4(b) −1 x −1 x
Consider sign of 3 − e 2 + 5 x − 4 or of 3 − e 2 − 5 x − 4 for 0.36 and 0.37 M1 but not for sign of 3 − e− 12 x − 5 x + 4 .
May be implied by −0.035... and 0.018... , or
equivalents.
Obtain −0.035... and 0.018... , or equivalents, and justify conclusion A1 AG necessary detail needed.
2
4(c) Use iteration process correctly at least once M1
Obtain final answer 1.295 A1 answer required to exactly 4 sf.
Show sufficient iterations to 6 sf to justify answer or show sign change in A1
interval [1.2945, 1.2955]
3
y
C
A B x
O
−1x
The diagram shows the curve with equation y = e 2 x2 − 5x + 4. The curve crosses the x-axis at the points A and B, and has a maximum at the point C.
(a) Find the exact gradient of the curve at B. [5]
(b) Find the exact coordinates of C. [4]
Mark scheme
5(a) Attempt use of product rule to find first derivative *M1
−1 x −1 x
Obtain − 12 e 2 ( x 2 − 5 x + 4) + e 2 (2 x − 5) A1 OE
Obtain x = 4 for point B B1
Substitute x = 4 to find the value of the derivative DM1
Obtain 3e −2 A1 or exact equivalent.
5
5(b) Equate their first derivative to zero and simplify as far as quadratic equation *M1 allow if it appears in part (a).
Obtain at least x 2 − 9 x + 14 = 0 A1 OE
Solve to find relevant x value and substitute to find the value of y DM1
Obtain x = 7 and y = 18e 2
−7 A1 or exact equivalent.
4
(a) Show that 4 sin 1 + 13 π cos 1 − 13 π 3 + 2 sin 21. [4]
(b) Find the exact value of 4 sin 17
24 π cos 24
1 π. [2]
1π
(c) Find the exact value of Ó 4 sin 2x + 13 π cos 2x − 13 π dx.
8
[4]
0
Mark scheme
6(a) Obtain at least either ( 12 sin θ + 12 3 cosθ ) or ( 12 cosθ + 12 3 sin θ ) B1 Allow if implied by decimal values.
Expand and simplify with correct use of sin 2 θ + cos 2 θ = 1 M1
Use sin θ cos θ = 12 sin 2θ M1
Confirm given result 3 + 2sin 2θ A1 AG necessary detail required.
4
6(b) Identify value of θ is 83 π *B1 OE
Obtain 3 + 2sin 34 π and conclude 3+ 2 DB1 or exact equivalent.
2
6(c) Identify integrand as 3 + 2sin 4 x B1
Integrate to obtain form k1 x + k2 cos 4 x M1 where k1k 2 ≠ 0 .
Obtain correct 3 x − 12 cos 4 x A1
Obtain 18 π 3 + 12 A1 or exact equivalent.
4
A curve has parametric equations
2t + 3 x= , y = t2 + at + 1, t+2 where a is a constant. It is given that, at the point P on the curve, the gradient is 1.
(a) Show that the value of t at P satisfies the equation
2t3 + a + 8t2 + 4a + 8t + 4a − 1 = 0. [4]
(b) It is given that t + 1 is a factor of
2t3 + a + 8t2 + 4a + 8t + 4a − 1.
Find the value of a. [2]
(c) Hence show that P is the only point on the curve at which the gradient is 1. [3]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) dx M1*
Attempt use of quotient rule (or equivalent) to find
dt
1 A1 or (unsimplified) equivalent.
Obtain
(t + 2) 2
dy DM1 dy
Equate to 1 Must be using their .
dx dx
Obtain (2t + a )(t + 2) 2 = 1 and expand to confirm given result A1 AG necessary detail required.
4
7(b) Substitute t = − 1 , equate to zero and attempt solution for a M1 Allow a complete method using algebraic long division
or synthetic division.
Obtain a = 3 A1
2
7(c) Divide their cubic by t + 1 at least as far as the x term M1 or equivalent (inspection, identity, …).
Obtain 2t 2 + 9t + 11 A1
Calculate discriminant, obtain −7 and conclude no further value of t A1 OE
3
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/22, May/June 2023 (9709_s23_ms_22), and the question paper (9709_s23_qp_22), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/22 for May/June 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.
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