Mathematics 9709/21 May/June 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The variables x and y satisfy the equation y = 42x−a , where a is an integer. As shown in the diagram, the graph of ln y against x is a straight line passing through the point 0, −20.8, where the second coordinate is given correct to 3 significant figures.
(a) Show that the gradient of the straight line is ln 16. [2]
(b) Determine the value of a. [2]
Mark scheme
1(a) State or imply equation is ln y = (2 x − a ) ln 4 B1 OE. Do not condone poor use of brackets.
State gradient is 2ln 4 and confirm ln16 B1 AG – necessary detail needed
2
1(b) Substitute for ln y and attempt value of a M1 Allow if ln y = 2 x − a ln 4
Obtain a = 15 A1 Integer answer required, but condone 15.0
2
(a) Express the equation 7 tan 1 + 4 cot 1 − 13 sec 1 = 0 in terms of sin 1 only. [3]
(b) Hence solve the equation 7 tan 1 + 4 cot 1 − 13 sec 1 = 0 for 0Å < 1 < 360Å. [3]
Mark scheme
2(a) Attempt to express left hand side of the equation in terms of sin θ and cosθ M1 with at least two of the terms correct and no missing
θs
Condone use of x instead of θ
Obtain 7 sin 2 θ + 4cos 2 θ − 13sin θ [ = 0] A1 SOI, OE
Obtain 3sin 2 θ − 13sin θ + 4 = 0 A1 Allow if missing θ s are recovered
4
SC Allow full marks for 3sin θ − 13 + =0
sin θ
Must be in terms of θ for final A mark
3
2(b) Attempt solution of 3-term quadratic equation for sin θ M1
Obtain sin θ = 13 and hence 19.5 A1 or greater accuracy
Obtain second value 160.5 A1 or greater accuracy; and no other values within the
given range
FT on 180o − their 19.5o
3
The diagram shows the curve with equation y = 3 sin x − 3 sin 2x for 0 ≤ x ≤ π. The curve meets the x-axis at the origin and at the points with x-coordinates a and π.
(a) Find the exact value of a. [3]
(b) Find the area of the shaded region. [4]
Mark scheme
3(a) Attempt to find x-value from 3sin x − 3sin 2 x = 0 using identity for sin 2x M1
Obtain at least cos x = 12 A1
Obtain 13 π A1 SC B3 can be spotted from sin x = sin 2 x
3
3(b) Integrate to obtain form k1 cos x + k2 cos 2 x *M1 non-zero constants k1 , k2
M0 for 3cos x ± 6cos 2 x
Obtain correct −3cos x + 32 cos 2 x A1
Attempt value of integral using their lower limit (in radians) and π correctly DM1 Allow one sign error
27 A1 OE
Obtain
4
4
A curve has equation x2 y + 2y3 = 48.
Find the equation of the normal to the curve at the point 4, 2, giving your answer in the form ax + by + c = 0 where a, b and c are integers. [7]
Mark scheme
4 Use product rule to differentiate x 2 y *M1 There must be evidence of implicit differentiation
dy A1
Obtain correct 2 xy + x 2
dx
dy dy *B1
Obtain 2 xy + x 2 + 6 y2 =0
dx dx
dy DM1 dy
Substitute x = 4, y = 2 to find value of dependent on at least one term involving
dx dx
2 A1 SOI, OE
Obtain −
5
Attempt equation of normal passing through ( 4, 2 ) M1 with numerical gradient correctly obtained from the
negative reciprocal of their derivative
Obtain 5 x − 2 y − 16 = 0 A1
7
(a) By sketching the graphs of y = 5 − 2x and y = 3 ln x on the same diagram, show that the equation 5 − 2x = 3 ln x has exactly two roots. [3]
(b) Show that the value of the larger root satisfies the equation x = 2.5 + 1.5 ln x. [1]
(c) Show by calculation that the value of the larger root lies between 4.5 and 5.0. [2]
(d) Use an iterative formula, based on the equation in part (b), to find the value of the larger root correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3]
9e2x + 16
A curve has equation y = ex − 1
Mark scheme
5(a) Draw correct sketch of y = 5 − 2 x *B1 with vertex on positive x-axis
Draw correct sketch of y = 3ln x *B1
Indicate the two roots either on the diagram or by a statement DB1
3
5(b) State 2 x − 5 = 3ln x and rearrange to confirm x = 2.5 + 1.5ln x B1 AG – necessary detail needed
1
5(c) Consider sign of x − 2.5 − 1.5ln x , or equivalent, for 4.5 and 5.0 M1
Obtain −0.25... and 0.08... or equivalents and justify conclusion A1 AG – necessary detail needed
Alternative method for question 5(c)
Consider sign of 5 − 2 x − 3ln x , or equivalent, for 4.5 and 5.0 M1
Obtain −0.51... and 0.17... or equivalents and justify conclusion A1 AG – necessary detail needed
2
5(d) Use iteration process correctly at least once M1
Obtain final answer 4.88 A1 Answer required to exactly 3 s.f.
Show sufficient iterations to 5 s.f. to justify answer or show sign change in A1
interval [4.875, 4.885]
3
.
(a) Show that the x-coordinate of any stationary point on the curve satisfies the equation ex 3ex − 8 3ex + 2 = 0. [4]
(b) Hence show that the curve has only one stationary point and find its exact coordinates. [4]
Mark scheme
6(a) Use quotient rule (or equivalent) to find first derivative *M1 Condone poor use of brackets if recovered later
18e 2 x (e x − 1) − e x (9e 2 x + 16) A1 OE
Obtain
(e x − 1) 2
Equate first derivative to zero and attempt factorisation DM1 Need to be working with 9e3 x − 18e2 x ± 16e x = 0
Obtain e x (3e x − 8)(3e x + 2) = 0 A1 AG – necessary detail needed
SC B3 If numerator is in incorrect order and given
result is obtained.
SC B3 If denominator is not squared and given result
is obtained.
4
6(b) Observe, using given result from part (a), at some stage, e x ≠ 0 and B1 Allow if discounted by crossing through OE
3e x + 2 ≠ 0 and hence one stationary point
Attempt exact solution for x and for y M1
Obtain x = ln( 83 ) or exact equivalent A1
Obtain y = 48 A1 Not from a rounded decimal
4
The polynomial p x is defined by
p x = 2x3 + 5x2 + ax + 2a, where a is an integer.
(a) Find, in terms of x and a, the quotient when p x is divided by x + 2, and show that the remainder is 4. [3]
1
(b) It is given that Ô dx = 22 + ln b, where b is an integer. p x x+2 3
−1
Find the values of a and b. [6]
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) Carry out algebraic long division at least as far as 2 x 2 + kx M1
Obtain quotient 2 x 2 + x + a − 2 A1
Confirm remainder is 4 A1 AG – necessary detail needed
SC B1 for use of remainder theorem to obtain 4
3
7(b) 4 B1 FT Following their quotient, may be implied
Identify integrand as 2 x 2 + x + a − 2 +
x+2
Integrate to obtain at least 2 terms from the form of k1 x3 + k2 x2 + k3 ln( x + 2) M1 for non-zero k1 , k2 , k3
Obtain correct 23 x 3 + 12 x 2 + ax − 2 x + 4ln( x + 2) A1
Apply limits correctly and attempt correct process to find a or b M1 Must have the correct form
Obtain − 83 + 2a = 223 or equivalent and hence a = 5 A1
Obtain 4 ln 3 and hence b = 81 B1
6
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/21, May/June 2022 (9709_s22_ms_21), and the question paper (9709_s22_qp_21), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/21 for May/June 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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