Mathematics 9709/32 February/March 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 3 Pure Mathematics, February/March 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Sketch the graph of y = x − 2. [1]
(b) Solve the inequality x − 2 < 3x − 4. [3]
Mark scheme
1(a) Make a recognisable sketch graph of y = x − 2 B1
1
1(b) Find x-coordinate of intersection with y = 3x – 4 M1
3 A1
Obtain x =
2
3 A1
State final answer x > only
2
Alternative method for question 1(b)
Solve the linear inequality 3x − 4 > 2 − x , or corresponding equation M1
3 A1
Obtain critical value x =
2
3 A1
State final answer x > only
2
Alternative method for question 1(b)
Solve the quadratic inequality ( x − 2 ) < ( 3x − 4 ) ,
2 2 M1
or corresponding equation
3 A1
Obtain critical value x =
2
3 A1
State final answer x > only
2
3
Solve the equation ln 3 + ln 2x + 5 = 2 ln x + 2. Give your answer in a simplified exact form. [4]
Mark scheme
2 Use law of logarithm of a power and sum and remove logarithms M1
Obtain a correct equation in any form, e.g. 3 ( 2 x + 5 ) = ( x + 2 )
2 A1
Use correct method to solve a 3-term quadratic, obtaining at least one root M1
Obtain final answer x = 1 + 2 3 or 1 + 12 only A1
4
(a) By sketching a suitable pair of graphs, show that the equation sec x = 2 − 12 x has exactly one root in the interval 0 ≤ x < 12 π. [2]
(b) Verify by calculation that this root lies between 0.8 and 1. [2]
A @
(c) Use the iterative formula xn+1 = cos−1
2 to determine the root correct to 2 decimal places.
4 − xn
Give the result of each iteration to 4 decimal places. [3]
1π
Find Ó
3
Mark scheme
3(a) Sketch the graph y = sec x M1
1 A1
Sketch the graph y = 2 − x , and justify the given statement
2
2
3(b) Calculate the values of a relevant expression or pair of expressions at x = 0.8 M1
and x = 1
Complete the argument correctly with correct calculated values A1
2
3(c) Use the iterative formula correctly at least once M1
Obtain final answer 0.88 A1
Show sufficient iterations to 4 d.p. to justify 0.88 to 2 d.p., or show there is a A1
sign change in the interval (0.875, 0.885)
3
x sec2 x dx. Give your answer in a simplified exact form. [7]
1π
6
cos 3x sin 3x
Mark scheme
4 Integrate by parts and reach ax tan x + b tan xdx M1*
Obtain x tan x − tan x dx A1
Complete the integration, obtaining a term ± lncos x , or equivalent M1
Obtain integral x tan x + lncos x , or equivalent A1
Substitute limits correctly, having integrated twice DM1
Use a law of logarithms M1
5 1 A1
Obtain answer 3π − ln 3 , or exact simplified equivalent
18 2
7
(a) Show that + = 2 cot 2x. [4]
sin x cos x
cos 3x sin 3x
(b) Hence solve the equation + = 4, for 0 < x < π. [3]
sin x cos x
Mark scheme
5(a) Express LHS correctly as a single fraction B1
Use cos ( A ± B ) formula to simplify the numerator M1
Use sin 2A formula to simplify the denominator M1
Obtain the given result. A1
4
5(b) Obtain an equation in tan2x and use correct method to solve for x M1
Obtain answer, e.g. 0.232 A1
Obtain second answer, e.g. 1.80 A1 Ignore answers outside the given interval.
3
The variables x and y satisfy the differential equation
dy 1 + 4y2
= . dx ex
It is given that y = 0 when x = 1.
(a) Solve the differential equation, obtaining an expression for y in terms of x. [7]
(b) State what happens to the value of y as x tends to infinity. [1]
Mark scheme
6(a) Separate variables correctly and attempt integration of at least one side B1
Obtain term of the form a tan −1 (2 y) M1
1 A1
Obtain term tan −1 ( 2 y )
2
Obtain term −e− x B1
Use x = 1, y = 0 to evaluate a constant or as limits in a solution containing M1
terms of the form atan −1 ( by ) and ce ± x
Obtain correct answer in any form A1
1 A1
Obtain final answer y =
2
( )
tan 2e −1 − 2e − x , or equivalent
7
6(b) 1 B1FT The FT is on correct work on a solution containing
State that y approaches
2
( )
tan 2e −1 , or equivalent
e− x .
1
The equation of a curve is x3 + 3xy2 − y3 = 5.
dy x2 + y2
(a) Show that = 2 . [4]
dx y − 2xy
(b) Find the coordinates of the points on the curve where the tangent is parallel to the y-axis. [5]
Mark scheme
7(a) dy B1
State or imply 3 y 2 + 6 xy as derivative of 3xy 2
dx
dy B1
State or imply 3y 2 as derivative of y 3
dx
dy M1 dy
Equate attempted derivative of LHS to zero and solve for Need to see factorised out prior to AG
dx dx
Obtain the given answer correctly A1 AG
4
7(b) Equate denominator to zero *M1
Obtain y = 2x, or equivalent A1
Obtain an equation in x or y DM1
Obtain the point (1, 2) A1
State the point ( 5,0)
3 B1 Alternatively (1.71, 0).
5
ρ
G N F
E
D
B
C
k
j M
O i A
In the diagram, OABCDEFG is a cuboid in which OA = 2 units, OC = 3 units and OD = 2 units.
Unit vectors i, j and k are parallel to OA, OC and OD respectively. The point M on AB is such that
MB = 2 AM . The midpoint of FG is N .
(a) Express the vectors OM and MN in terms of i, j and k. [3]
(b) Find a vector equation for the line through M and N . [2]
(c) Find the position vector of P, the foot of the perpendicular from D to the line through M and N .
[4]
2 + 11x − 10x2
Mark scheme
8(a) Obtain OM = 2i + j B1
ρ
Use a correct method to find MN M1
ρ
Obtain MN = – i + 2j + 2k A1
3
8(b) Use a correct method to form an equation for MN M1
Obtain r = 2i + j + λ(– i + 2j + 2k), or equivalent A1
2
ρ
8(c) Find DP for a point P on MN with parameter λ, e.g. ( 2 − λ ,1 + 2λ , −2 + 2λ ) B1
ρ
Equate scalar product of DP and a direction vector for MN to zero and solve M1
for λ
4 A1
Obtain λ =
9
14 17 8 A1
State that the position vector of P is i + j+ k
9 9 9
4
Let f x = .
1 + 2x 1 − 2x 2 + x
(a) Express f x in partial fractions. [5]
(b) Hence obtain the expansion of f x in ascending powers of x, up to and including the term in x2 .
[5]
Mark scheme
9(a) A B C B1
State or imply the form + +
1 + 2x 1 − 2x 2 + x
Use a correct method for finding a constant M1
Obtain one of A = – 2, B = 1 and C = 4 A1
Obtain a second value A1
Obtain the third value A1
5
9(b) Use correct method to find the first two terms of the expansion of (1 + 2 x ) ,
−1 M1
−1
−1 −1 1
(1 − 2 x ) , ( 2 + x ) or 1 + x
2
Obtain correct unsimplified expansions up to the term in x 2 of each partial A1FT The FT is on A, B and C.
fraction + A1FT
+ A1FT
7 2 A1
Obtain final answer 1 + 5 x − x
2
5
(a) The complex numbers v and w satisfy the equations v + iw = 5 and 1 + 2iv − w = 3i.
Solve the equations for v and w, giving your answers in the form x + iy, where x and y are real.
[6]
(b) (i) On an Argand diagram, sketch the locus of points representing complex numbers z satisfying z − 2 − 3i = 1. [2]
(ii) Calculate the least value of arg z for points on this locus. [2]
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
10(a) Solve for v or w M1
Use i2 =−1 M1
2i 5 + 7i A1
Obtain v = – or w =
1+ i −1 + i
Multiply numerator and denominator by the conjugate of the denominator M1
Obtain v = – 1 – i A1
Obtain w = 1 – 6i A1
6
10(b)(i) Show a circle with centre 2 + 3i B1
Show a circle with radius 1 and centre not at the origin B1
2
10(b)(ii) Carry out a complete method for finding the least value of arg z M1
Obtain answer 40.2° or 0.702 radians A1
2
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/32, February/March 2020 (9709_m20_ms_32), and the question paper (9709_m20_qp_32), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/32 for February/March 2020 is out of 75 marks, with 1 hour 50 minutes to answer it.
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