Mathematics 9709/33 October/November 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 3, October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
1 State or imply non-modular inequality ( x + 2 ) > ( 3x − 1) , or corresponding B1
2 2
quadratic equation, or pair of linear equations 2(x + 2) = ± (3x – 1)
Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations M1
for x
3 A1
Obtain critical values x = − and x = 5
5
3 A1
State final answer − < x < 5
5
Alternative method for question 1
Obtain critical value x = 5 from a graphical method, or by inspection, or by solving B1
a linear equation or an inequality
3 B2
Obtain critical value x = − similarly
5
3 B1
State final answer − < x < 5
5
4
The polynomial 6x3 + ax2 + bx − 2, where a and b are constants, is denoted by p x. It is given that
2x + 1 is a factor of p x and that when p x is divided by x + 2 the remainder is −24. Find the values of a and b. [5]
32x + 3−x
Mark scheme
2 1 B1
Substitute x = − , equate result to zero and obtain a correct equation, e.g.
2
6 1 1
− + a− b−2=0
8 4 2
Substitute x = – 2 and equate result to – 24 *M1
Obtain a correct equation, e.g. −48 + 4a − 2b − 2 = −24 A1
Solve for a or for b DM1
Obtain a = 5 and b = – 3 A1
5
Showing all necessary working, solve the equation = 4. Give your answer correct to
32x − 3−x
3 decimal places. [4]
Mark scheme
3 Reduce the equation to a horizontal equation in 33 x , 33x+1 or 27 x M1
( )
Simplify and reach 3( 33 x ) = 5 , 3 27 x = 5 , or equivalent A1
Use correct method for finding x from a positive value of 33 x , 33 x +1 or 27 x M1
Obtain answer x = 0.155 A1
4
(i) By first expanding tan 2x + x, show that the equation tan 3x = 3 cot x can be written in the form tan4 x − 12 tan2 x + 3 = 0. [4]
(ii) Hence solve the equation tan 3x = 3 cot x for 0Å < x < 90Å. [3]
Mark scheme
4(i) Use tan (A + B) formula to express the LHS in terms of tan 2x and tan x M1
Using the tan 2A formula, express the entire equation in terms of tan x M1
Obtain a correct equation in tan x in any form A1
Obtain the given form correctly A1 AG
4
4(ii) Use correct method to solve the given equation for x M1
Obtain answer, e.g. x = 26.8 ° A1
Obtain second answer, e.g. x = 73.7 ° and no other A1 Ignore answers outside the given interval
3
(i) By sketching a suitable pair of graphs, show that the equation ln x + 2 = 4e−x has exactly one real root. [2]
(ii) Show by calculation that this root lies between x = 1 and x = 1.5. [2]
@ A
4
(iii) Use the iterative formula xn+1 = ln to determine the root correct to 2 decimal places. ln xn + 2
Give the result of each iteration to 4 decimal places. [3]
Mark scheme
5(i) Sketch a relevant graph, e.g. y = ln ( x + 2 ) B1
Sketch a second relevant graph, e.g. y = 4e − x , and justify the given statement B1 Consideration of behaviour for x < 0 is
needed for the second B1
2
5(ii) Calculate the values of a relevant expression or pair of expressions at x = 1 and x = 1.5 M1
Complete the argument correctly with correct calculated values A1
2
5(iii) Use the iterative formula correctly at least twice using output from a previous M1
iteration
Obtain final answer 1.23 A1
Show sufficient iterations to 4 d.p. to justify 1.23 to 2 d.p., or show there is a sign A1
change in the interval (1.225, 1.235)
3
Throughout this question the use of a calculator is not permitted.
The complex number with modulus 1 and argument 13 0 is denoted by w.
(i) Express w in the form x + iy, where x and y are real and exact. [1]
The complex number 1 + 2i is denoted by u. The complex number v is such that v = 2 u and arg v = arg u + 13 0 .
(ii) Sketch an Argand diagram showing the points representing u and v. [2]
(iii) Explain why v can be expressed as 2uw. Hence find v, giving your answer in the form a + ib, where a and b are real and exact. [4]
Mark scheme
6(i) 1 3 B1
Obtain answer w = + i
2 2
1
6(ii) Show point representing u B1
Show point representing v in relatively correct position B1
2
6(iii) Explain why the moduli are equal B1
Explain why the arguments are equal B1
Use i 2 = −1 and obtain 2uw in the given form M1
(
Obtain answer 1 − 2 3 + 2 + 3 i ) A1
4
The plane m has equation x + 4y − 8z = 2. The plane n is parallel to m and passes through the point
P with coordinates 5, 2, −2.
(i) Find the equation of n, giving your answer in the form ax + by + cz = d . [2]
(ii) Calculate the perpendicular distance between m and n. [3]
(iii) The line l lies in the plane n, passes through the point P and is perpendicular to OP, where O is the origin. Find a vector equation for l. [4]
Mark scheme
7(i) Substitute coordinates (5, 2, – 2) in x + 4y – 8z = d M1
Obtain plane equation x + 4y – 8z = 29, or equivalent A1
2
7(ii) Attempt to use perpendicular formula to find perpendicular from (5, 2, – 2) to m M1
5 + 8 + 16 − 2 A1
Obtain a correct unsimplified expression, e.g.
(1 + 16 + 64 )
Obtain answer 3 A1
Alternative method 1 for question 7(ii)
2 29 B1
State or imply perpendicular from O to m is or from O to n is
9 9
Find difference in perpendiculars M1
Obtain answer 3 A1
Alternative method 2 for question 7(ii)
Obtain correct parameter value, or position vector or coordinates of the foot of the B1
1 14 2 2
perpendicular from (5, 2, – 2) to m, e.g. µ = ± ; , ,
3 3 3 3
Calculate the length of the perpendicular M1
Obtain answer 3 B1
3
7(iii) Calling the direction vector ai + bj + ck, use a scalar product to form a relevant B1
equation in a, b and c, e.g. a + 4b – 8c = 0 or 5a + 2b – 2z = 0
Solve two relevant equations for the ratio a : b : c M1
Obtain a : b : c = 4 : – 19: – 9 A1 OE
State answer r = 5i + 2j – 2k + λ(4i – 19j – 9k) A1 OE
Alternative method for question 7(iii)
Attempt to calculate vector product of two relevant vectors, e.g. (i + 4j – 8k)×(5i + M1
2j – 2k)
Obtain two correct components A1
Obtain 8i – 38j – 18k A1 OE
State answer r = 5i + 2j – 2k + λ(4i – 19j – 9k) A1 OE
4
y
x
O 1
20
The diagram shows the graph of y = sec x for 0 ≤ x < 12 0.
1.2
(i) Use the trapezium rule with 2 intervals to estimate the value of Ó sec x dx, giving your answer
0 correct to 2 decimal places. [3]
(ii) Explain, with reference to the diagram, whether the trapezium rule gives an overestimate or an underestimate of the true value of the integral in part (i). [1]
(iii) P is the point on the part of the curve y = sec x for 0 ≤ x < 12 0 at which the gradient is 2. By first
1 differentiating , find the x-coordinate of P, giving your answer correct to 3 decimal places. cos x
[6]
dx
Mark scheme
8(i) State or imply ordinates 1, 1.2116…, 2.7597... B1
Use correct formula, or equivalent, with h = 0.6 M1
Obtain answer 1.85 A1
3
8(ii) Explain why the rule gives an overestimate B1
1
8(iii) Differentiate using quotient or chain rule M1
Obtain correct derivative in terms of sin x and cos x A1
Equate derivative to 2, use Pythagoras and obtain an equation in sin x M1
2
Obtain 2sin x + sin x − 2 = 0 A1 OE
Solve a 3-term quadratic for x M1
Obtain answer x = 0.896 only A1
6
The variables x and t satisfy the differential equation 5 = 20 − x 40 − x. It is given that x = 10 dt when t = 0.
(i) Using partial fractions, solve the differential equation, obtaining an expression for x in terms of t.
[9]
(ii) State what happens to the value of x when t becomes large. [1]
Mark scheme
9(i) Separate variables correctly and integrate one side B1
Obtain term 0.2t, or equivalent B1
Carry out a relevant method to obtain A and B such that *M1 OE
1 A B
≡ +
( 20 − x )( 40 − x ) 20 − x 40 − x
1 1 A1
Obtain A = and B = −
20 20
1 1 A1FT The FT is on A and B
Integrate and obtain terms − ln ( 20 − x ) + ln ( 40 − x ) OE +A1FT
20 20
Use x = 10, t = 0 to evaluate a constant, or as limits DM1
Obtain correct answer in any form A1
60e 4t − 40 A1 OE
Obtain final answer x =
3e 4t − 1
9
9(ii) State that x approaches 20 B1
1
y
M
R
x
O 0
The diagram shows the graph of y = ecos x sin3 x for 0 ≤ x ≤ 0, and its maximum point M . The shaded region R is bounded by the curve and the x-axis.
(i) Find the x-coordinate of M . Show all necessary working and give your answer correct to 2 decimal places. [5]
(ii) By first using the substitution u = cos x, find the exact value of the area of R. [7]
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(i) Use product rule and chain rule at least once M1
Obtain correct derivative in any form A1
Equate derivative to zero, use Pythagoras and obtain an equation in cos x M1
Obtain cos2 x + 3cos x − 1 = 0 , or 3-term equivalent A1
Obtain answer x = 1.26 A1
5
10(ii) Using du = ± sin x dx express integrand in terms of u and du M1
(
Obtain integrand eu u 2 − 1 ) A1 OE
( )
Commence integration by parts and reach aeu u 2 − 1 + b ∫ ueu du *M1
( )
Obtain eu u 2 − 1 − 2 ∫ ueu du A1 OE
(
Complete integration, obtaining eu u 2 − 2u + 1 ) A1 OE
Substitute limits u = 1 and u = – 1 (or x = 0 and x = π ), having integrated DM1
completely
4 A1
Obtain answer , or exact equivalent
e
7
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Where can I find the 9709/33 October/November 2019 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/33, October/November 2019 (9709_w19_ms_33), and the question paper (9709_w19_qp_33), to read or download, with each question's marking points written out underneath.
How many marks is 9709/33 out of?+
Mathematics 9709/33 for October/November 2019 is out of 75 marks, with 1 hour 45 minutes to answer it.
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