Mathematics 9709/22 February/March 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, February/March 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Sketch, on the same diagram, the graphs of y = 3x − 5 and y = x + 2. [2]
(b) Solve the equation 3x − 5 = x + 2. [3]
Mark scheme
1(a) Draw V-shaped graph with vertex on positive x-axis B1
Draw correct graph of y = x + 2 with smaller positive gradient B1 Crossing y-axis between 0 and y-intercept of first
graph.
2
1(b) 7 B1
Solve 3 x − 5 = x + 2 to obtain x =
2
Attempt solution of linear equation where signs of 3x and x are different. M1
3 A1
Obtain x =
4
Alternative method for question 1(b)
State or imply non-modulus equation (3x − 5)2 = ( x + 2)2 B1
Attempt solution of 3-term quadratic equation M1
3 7 A1
Obtain and
4 2
3
Solve the equation sec2 1 cot 1 = 8 for 0 < 1 < π. [5]
Mark scheme
2 1 cosθ B1 OE involving sin θ and cos θ only.
State × =8
cos θ sin θ
2
Attempt use of sin 2θ identity to obtain sin 2θ = k M1
1 A1
Obtain sin 2θ =
4
Use correct process to find two values of θ between 0 and π M1 Allow if working in degrees.
Obtain 0.126 and 1.44 A1 AWRT
Alternative method for question 2
1 + tan 2 θ B1
State =8
tan θ
Attempt solution of 3-term quadratic equation to find values of tan θ M1 OE involving tan θ only.
8 ± 60 A1 OE
Obtain tan θ =
2
Solve tan θ = ... to find two values of θ between 0 and π M1 Allow if working in degrees.
Obtain 0.126 and 1.44 A1 AWRT
5
The parametric equations of a curve are
x = e2t cos 4t, y = 3 sin 2t.
Find the gradient of the curve at the point for which t = 0. [5]
Mark scheme
3 dx M1
Attempt use of product rule to find
dt
Obtain 2e 2t cos 4t − 4e 2t sin 4t A1
dy B1
Obtain = 6cos 2t
dt
dy M1
Divide to find with t = 0 substituted
dx
Obtain 3 A1 CWO
5
y
x
O 1 3
The diagram shows part of the curve with equation y =
5x
4x3 + 1
. The shaded region is bounded by the curve and the lines x = 1, x = 3 and y = 0.
dy
(a) Find and hence find the x-coordinate of the maximum point. [4]
dx
(b) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures. [3]
(c) State, with a reason, whether your answer to part (b) is an over-estimate or under-estimate of the exact area of the shaded region. [1]
_
(a) Given that 2 ln x + 1 + ln x = ln x + 9, show that x =
9 x+2
Mark scheme
4(a) Differentiate using quotient rule (or product rule) M1*
5(4 x 3 + 1) − 60 x 3 A1 OE
Obtain
(4 x 3 + 1) 2
Equate first derivative to zero and attempt solution DM1
1 A1
Obtain x =
2
4
4(b) 5 10 15 B1
Use y values , , or decimal equivalents
5 33 109
Use correct formula, or equivalent, with h = 1 M1
1 20 15 A1
Obtain 1 + + or equivalent and hence 0.87
2 33 109
3
4(c) State over-estimate with reference to top of each trapezium above curve B1 Or clear equivalent.
1
. [3]
_
(b) It is given that the equation x =
9 x+2 has a single root.
Show by calculation that this root lies between 1.5 and 2.0. [2]
(c) Use an iterative formula, based on the equation in part (b), to find the root correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3]
Mark scheme
5(a) Use the power law correctly *M1
Use correct process to obtain equation with no logarithms DM1
9 A1 AG; condone absence of justification for choice of
Confirm x = positive root.
x+2
3
5(b) 9 M1
Consider sign of x − or equivalent for 1.5 and 2
x+2
Obtain −0.1... and 0.5 or equivalents and justify conclusion A1 AG
2
5(c) Use iteration process correctly at least once M1
Obtain final answer 1.58 A1 Final answer required to exactly 3 sf.
Show sufficient iterations to 5 sf to justify answer or show a sign change in A1
interval [1.575, 1.585]
3
The polynomial p x is defined by
p x = x3 + ax + b,
where a and b are constants. It is given that x + 2 is a factor of p x and that the remainder is 5 when p x is divided by x − 3.
(a) Find the values of a and b. [5]
(b) Hence find the exact root of the equation p e2y = 0.
[5]
(a) Express 5 3 cos x + 5 sin x in the form R cos x − !, where R > 0 and 0 < ! < 12 π.
Mark scheme
6(a) Substitute x = − 2 and equate to zero *M1
Substitute x = 3 and equate to 5 *M1
Obtain −8 − 2 a + b = 0 and 27 + 3a + b = 5 or equivalents A1
Solve a pair of relevant linear simultaneous equations for a or b DM1 Dependent at least one M mark.
Obtain a = − 6 and b = −4 A1
5
6(b) Attempt division by x + 2 at least as far as x 2 + kx M1
Obtain x 2 − 2 x − 2 A1
2 + 12 A1
Obtain (at least) the positive root or exact equivalent
2
Equate e 2 y to positive root, apply logarithms and use power law M1
1 2 + 12 1 A1
Obtain ln or ln(1 + 3) or exact equivalent
2 2 2
5
[3]
(b) As x varies, find the least possible value of
4 + 5 3 cos x + 5 sin x,
and determine the corresponding value of x where −π < x < π. [3]
(c) Find Ô d1.
1
5 3 cos 31 + 5 sin 312
[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) State R = 10 B1
Use appropriate trigonometry to find α M1
1 A1
Obtain α = π
6
3
7(b) State −6 B1 FT Following their value of R.
Attempt to find x from their cos( x − α ) = −1 M1
1 5 A1
Obtain x − π = −π and hence − π
6 6
3
7(c) 1 *M1
State integrand of form k1 sec 2 3θ − π
6
1 DM1
Integrate to obtain form k 2 tan 3θ − π
6
1 1 A1
Obtain tan 3θ − π + c
300 6
3
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/22, February/March 2021 (9709_m21_ms_22), and the question paper (9709_m21_qp_22), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/22 for February/March 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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