Mathematics 9709/21 October/November 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, October/November 2024. 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 a 2y = e 3x + k , where a and k are constants.
The graph of y against x is a straight line.
3
(a) Use logarithms to show that the gradient of the straight line is . [1]
2 ln a
(b) Given that the straight line passes through the points (0.4, 0.95) and (3.3, 3.80) , find the values of a and k. [4]
Mark scheme
1(a) 3 B1 AG – necessary detail needed.
State or imply 2 y ln a = 3x + k and conclude that gradient is
2ln a
1
1(b) 3 M1
Equate to gradient of line
2ln a
3 2.85 29 A1 Allow greater accuracy.
Obtain = or equivalent and hence obtain a = 4.6 or a = e 19
2ln a 2.9
Substitute appropriate values to find value of k M1
Obtain k = 1.7 A1
Alternative Method for Question 1(b)
Obtain 0.95 ( 2ln a ) = 3 ( 0.4 ) + k M1 OE
or a1.9 = e1.2+ k
Obtain 3.80 ( 2ln a ) = 3 ( 3.3) + k M1 OE
or a 7.6 = e9.9+ k
29
Obtain a = 4.6 or a = e 19 A1 Allow greater accuracy.
Obtain k = 1.7 A1
4
2 Attempt solution of equation or inequality, where signs of x and 4x are different M1
Obtain 54 … A1 OE
… and finally no other value A1
Conclude x < 54 A1 4
Allow −∞, .
5
Alternative Method for Question 2
State or imply non-modulus equation ( x − 7) 2 = (4 x + 3) 2 or inequality B1
Attempt solution of three-term quadratic equation or inequality M1
Obtain finally 54 only A1
Conclude x < 54 A1 4
Allow −∞,
5
4
The function f is defined by f (x) = tan 2 a 12 xk for 0 G x 1 r .
(a) Find the exact value of f la 23 rk. [3]
1
y `f (x) + sin xj dx .
2r
(b) Find the exact value of [4]
0
Mark scheme
3(a) Differentiate to obtain form k tan 12 x sec2 12 x M1 OE. May use identities before differentiation.
Obtain correct tan 12 x sec2 12 x A1 OE. Allow unsimplified.
Substitute 23 π to obtain 4 3 A1
3
3(b) Express integrand as sec2 12 x − 1 + sin x B1
Integrate to obtain k1 tan 12 x − x + k2 cos x M1 Where k1k2 ≠ 0.
Obtain correct 2 tan 12 x − x − cos x A1
Apply limits correctly to obtain 3 − 12 π or exact equivalent A1
4
The polynomial p (x) is defined by
p (x) = ax 3 - ax 2 - 15x + 18 ,
where a is a constant. It is given that (x + 2) is a factor of p (x) .
(a) Find the value of a. [2]
(b) Hence factorise p (x) completely. [3]
(c) Solve the equation p (cosec 2 i) = 0 for - 90° 1 i 1 90° . [3]
a3
Mark scheme
4(a) Substitute x = −2, equate to zero and attempt solution M1
Obtain a = 4 A1
2
4(b) Divide by x + 2 at least as far as k1 x 2 + k2 x M1
Obtain 4 x 2 − 12 x + 9 A1
Obtain ( x + 2)(2 x − 3) 2 or equivalent with integer coefficients only A1
3
4(c) Equate sin 2 θ to appropriate value from factorised form and attempt solution M1
Using their
2
.
3
Obtain 54.7 A1 Or greater accuracy.
Obtain –54.7 A1 Or greater accuracy.
No others in −90° < θ < 90°.
3
It is given that y 2x10+ 1 dx = 7, where a is a constant greater than 1. a
(a) Show that a = 3 0.5e 1.4 (2a + 1) - 0.5 . [5]
(b) Use an iterative formula, based on the equation in part (a), to find the value of a correct to
3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures. [3]
Mark scheme
5(a) Obtain integral of form k ln(2 x + 1) *M1
Obtain correct 5ln(2 x + 1) A1
Apply limits correctly and equate to 7 DM1
Apply appropriate logarithm property to reach at least a3 = ... DM1
Confirm a = 3 0.5e1.4 (2a + 1) − 0.5 A1 AG – necessary detail needed.
5
5(b) Use iterative process correctly at least once M1
Obtain final answer 2.18 A1 Answer required to exactly 3 sf.
Show sufficient iterations to 5 sf to justify answer or show a sign change in the A1
interval [2.175, 2.185]
3
A curve has parametric equations e 2t - 2 x= , y = e 3t + 1. e 2t + 1 dy
(a) Find an expression for in terms of t. [4]
dx
(b) Find the exact gradient of the curve at the point where the curve crosses the y-axis. [3]
Mark scheme
6(a) Differentiate x using quotient rule or correct equivalent *M1
(e2t + 1) 2e2t − (e2t − 2) 2e2t A1
Obtain or equivalent
(e2t + 1)2
dy DM1
Attempt expression for in terms of t
dx
Obtain 12 et (e2t + 1)2 or (unsimplified) equivalent A1 No fractions within fractions.
Attempt to simplify (e 2t + 1) 2e 2t − (e 2t − 2) 2e 2t
must be seen.
4
6(b) Identify t = 12 ln 2 at point where curve crosses y-axis B1
dy M1
Substitute non-zero value of t in their expression for and attempt simplification
dx
Obtain 92 2 or exact equivalent A1
3
(a) Prove that cos (i + 30°) cos (i + 60°) / 14 3 - 12 sin 2i . [4]
(b) Solve the equation 5 cos (2a + 30°) cos (2a + 60°) = 1 for 0° 1 a 1 90° . [4]
(c) Show that the exact value of cos 20° cos 50° + cos 40° cos 70° is 12 3 . [3]
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
7(a) State (cosθ cos30 − sin θ sin30)(cosθ cos60 − sin θ sin 60) B1
Expand and use correct exact values M1
Obtain 14 3(cos2 θ + sin 2 θ ) − sin θ cosθ or similarly simplified equivalent A1
Conclude 14 3 − 12 sin 2θ A1 AG – necessary detail needed.
4
7(b) Use identity to obtain value for sin 4α *M1
Obtain sin 4α = 12 3 − 52 or 0.466… A1
Show correct process to obtain one value of α DM1
Obtain 6.9 and 38.1 A1 Or greater accuracy; and no others between
0° and 90°.
4
7(c) Substitute θ = −10 to obtain cos20cos50 = 14 3 − 12 sin(−20) B1
3 1 B1
Substitute θ = 10 to obtain cos 40cos 70 = − sin 20
4 2
Add and confirm 12 3 with clear indication that sin(−20) = − sin 20 B1 AG – necessary detail needed.
Alternative solution for Question 7(c)
Rewrite as sin70cos50 + sin50cos70 or cos20sin 40 + cos40sin 20 or B1
sin70sin 40 + cos70cos40
Obtain sin120 or sin 60 or cos30 B1
Confirm 12 3 B1 AG – necessary detail needed.
3
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9709/21 October/November 2024 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/21, October/November 2024 (9709_w24_ms_21), and the question paper (9709_w24_qp_21), to read or download, with each question's marking points written out underneath.
How many marks is 9709/21 out of?+
Mathematics 9709/21 for October/November 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/21 answers marked?+
Yes. Upload your answers to 9709/21/O/N/24 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.