Mathematics 9709/23 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
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Mathematics 9709/23 for October/November 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
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