Mathematics 9709/22 October/November 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The solutions of the equation 4x − 1 = x + 3 are x = p and x = q, where p < q.
Find the exact values of p and q, and hence determine the exact value of p − 2 − q − 1. [5]
Mark scheme
2 Solve 4x −1 = x + 3 to obtain x = 43 B1
Attempt solution of linear equation where signs of 4x and x are different M1
Obtain final value x = − 52 A1
Substitute numerical values and apply modulus signs correctly to obtain M1 Allow their p and q, p < q.
− 125 − 13 or equivalent, retaining exactness and with no subsequent squaring
31
Obtain 15 A1 or exact equivalent.
Alternative method for Question 2
State or imply non-modulus equation (4 x − 1) 2 = ( x + 3) 2 B1
Attempt solution of 3-term quadratic equation M1
Obtain final values − 52 and 43 A1
2 Substitute numerical values and apply modulus signs correctly to obtain M1 Allow their p < q .
− 125 − 13 or equivalent, retaining exactness and with no subsequent squaring
31
Obtain 15 A1 or exact equivalent.
5
ln y
0.81, 3.47
0.56, 2.87
ln x
O
The variables x and y satisfy the equation y = Axk , where A and k are constants. The graph of ln y against ln x is a straight line passing through the points 0.56, 2.87 and 0.81, 3.47, as shown in the diagram.
Find the value of k, and the value of A correct to 2 significant figures. [5]
Mark scheme
3 State or imply equation is ln y = ln A + k ln x B1
Equate k to gradient of line M1 or eliminate ln A from simultaneous equations using
appropriate values.
Obtain k = 2.4 A1
Substitute appropriate values to find ln A M1
Obtain ln A = 1.526 and hence A = 4.6 A1 AWRT
Alternative method for Question 3
State or imply e3.47 = Ae0.81k and e2.87 = Ae0.56 k B1
Eliminate A M1
Obtain k = 2.4 A1
Substitute appropriate values to find A M1
Obtain A = e1.526 and hence A = 4.6 A1
5
The polynomial p x is defined by
p x = ax3 + 23x2 − ax − 8,
where a is a constant. It is given that 2x + 1 is a factor of p x.
(a) Find the value of a and hence factorise p x completely. [5]
(b) Hence solve the equation p e4y = 0, giving your answer correct to 3 significant figures. [2]
Mark scheme
4(a) Substitute x = − 12 , equate to zero and attempt solution M1
Obtain a = 6 A1
Divide by 2 x + 1 at least as far as the x term M1 or use of identity or by inspection.
Obtain 3x 2 + 10 x − 8 A1
Conclude (2 x + 1)(3x − 2)( x + 4) A1
5
4(b) Apply logarithms and use power law for e 4 y = k where k > 0 M1 Using their answer to (a).
Obtain y = −0.101 and no other answers A1 or greater accuracy.
2
The curve with equation y = x ln 4x + 1 − 3x has one stationary point P.
(a) Show that the x-coordinate of P satisfies the equation
2x + 0.75 x= − 0.25. [4]
ln 4x + 1
(b) Show by calculation that the x-coordinate of P lies between 1.8 and 1.9. [2]
(c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of P correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3]
Mark scheme
5(a) Attempt use of product rule to differentiate x ln(4 x + 1) *M1
4x A1
Obtain ln(4 x + 1) + −3
4x + 1
Equate first derivative to zero and attempt correct rearrangement to obtain DM1 OE
ax + b
the form 4 x + 1 =
ln ( 4 x + 1)
2 x + 0.75 A1 AG – necessary detail needed.
Confirm x = − 0.25
ln(4 x + 1) Allow verification.
4
5(b) 2 x + 0.75 M1
Consider sign of x − + 0.25 or equivalent for 1.8 and 1.9
ln(4 x + 1)
Obtain −0.017... and 0.035... or equivalents and justify conclusion A1 AG – necessary detail needed.
2
5(c) Use iteration process correctly at least once M1
Obtain final answer 1.83 A1 answer required to exactly 3 s.f.
Show sufficient iterations to 5 sf to justify answer or show sign change in the A1
interval [1.825, 1.835]
3
y
x
O 4
6
The diagram shows the curves y = and y = 3e−x − 3 for values of x between 0 and 4. The
3x + 2 shaded region is bounded by the two curves and the lines x = 0 and x = 4.
Find the exact area of the shaded region, giving your answer in the form ln a + b + ced . [9]
Mark scheme
6 6 *M1 for any constant k1 .
Integrate to obtain form k1 ln(3x + 2)
3x + 2
Obtain correct 2ln(3x + 2) A1
Apply limits correctly DM1
Obtain 2ln14 − 2ln2 and hence ln 49 A1 at this stage or later.
Integrate 3e− x − 3 to obtain form k2 e − x + k3 x M1 for any non-zero constants k 2 , k 3 .
Obtain correct −3e− x − 3x A1
Apply limits to obtain −3e−4 − 12 + 3 A1 OE; implied if ( y − y )dx approach used.
1 2
Use correct procedure to find exact total area M1
Obtain ln 49 + 9 + 3e−4 A1
9
y
x
O
P
Q
The diagram shows the curve with parametric equations x = 3 cos 21, y = 4 sin 1, for π ≤ 1 ≤ 32 π. Points P and Q lie on the curve. The gradient of the curve at P is 2. The straight line
3x + y = 0 meets the curve at Q.
(a) Find the value of 1 at P, giving your answer correct to 3 significant figures. [5]
(b) Find the gradient of the curve at Q, giving your answer correct to 3 significant figures. [6]
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) dx dy B1
State = −6sin 2θ and = 4cosθ
dθ dθ
dy dy dx M1
Use = / and equate to 2
dx dθ dθ
Use sin 2θ = 2sinθ cosθ and attempt value of sinθ M1
Obtain sin θ = − 16 A1
7(a) Obtain θ = 3.31 only A1 AWRT; and no second answer.
Alternative method for Question 7(a)
( ) ( ) ( )
Using x = 3 2cos2 θ − 1 , x = 3 1 − 2sin 2 θ or x = 3 cos2 θ − sin 2 θ to M1
dx
obtain = a sin θ cosθ
dθ
dy 4cos θ A1
Obtain = =2
dx −12sin θ cosθ
Attempt value of sinθ M1
Obtain sin θ = − 16 A1
Obtain θ = 3.31 only in the given range A1
5
7(b) State or imply 9cos2θ + 4sinθ = 0 and use identity to obtain quadratic in M1
sinθ
Obtain 18sin 2 θ − 4sin θ − 9 = 0 A1 OE
Attempt solution to find negative value of sinθ DM1
Obtain sinθ = −0.604... A1 2 − 166
Or , θ = 3.79...
18
Substitute value of sinθ (or their θ between π and 32 π ) in expression for M1
first derivative
Obtain 0.551 A1 AWRT
7(b) Alternative method for question 7(b)
y2 x M1 Must be a complete method, allow unsimplified.
Cartesian equation of curve 1 − = oe
8 3
Intersection of line and curve 27 x 2 + 8 x − 24 = 0 oe M1
x = 0.8062... A1
θ = 3.791... A1
Substitute value of θ (or their θ between π and 32 π ) in expression for first M1
derivative
Obtain 0.551 A1 AWRT
6
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/22, October/November 2022 (9709_w22_ms_22), and the question paper (9709_w22_qp_22), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/22 for October/November 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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