Mathematics 9709/21 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.
1 Solve 2x − 5 = x to obtain x = 5 B1
Attempt solution of linear equation where signs of 2x and x are different M1
Obtain x = 53 A1
Conclude x < 53 , x > 5 A1 Must be 2 separate inequalities.
5
Allow equivalents −∞, ( 5, ∞ ) .
3
Alternative method for question 1
State or imply non-modulus equation (2 x − 5) 2 = x 2 B1
Attempt solution of 3-term quadratic equation M1
Obtain 53 and 5 A1
Conclude x < 53 , x > 5 A1 Must be 2 separate inequalities.
5
Allow equivalents −∞, ( 5, ∞ ) .
3
4
Use logarithms to solve the equation 14e−2x = 5x+1 , giving your answer correct to 3 significant figures.
[4]
It is given that sec 1 = 17 where 0 < 1 < 12 π.
Mark scheme
2 Apply logarithms correctly to both sides and apply power law at least once *M1
Obtain ln14 − 2 x = ( x + 1)ln5 A1 OE with x no longer part of a power.
Attempt solution of linear equation DM1 Must have ln14 − ln 5 = x ( 2 + ln 5) .
Obtain 0.285 A1
4
3 Use identity sec2 θ = 1 + tan 2 θ to find value of tanθ M1
OE using cos θ =
1
and sin θ =
4
values.
17 17
Obtain tan θ = 4 A1 Condone inclusion of tanθ = −4 .
tan θ + tan 14 π M1 OE
State or imply tan(θ + 14 π) = and substitute value of tanθ
1 − tan θ tan 14 π
Obtain − 53 or exact equivalent only A1
4
(a) By sketching a suitable pair of graphs on the same diagram, show that the equation
−1x e 2 = x5 has exactly one real root. [2]
?
5 −1x
(b) Use the iterative formula xn+1 = e 2 n to determine the root correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]
Mark scheme
4(a) −1 x B1 with some curve in second quadrant as well as first.
Draw approximately correct sketch of y = e 2
Draw approximately correct sketch of y = x 5 and confirm one root B1 with some curve in third quadrant as well as first.
Alternative method for question 4(a)
Draw approximately correct sketch of y = 5ln x or y = ln x B1
5
x B1 Must have intersection in the 4th quadrant.
Draw approximately correct sketch of y = − and confirm one root
2
2
4(b) Use iteration process correctly at least once M1
Obtain final answer 0.9128 A1 answer required to exactly 4 s.f.
Show sufficient iterations to 6 s.f. to justify answer or show sign change in A1
the interval [0.91275, 0.91285]
3
A curve has equation 4e2x y + y2 = 21.
Find the gradient of the curve at the point 0, −7. [5]
Mark scheme
5 Use product rule to differentiate 4e 2 x y M1
dy A1
Obtain correct 8e2 x y + 4e2 x
dx
dy dy B1
Obtain 8e 2 x y + 4e 2 x + 2y = 0
dx dx
dy M1 dy
Substitute x = 0 and y = −7 to find value of dependent at least one term involving from
dx dx
implicit differentiation.
Obtain − 28
5
A1 OE
5
The polynomial p x is defined by
p x = 12x3 − 9x2 + 8x − 4.
(a) Find the quotient when p x is divided by 4x − 3 and show that the remainder is 2. [3]
12 @ A
(b) Hence find Ô − 3x2 dx, giving your answer in the form a + ln b. p x
4x − 3
[6]
2
Mark scheme
6(a) Carry out division at least as far as the 3x 2 + kx stage M1 where k should be zero.
Obtain quotient 3x 2 + 2 A1
Confirm remainder is 2 A1 AG – necessary detail needed.
SC If M0A0 obtained then allow B1 if remainder
theorem is used to obtain 2. Must see sufficient detail.
3
6(b) 2 B1 FT FT their quotient.
Identify integrand as 2 +
4x − 3
Integrate to obtain at least k ln(4 x − 3) term *M1
Obtain correct 2x + 12 ln(4x − 3) A1
Apply limits correctly DM1
Use relevant logarithm properties M1 Must be in the form k ln ( 4 x − 3) .
Obtain 20 + ln3 A1
6
y
B
A x
O 3
The diagram shows the curve with equation y =
2 ln x
3x + 1
. The curve crosses the x-axis at the point A and has a maximum point B. The shaded region is bounded by the curve and the lines x = 3 and y = 0.
(a) Find the gradient of the curve at A. [3]
(b) Show by calculation that the x-coordinate of B lies between 3.0 and 3.1. [3]
(c) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places. [3]
Mark scheme
7(a) Differentiate using quotient rule (or product rule) M1
2 A1 OE
(3x + 1) − 6ln x
Obtain x
(3x + 1)2
Substitute x = 1 to obtain 12 A1 OE
3
7(b) Equate numerator of first derivative to zero M1 May be implied.
2 M1 OE
Consider sign of (3x + 1) − 6ln x for 3.0 and 3.1
x
Obtain 0.074… and –0.14… or equivalents and justify conclusion A1 AG – necessary detail needed.
0.00075 and – 0.001275 .
3
7(c) Use y-values [0], 72 ln 2 or 0.1980 and 10
2
ln 3 or 0.2197 B1
Use correct formula, or equivalent, with h = 1 M1
Obtain 0.31 A1
3
The expression f 1 is defined by f 1 = 12 sin 1 cos 1 + 16 cos2 1.
(a) Express f 1 in the form R cos 21 − ! + k, where R > 0, 0 < ! < 12 π and k is a constant. State the values of R and k, and give the value of ! correct to 4 significant figures. [5]
(b) Find the smallest positive value of 1 satisfying the equation f 1 = 17. [3]
(c) Find Ó f 1 d1. [2]
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
8(a) Use correct identity for sin 2θ or cos2θ (or both) M1
Obtain 6sin 2θ + 8cos 2θ ( +8) A1
State R = 10 B1 FT FT their a sin 2θ + b cos2θ form.
Use appropriate trigonometry to find α using their a sin 2θ + b cos2θ form M1 Allow 0.927 .
Obtain 10cos(2θ − 0.6435) + 8 A1
5
8(b) State or imply cos(2θ − 0.6435) = 0.9 B1 FT FT their R, α and k provided RHS less than 1.
Carry out correct process to find any positive value of θ M1 0.547
Obtain 0.0962 A1 AWRT
3
8(c) Integrate trigonometry term from part (a) to obtain k1 sin(2θ − their 0.6435) M1 any non-zero constant k1 .
Obtain 5sin(2θ − 0.6435) + 8θ A1 condone absence of ... + c .
Alternative method for question 8(c)
Integrate to obtain at least form k2 cos 2θ + k3 sin 2θ M1 any non-zero constants k2 , k3 .
Obtain −3cos2θ + 4sin 2θ + 8θ A1 condone absence of ... + c .
2
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/21, October/November 2022 (9709_w22_ms_21), and the question paper (9709_w22_qp_21), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/21 for October/November 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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