Mathematics 9709/22 May/June 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
1 Solve 5x + 7 = 2 x − 3 to obtain − 103 B1 Or inequality.
Attempt solution of linear equation where 5x and 2x have M1 Or inequality.
different signs
Obtain − 74 A1
State x < − 103 , x > − 74 A1 A0 if ‘… and …’ used.
Alternative Method for Question 1
State or imply non-modulus equation (5 x + 7) 2 = (2 x − 3) 2 (B1) Or inequality.
Attempt solution of three-term quadratic equation (M1) Or inequality.
Obtain − 103 and − 74 (A1)
State x < − 103 , x > − 74 (A1) A0 if ‘… and …’ used.
4
y
A
B
O x
The diagram shows the curve with equation y = 8e -x - e 2x . The curve crosses the y-axis at the point A and the x-axis at the point B. The shaded region is bounded by the curve and the two axes.
(a) Find the gradient of the curve at A. [3]
(b) Show that the x-coordinate of B is ln 2 and hence find the area of the shaded region. [5]
Mark scheme
3(a) Differentiate to obtain form k1e − x + k2 e 2 x M1 Where k1k 2 ≠ 0 , k1 ≠ 8 and k 2 ≠ −1 .
Obtain −8e − x − 2e 2 x A1
Substitute x = 0 to obtain –10 A1
3
3(b) Attempt to find x-coordinate of B M1 8e − x − e 2 x = 0 .
Obtain e3 x = 8 and hence x = ln 2 A1 AG so necessary detail needed.
A0 if decimals used.
Integrate to obtain −8e − x − 12 e 2 x B1
Use limits 0 and ln 2 correctly to find area M1 For integral of form k e − x + k e 2 x where k k ≠ 0 .
3 4 3 4
k1 ≠ 8 and k 2 ≠ −1 .
Obtain 52 A1 OE
5
A curve is defined by the parametric equations
x = 4 cos 2 t , y = 3 sin 2t ,
for values of t such that 0 1 t 1 12 r .
Find the equation of the normal to the curve at the point for which t = 16 r . Give your answer in the form ax + by + c = 0 where a, b and c are integers. [7]
Mark scheme
4 dx dy *M1
Obtain forms = k1 cos t sin t or = k2 cos 2t
dt dt
Obtain correct −8cos t sin t or −4sin 2t and 2 3 cos 2t A1
dy *DM1 Need to see attempt at substitution.
Attempt value of when t = 16 π
dx
dy 1 A1
Obtain =−
dx 2
State or imply gradient of normal is 2 **M1FT Following their value of the first derivative.
Attempt equation of normal **DM1 3
Not tangent and with attempt to find coordinates 3, .
2
Obtain 4 x − 2 y − 9 = 0 A1 Or equivalent of requested form.
7
The polynomial p (x) is defined by p (x) = 9x 3 + 18x 2 + 5x + 4 .
(a) Find the quotient when p (x) is divided by (3x + 2) , and show that the remainder is 6. [3]
y 3px(+x)2 dx , giving your answer in the form a + ln b where a and b are integers.
2
(b) Find the value of
0 [5]
Mark scheme
5(a) Carry out division at least as far as 3x 2 + k1 x M1 Or equivalent (inspection, …).
Obtain quotient 3 x 2 + 4 x − 1 A1
Confirm remainder is 6 A1 Answer given – necessary detail needed.
SC B1 for use of the factor theorem to show remainder is 6 if no
other marks are awarded.
Alternative Method for Question 5(a)
Synthetic division (M1)
–2/3 9 18 5 4
–6 8 –2
9 12 –3 6
Obtain quotient 3 x 2 + 4 x − 1 (A1)
Confirm remainder is 6 (A1)
3
5(b) 6 B1FT Following their quotient.
Identify integrand as 3x 2 + 4 x − 1 +
3x + 2
Integrate to obtain at least x 3 and k2 ln(3 x + 2) terms *M1
Obtain x 3 + 2 x 2 − x + 2ln(3 x + 2) A1
Apply limits and appropriate logarithm properties DM1
Obtain 14 + ln16 A1
5
y
M
O x
ln (2x + 1)
The diagram shows the curve with equation y = . The curve has a maximum point M. x+3 dy
(a) Find an expression for . [2]
dx
x+3
(b) Show that the x-coordinate of M satisfies the equation x = - 0.5 . [2]
ln (2x + 1)
(c) Show by calculation that the x-coordinate of M lies between 2.5 and 3.0 . [2]
(d) Use an iterative formula based on the equation in part (b) to find the x-coordinate of M correct to
4 significant figures. Give the result of each iteration to 6 significant figures. [3]
Mark scheme
6(a) Attempt use of quotient rule M1 Or equivalent method.
( x + 3) 2 x2+1 − ln(2 x + 1) A1 OE
Obtain 2
( x + 3)
2
6(b) Equate first derivative to zero and arrange as far as 2 x + 1 = ... M1
x+3 A1 Answer given – necessary detail needed.
Confirm x = − 0.5
ln(2 x + 1)
2
6(c) x+3 M1
Consider sign of x − + 0.5 or equivalent for 2.5 and 3.0
ln(2 x + 1)
Obtain −0.07 ( 0.0696...) and 0.4 ( 0.4166...) and justify A1 Answer given – necessary detail needed.
conclusion
2
Alternative Method for Question 6(c)
x+3 (M1)
Consider the values of f ( x ) = − 0.5 and obtain
ln ( 2 x + 1)
f ( 2.5 ) = 2.57 (2.5696…) and f ( 3) = 2.58 (2.58339…)
Conclude f ( 2.5 ) < 3 and f ( 3) > 2.5 so root lies in given interval (A1) Answer given – necessary detail needed.
2
6(d) Use iterative process correctly at least once M1
Obtain final answer 2.569 A1 Answer required to exactly 4sf.
Show sufficient iterations to 6 sf to justify answer or show sign A1
change in interval [2.5685, 2.5695]
3
(a) Prove that 2 sin i cosec 2i / sec i . [2]
(b) Solve the equation tan 2 i + 7 sin i cosec 2i = 8 for - r 1 i 1 r . [5]
(c) Find y 8 sin x cosec x dx .
21
2
2
[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) Express left-hand side in terms of sin θ and cosθ using M1
1
cosec 2θ =
sin 2θ
1 A1 Answer given – necessary detail needed.
Obtain and confirm secθ
cosθ
2
7(b) Attempt to obtain quadratic equation in secθ or cosθ only *M1
Obtain sec 2 θ − 1 + 72 secθ = 8 involving one trigonometric ratio A1 Or equivalent, may be unsimplified, but reduce to
2sec 2 θ + 7 sec θ − 18 = 0
18cos 2 θ − 7 cos θ − 2 = 0 .
Attempt to solve 3-term quadratic equation for secθ , using a DM1 Or equivalent using cosθ .
correct method, to find at least one value of θ
Obtain any two of the four correct solutions ±0.952, ± 1.76 A1 Or greater accuracy.
Obtain remaining two correct solutions A1 Or greater accuracy; and no others between − π and π .
5
7(c) Identify integrand as 2sec 2 12 x B1
Integrate k sec 2 12 x to obtain 2k tan 12 x M1
Obtain correct 4 tan 12 x A1 Condone omission of ... + c .
3
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Mathematics 9709/22 for May/June 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
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