Mathematics 9709/22 October/November 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The polynomial p x is defined by
p x = ax3 + bx − 10,
where a and b are constants. It is given that x + 2 is a factor of p x and that the remainder is −55 when p x is divided by x + 3.
(a) Find the values of a and b. [5]
(b) Hence factorise p x completely. [3]
Mark scheme
1(a) Substitute x = −2 and equate to zero *M1
Substitute x = −3 and equate to −55 *M1
Obtain −8a − 2b − 10 = 0 and −27a − 3b − 10 = −55 A1 OE
Solve a pair of relevant simultaneous linear equations to find a or b DM1 Dependent at least one M1.
Obtain a = 4 and b = −21 A1
5
1(b) Divide by x + 2 at least as far as the x term M1 Or using identity or inspection.
Obtain 4 x 2 − 8 x − 5 A1
Conclude ( x + 2)(2 x + 1)(2 x − 5) A1 Condone inclusion of = 0.
3
(a) Sketch, on the same diagram, the graphs of y = x + 3 and y = 2x − 1. [2]
(b) Solve the equation x + 3 = 2x − 1. [3]
. .
(c) Find the value of y such that 5 2 + 3 = 2 × 5 2 − 1 . Give your answer correct to 3 significant
1y 1y
figures. [2]
Mark scheme
2(a) Draw V-shaped graph with vertex on positive x-axis *B1 Must be straight lines.
Draw (more or less) correct graph of y = x + 3 with smaller gradient together DB1 And crossing y-axis above y-intercept of first graph.
with a V shaped graph Intersection in the first quadrant may be implied.
2
2(b) Solve x + 3 = 2x − 1 to obtain x = 4 B1
Attempt solution of linear equation where signs of 2x and x are different M1
Obtain x = − 23 A1
Alternative method for question 2(b)
State or imply non-modulus equation ( x + 3)2 = (2 x − 1)2 B1
Attempt solution of 3-term quadratic equation obtained from squaring both M1 Must have B1.
terms.
Obtain − 23 and 4 A1
3
2(c) 1
Apply logarithms and use power law for 5 2 y = k where k > 0 M1 Using their positive root from part (b).
Allow M1 for y = 2log5 4 .
Obtain y = 1.72 A1 AWRT; and no other values.
2
The curve with equation y = 5x − 2 tan 2x has exactly one stationary point in the interval 0 ≤ x < 14 π.
Find the coordinates of this stationary point, giving each coordinate correct to 3 significant figures.
[6]
a+14
Given that Ô dx = ln 2, find the value of the positive constant a.
1
Mark scheme
3 Differentiate to obtain first derivative of form k1 + k2 sec2 2 x M1 Where k1k2 ≠ 0 .
Obtain correct 5 − 4sec2 2 x A1
Equate first derivative of form k1 + k2 sec2 2 x to zero and obtain value for M1 OE Or value for tan 2 2x which must be > 0.
cos 2 2x –1 ⩽ cos 2x ⩽ 1.
2 A1 Or tan 2 x = 1
Obtain cos 2 x = or (decimal) equivalent 2
5
Obtain x = 0.232 A1 AWRT
Obtain y = 0.159 A1 AWRT
6
4 Integrate to obtain form k ln x or k ln(3 x ) M1
Apply limits and obtain A1
1
ln(a + 14) − 13 ln a = ln 2 or 1
ln(3a + 42) − 13 ln(3a ) = ln 2 OE
3 3
Use one relevant logarithm property correctly M1
Apply correct process to obtain equation without logarithms M1 OE. M0 if incorrect logarithm property used.
a + 14 A1
Obtain = 8 or equivalent and hence a = 2
a
5
A curve has equation x2 + 4x cos 3y = 6.
2, 12 π .
1
Find the exact value of the gradient of the normal to the curve at the point [6]
Mark scheme
5 Use product rule to differentiate 4 x cos3 y *M1 k1k2 ≠ 0
dy
to obtain k1 cos3 y + k2 x sin 3 y
dx
dy A1 Allow unsimplified.
Obtain 4cos3 y − 12 x sin 3 y
dx
dy A1 Allow unsimplified.
Obtain 2 x + 4cos3 y − 12 x sin 3 y =0
dx
Substitute x- and y-values to find value of first derivative DM1
dy M1
Apply −1 / to find gradient of normal
dx
Obtain − 32 or − 32 2 or exact equivalent A1
6
(a) By sketching a suitable pair of graphs on the same diagram, show that the equation ln x = 2e−x has exactly one root. [2]
(b) Verify by calculation that the root lies between 1.5 and 1.6. [2]
(c) Show that if a sequence of values given by the iterative formula
−xn xn+1 = e2e converges, then it converges to the root of the equation in part (a). [1]
(d) Use the iterative formula in part (c) to determine the root correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3]
(a) Prove that 4 sin x sin x + 16 π 3 − 3 cos 2x + sin 2x.
Mark scheme
6(a) Draw correct sketch of y = ln x or y = 2e − x B1 y = ln x must extend into 1st and 4th quadrants.
y = 2e − x must extend into 1st and 2nd quadrants.
Draw correct sketch of second curve and indicate one root B1 Point of intersection must be circled/identified or a
statement such that ‘there is only one point of intersection
so one root only’ or similar.
2
6(b) Consider sign of ln x − 2e − x , or equivalent, for 1.5 and 1.6 M1
Obtain −0.04... and 0.06... or equivalents and justify conclusion A1
2
6(c) Replace xn+1 and xn by x and apply logarithms to confirm result B1 AG
Allow if done ‘in reverse’ but xn and xn+1 need to be
seen in the final statement.
1
6(d) Use iteration process correctly at least once M1 Need to see 3 correct values.
Obtain final answer 1.54 A1 Answer required to exactly 3sf
Show sufficient iterations to 5sf to justify answer or show sign change in A1
interval [1.535, 1.545]
3
[3]
5π
(b) Find the exact value of Ó 4 sin x sin x + 16 π dx.
6
[4]
0
(c) Find the smallest positive value of y satisfying the equation
4 sin 2y sin 2y + 16 π = 3.
Give your answer in an exact form. [3]
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) State 4sin x(sin x cos 16 π + cos x sin 16 π) B1 Allow marks if completed correctly, starting with RHS.
Attempt relevant identities to express at least one term in terms of cos2 x and M1
sin 2x
Confirm 3 − 3 cos 2 x + sin 2 x with sufficient detail A1 AG
3
7(b) Integrate to obtain form k1 x + k2 sin 2 x + k3 cos 2 x *M1 Where k1k2 k3 ≠ 0
Obtain correct 3x − 12 3 sin 2 x − 12 cos 2 x A1
Apply limits correctly and attempt simplification, retaining exactness DM1 Condone one sign slip.
Obtain 5 6 3 π + 1 or exact equivalent A1
4
7(c) Apply identity with x replaced by 2 y M1 May be implied later.
Attempt solution of tan 4 y = 3 M1
Obtain 4 y = 13 π and hence y = 121 π A1
3
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/22, October/November 2021 (9709_w21_ms_22), and the question paper (9709_w21_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 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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