Mathematics 9709/21 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
1 State or imply non-modulus inequality (3x − 7) 2 < (4 x + 5) 2 or B1
corresponding equation or pair of linear equations
Attempt solution of 3-term quadratic equation/inequality or of two M1
linear equations
2 A1 May be seen in a number line.
Obtain critical values −12 and
7
2 A1 OE
State answer x < −12, x >
7 2
−12 > x > or similar would get A0
or 7
Mark the final answer.
( −∞. − 12 ) ∪ , ∞ or ( −∞. − 12 ) , , ∞
2 2
7 7
4
By first expanding sin 1 + 30Å, solve the equation sin 1 + 30Å cosec 1 = 2 for 0Å < 1 < 360Å. [6]
Mark scheme
2 Express sin(θ + 30) as sin θ cos 30 + cos θ sin 30 B1
1 B1 SOI
Use cosecθ =
sin θ
Correctly obtain a linear equation in tan θ or cot θ M1 Allow unsimplified
1 4+ 3 1 A1 OE
Obtain tan θ = , , or 0.440… May be implied by a correct answer.
4− 3 13 2.26795..
Obtain 23.8 A1 AWRT, e.g. 23.793…
Obtain 203.8 A1 FT AWRT; following their first value and no other solutions within
the range.
6
(a) Show that sec x + cos x2 can be expressed as sec2 x + a + b cos 2x, where a and b are constants to be determined. [2]
1π
(b) Hence find the exact value of Ó sec x + cos x2 dx.
4
[4]
0
Mark scheme
3(a) Expand to obtain integrand of form sec 2 x + k1 + k2 cos 2 x M1 With k1k2 ≠ 0 and k1 ≠ 2
5 1 A1 OE, allow unsimplified.
Obtain correct sec2 x + + cos 2 x
2 2
2
3(b) Integrate to obtain at least terms of form k3 tan x and k4 sin 2 x *M1 With k3k4 ≠ 0
Allow in terms of a and b
Condone omission of term in x
5 1 A1 OE, allow unsimplified.
Obtain correct tan x + x + sin 2 x
2 4
Apply limits correctly to integral involving at least two terms DM1 Need to see at least 2 correct substitutions using their integral.
5 5 A1 CWO
Obtain + π or exact equivalent
4 8
4
A curve has parametric equations x = ln 2t + 6 − ln t, y = t ln t.
(a) Find the value of t at the point P on the curve for which x = ln 4. [3]
(b) Find the exact gradient of the curve at P. [5]
Mark scheme
4(a) Equate x to ln 4 and use relevant logarithm property M1
2t + 6 A1 OE
Obtain equation with no logarithm present, =4
t
Obtain t = 3 A1
3
4(b) dx 2 1 B1
Obtain = −
dt 2t + 6 t
dy M1
Use product rule to find
dt
1 A1
Obtain ln t + t ×
t
dy dy dx DM1 Must have at least B1 or M1.
Divide to obtain using their and correctly Do not condone incorrect inverting of terms unless a correct
dx dt dt
statement is seen initially.
Obtain −6(ln 3 + 1) A1 or exact equivalents
5
y
M
x
O
3x + 2
The diagram shows the curve with equation y = . The curve has a minimum point M . ln x
3x + 2 and show that the x-coordinate of M satisfies the equation x = dy
(a) Find an expression for . dx 3 ln x
[3]
(b) Use the equation in part (a) to show by calculation that the x-coordinate of M lies between
3 and 4. [2]
(c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct to 5 significant figures. Give the result of each iteration to 7 significant figures. [3]
4
(a) Use the trapezium rule with three intervals to find an approximation to Ô
6
1+ x
Mark scheme
5(a) Use quotient rule (or equivalent) to find first derivative M1
1 A1 OE
3ln x − (3x + 2)
dy x
Obtain =
dx (ln x) 2
3x + 2 A1 AG
Equate first derivative to zero and confirm x =
3ln x
3
5(b) 3x + 2 M1 dy
Consider x − or equivalent for values 3 and 4 M0 if using
3ln x dx
Obtain −0.33... and 0.63... or equivalents and justify conclusion A1 AG
2
5(c) Use iteration process correctly at least once M1
Obtain final answer 3.3223 A1 Answer required to exactly 5 s.f.
Show sufficient iterations to 7 s.f. to justify answer or show sign A1
change in the interval [3.32225, 3.32235]
3
dx. Give your
1 answer correct to 5 significant figures. [3]
1 x−2
(b) Find the exact value of Ó 2e 2
4 dx. [3]
1
(c) y
1 y = 2e 2 x−2
6 y=
1+ x
x
O 1 4
1 x−2
The diagram shows the curves y = and y = 2e 2 which meet at a point with x-coordinate 4.
6
1+ x
The shaded region is bounded by the two curves and the line x = 1.
Use your answers to parts (a) and (b) to find an approximation to the area of the shaded region.
Give your answer correct to 3 significant figures. [2]
(d) State, with a reason, whether your answer to part (c) is an over-estimate or under-estimate of the exact area of the shaded region. [1]
Mark scheme
6(a) 6 6 B1
Use y-values 3, , , 2
1+ 2 1+ 3
or 3, 6 2 − 6, 3 3 − 3, 2
or 3, 2.4853..., 2.19615..., 2
Use correct formula, OE, with h = 1 M1 Allow 3 separate trapezia of width 1.
Obtain 7.1814 A1 AWRT
3
6(b) 1
x−2 M1 k ≠1
Integrate to obtain form ke 2 If k = 2 , integration must be implied by use of square bracket
notation or by substitution of limits.
1
x −2 A1 or exact equivalents
Obtain correct 4e 2
−
3 A1 or exact equivalents
Obtain 4 − 4e 2
3
6(c) Evaluate answer to part (a) minus answer to part (b) M1
Obtain 4.07 A1
2
6(d) State over-estimate with reference to top of each trapezium being B1
above the [first] curve, or clear equivalent, e.g. concave up so over-
estimate or convex down so over-estimate.
1
The polynomial p x is defined by
p x = ax3 − 11x2 − 19x − a,
where a is a constant. It is given that x − 3 is a factor of p x.
(a) Find the value of a. [2]
(b) When a has this value, factorise p x completely. [3]
(c) Hence find the exact values of y that satisfy the equation p ey + e−y = 0.
[4]
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) Substitute x = 3 , equate to zero and attempt solution M1
Obtain a = 6 A1
2
7(b) Divide by x − 3 at least as far as the x term M1 Or use of identity or by inspection
Obtain 6 x 2 + 7 x + 2 A1
Conclude ( x − 3)(3x + 2)(2 x + 1) A1
3
7(c) Equate e y + e− y to positive value resulting from part (b) B1 Condone if seen with equating to negative values as well.
Multiply by e y and use quadratic formula M1
3± 5 A1 or exact equivalents
Obtain e y =
2
3± 5 A1 or exact equivalents
Obtain ln and no other values which are undefined.
2
4
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/21, May/June 2021 (9709_s21_ms_21), and the question paper (9709_s21_qp_21), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/21 for May/June 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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