Mathematics 9709/22 February/March 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics, February/March 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Solve the equation sec2 1 + tan2 1 = 5 tan 1 + 4 for 0Å < 1 < 180Å. Show all necessary working. [4]
Mark scheme
1 Use identity sec 2 θ = 1 + tan 2 θ B1
Attempt solution of quadratic equation to find two values of tan θ M1
Obtain tan θ = − 12 , 3 A1
Obtain 71.6 and 153.4 and no others between 0 and 180 A1
4
Given that x satisfies the equation 2x + 3 = 2x − 1, find the value of
4x − 3 − 6x . [4]
Mark scheme
2 Solve non-modular equation (2 x + 3) 2 = (2 x − 1) 2 or linear equation with signs of 2x different M1
Obtain x = − 12 A1
Substitute negative value into expression and show correct evaluation of modulus at least once M1
Obtain 5 − 3 = 2 with no errors seen A1
4
ln y
6, 6.585
1, 2.835
x
O
The variables x and y satisfy the equation y = Ae px+p , where A and p are constants. The graph of ln y against x is a straight line passing through the points 1, 2.835 and 6, 6.585, as shown in the diagram. Find the values of A and p. [5]
Mark scheme
3 State or imply equation is ln y = ln A + px + p B1
Equate gradient of line to p M1
Obtain p = 0.75 A1
Substitute appropriate values to find ln A M1
Obtain ln A = 1.335... and hence A = 3.8 A1
5
(i) Find the quotient when 4x3 + 8x2 + 11x + 9 is divided by 2x + 1, and show that the remainder is 5. [3]
(ii) Show that the equation 4x3 + 8x2 + 11x + 4 = 0 has exactly one real root. [3]
The equation of a curve is y = e 2x
4x + 1
Mark scheme
4(i) Carry out division at least as far as 2x 2 + kx M1
Obtain quotient 2 x 2 + 3 x + 4 A1
Confirm remainder is 5 A1 Answer given; necessary
detail needed
3
4(ii) State or imply equation is (2 x + 1)(2 x 2 + 3x + 4) = 0 B1 FT their quotient from part
(i)
Calculate discriminant of 3-term quadratic expression or equivalent M1
Obtain –23 or equiv and conclude appropriately A1
3
and the point P on the curve has y-coordinate 10.
(i) Show that the x-coordinate of P satisfies the equation x = 12 ln 40x + 10. [2]
(ii) Use the iterative formula xn+1 = 12 ln 40xn + 10 with x1 = 2.3 to find the x-coordinate of P correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]
(iii) Find the gradient of the curve at P, giving the answer correct to 3 significant figures. [4]
4@ A
(a) Show that Ô + dx = ln 48.
2 2 x 2x + 1
Mark scheme
5(i) e2 x M1
Attempt rearrangement of = 10 to x = ... involving ln
4x + 1
Confirm x = 12 ln(40 x + 10) A1 Answer given; necessary
detail needed
2
5(ii) Use iteration process correctly at least once M1
Obtain final answer 2.316 A1
Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval [2.3155, 2.3165] A1
3
5(iii) Use quotient rule (or product rule) to find derivative M1
2e 2 x (4 x + 1) − 4e 2 x A1
Obtain or equivalent
(4 x + 1)2
Substitute answer from part (ii) (or more accurate value) into attempt at first derivative M1
Obtain 16.1 A1
4
6(a) Integrate to obtain form k1 ln x + k 2 ln(2 x + 1) M1
Obtain correct 2ln x + ln(2 x + 1) A1
Use logarithm addition/subtraction property correctly M1
Use logarithm power property correctly M1
Confirm ln 48 with no errors seen A1 Answer given; necessary
detail needed
5
6(b) Use identity sin 2 x = 2sin x cos x B1
cos x 2 B1
State or imply cot x + 2cosec x = +
sin x sin x
Attempt to express integrand in terms of cos 2x and cos x M1
Obtain correct integrand 1 + cos 2 x + 4 cos x A1
Integrate to obtain at least terms k3 sin 2 x and k 4 sin x M1
Obtain correct x + 12 sin 2 x + 4sin x + c A1
6
The parametric equations of a curve are x = 2t − sin 2t, y = 5t + cos 2t, for 0 ≤ t ≤ 12 0. At the point P on the curve, the gradient of the curve is 2.
(i) Show that the value of the parameter at P satisfies the equation 2 sin 2t − 4 cos 2t = 1. [4]
(ii) By first expressing 2 sin 2t − 4 cos 2t in the form R sin 2t − !, where R > 0 and 0 < ! < 12 0, find the coordinates of P. Give each coordinate correct to 3 significant figures. [7]
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(i) Obtain ddxt = 2 − 2cos 2t B1
Obtain ddyt = 5 − 2sin 2t B1
Equate attempt at ddyx to 2 and rearrange M1
Confirm equation 2sin 2t − 4cos 2t = 1 A1 Answer given; necessary
detail needed
4
7(ii) State R = 20 or 4.47 B1
Use appropriate trigonometry to find α M1
Obtain α = 1.107 with no errors seen A1
Carry out correct method to find value of t M1
Obtain t = 0.666 A1
Substitute value of t between 0 and 12 π into expressions for x and y M1
Obtain x = 0.361, y = 3.57 A1
7
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/22, February/March 2019 (9709_m19_ms_22), and the question paper (9709_m19_qp_22), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/22 for February/March 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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