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Cambridge AS & A Level Mathematics (9709), Paper 1, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s19_qp_119709_s19_ms_119709/11/M/J/19
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
The term independent of x in the expansion of 2x + , where k is a constant, is 540. x (i) Find the value of k. [3] (ii) For this value of k, find the coefficient of x2 in the expansion. [2]
Mark scheme
1(i) k 3 B2,1,0 Term must be isolated Ind term = ( 2 x ) × × 6C3 3 x = 540 → k = 1½ B1 3 1(ii) k 2 B1 All correct – even if k incorrect. Term, in x² is ( 2 x ) × × 6C2 4 x 15 × 16 × k² = 540 (or 540 x 2 ) B1 FT For 240k² or 240 k 2 x 2 2
Question
The line 4y = x + c, where c is a constant, is a tangent to the curve y2 = x + 3 at the point P on the curve. (i) Find the value of c. [3] (ii) Find the coordinates of P. [2]
Mark scheme
2(i) Eliminates x or y → y ² − 4 y + c − 3 = 0 M1 Eliminates x or y completely to a quadratic or x² + ( 2c − 16) x + c² − 48 = 0 Uses b ² = 4 ac → 4c – 28 = 0 M1 Uses discriminant = 0. (c the only variable) Any valid method (may be seen in part (i)) c=7 A1 Alternative method for question 2(i) dy 1 1 M1 = = dx 2 ( x + 3) 4 Solving M1 c=7 A1 3 2(ii) Uses c = 7, y² − 4y + 4 = 0 M1 Ignore (1,–2), c=-9 (1, 2) A1 2
Question
A sector of a circle of radius r cm has an area of A cm2. Express the perimeter of the sector in terms of r and A. [4]
Mark scheme
3 Uses A = ½r²θ M1 Uses area formula. 2A A1 θ= r² P = r + r + rθ B1 2A A1 Correct simplified expression for P. P = 2r + r 4
Question
y C h, 3h D B 0, 2 x O A 4, 0 The diagram shows a trapezium ABCD in which the coordinates of A, B and C are 4, 0, 0, 2 and h, 3h respectively. The lines BC and AD are parallel, angle ABC = 90Å and CD is parallel to the x-axis. (i) Find, by calculation, the value of h. [3] (ii) Hence find the coordinates of D. [3]
Mark scheme
4(i) Gradient of AB = −½ → Gradient of BC = 2 M1 Use of m1.m2 = −1 for correct lines 3h − 2 M1 Uses normal line equation or gradients for h. Forms equation in h =2 h h=2 A1 Alternative method for question 4(i) Vectors AB.BC=0 M1 Use of vectors AB and BC Solving M1 h=2 A1 Alternative method for question 4(i) Use of Pythagoras to find 3 lengths M1 Solving M1 h=2 A1 3 4(ii) y coordinate of D is 6, (3 × ‘their’ h) B1 FT 6−0 = 2 → x = 7 → D (7, 6) x−4 Vectors: AD.AB=0 M1 A1 Must use y = 6 Realises the y values of C and D are equal. Uses gradient or line equation to find x. 3
Question
The function f is defined by f x = −2x2 + 12x − 3 for x ∈ >. (i) Express −2x2 + 12x − 3 in the form −2 x + a2 + b, where a and b are constants. [2] (ii) State the greatest value of f x. [1] The function g is defined by g x = 2x + 5 for x ∈ >. (iii) Find the values of x for which gf x + 1 = 0. [3] @ A2 1 1 − sin x
Mark scheme
5(i) −2 ( x − 3) + 15 (a = − 3, b = 15) B1 B1 Or seen as a = − 3, b = 15 B1 for each value 2 2 5(ii) (f(x) ⩽) 15 B1 FT for (⩽) their “b” Don’t accept (3,15) alone 1 5(iii) gf(x) = 2( −2 x ² + 12 x − 3 ) + 5 = −4x² + 24x – 6 + 5 B1 gf(x) + 1 = 0 → −4x² + 24x = 0 M1 x = 0 or 6 A1 Forms and attempts to solve a quadratic Both answers given. 3
Question
(i) Prove the identity − tan x . [4] cos x 1 + sin x @ A2 1 (ii) Hence solve the equation − tan 2x = 13 for 0 ≤ x ≤ 0. [3] cos 2x
Mark scheme
6(i) 1 s LHS = − = (1 − s )(1 − s ) (1 − s )(1 − s ) 2 B1 Expresses tan in terms of sin and cos = c c c² 1 − s2 B1 correctly 1– s2 as the denominator (1 − s )(1 − s ) M1 Factors and correct cancelling www = (1 − s )(1 + s ) 1 − sinx A1 AG 1 + sinx 4 6(ii) 1 − sin2 x 1 M1 Realises use of 2x and makes sin2x the subject Uses part (i) to obtain = → sin 2x = ½ 1 + sin2 x 3 π A1 Allow decimal (0.262) x= 12 5π A1 FT for ½π – 1st answer. (or) x = Allow decimal (1.31) 12 π 5π and only, and no others in range. 12 12 π 5π SC sinx=½ → B1 6 6 3 JJJJG
Question
G F D E M C B k j O i A The diagram shows a three-dimensional shape in which the base OABC and the upper surface DEFG are identical horizontal squares. The parallelograms OAED and CBFG both lie in vertical planes. The point M is the mid-point of AF. Unit vectors i and j are parallel to OA and OC respectively and the unit vector k is vertically upwards. −−→ −−→ The position vectors of A and D are given by OA = 8i and OD = 3i + 10k. −−→ −−−→ (i) Express each of the vectors AM and GM in terms of i, j and k. [3] (ii) Use a scalar product to find angle GMA correct to the nearest degree. [4]
Mark scheme
7(i) AM = 1.5i + 4j + 5k B3,2,1 Loses 1 mark for each error. JJJJG GM = 6.5i – 4j − 5k 3 JJJJG JJJJG 7(ii) AM . GM = 9.75 −16 – 25 = −31.25 M1 Use of x1 x2 + y1 y2 + z1 z2 on AM and GM JJJJG JJJJG AM . GM = √(1.5²+4²+5²) × √( 6.5²+4²+5²) cos GMA M1 M1 M1 for product of 2 modulii M1 all correctly connected Equating → Angle GMA = 121° A1 4
Question
(a) The third and fourth terms of a geometric progression are 48 and 32 respectively. Find the sum to infinity of the progression. [3] (b) Two schemes are proposed for increasing the amount of household waste that is recycled each week. Scheme A is to increase the amount of waste recycled each month by 0.16 tonnes. Scheme B is to increase the amount of waste recycled each month by 6% of the amount recycled in the previous month. The proposal is to operate the scheme for a period of 24 months. The amount recycled in the first month is 2.5 tonnes. For each scheme, find the total amount of waste that would be recycled over the 24-month period. [5] Scheme A …… Scheme B ……
Mark scheme
8(a) ar² = 48, ar³ = 32, r = ⅔ or a = 108 M1 Solution of the 2 eqns to give r (or a). A1 (both) r = ⅔ and a = 108 A1 108 A1 FT Needs correct formula and r between −1 and 1. S∞ = = 324 1 3 3 8(b) Scheme A a = 2.50, d = 0.16 M1 Correct use of either AP Sn formula. Sn = 12(5 + 23×0.16) Sn = 104 tonnes. A1 Scheme B a = 2.50, r = 1.06 B1 Correct value of r used in GP. = ( 2.5 1.0624 − 1) M1 Correct use of either Sn formula. 1.06 − 1 Sn = 127 tonnes. A1 5
Question
The function f is defined by f x = 2 − 3 cos x for 0 ≤ x ≤ 20. (i) State the range of f. [2] (ii) Sketch the graph of y = f x. [2] The function g is defined by g x = 2 − 3 cos x for 0 ≤ x ≤ p, where p is a constant. (iii) State the largest value of p for which g has an inverse. [1] (iv) For this value of p, find an expression for g−1 x. [2] d2 y
Mark scheme
9(i) −1 ⩽ f(x) ⩽ 5 or [–1, 5] (may use y or f instead of f(x)) B1 B1 –1 < f ( x ) ⩽ 5 or −1 ⩽ x ⩽ 5 or (–1,5) or [5,–1] B1 only 2 9(ii) *B1 Start and end at –ve y, symmetrical, centre +ve. g(x) = 2 − 3cosx for 0 ⩽ x ⩽ p DB1 Shape all ok. Curves not lines. One cycle [0,2π] Flattens at each end. 2 9(iii) (greatest value of p =) π B1 1 9(iv) x = 2 − 3cosx → cosx = ⅓(2 – x) M1 Attempt at cosx the subject. Use of cos −1 2− x A1 Must be a function of x, g−1(x) = cos−1 (may use ‘y =’) 3 2
Question
A curve for which = 2x − 5 has a stationary point at 3, 6. dx2 (i) Find the equation of the curve. [6] (ii) Find the x-coordinate of the other stationary point on the curve. [1] (iii) Determine the nature of each of the stationary points. [2]
Mark scheme
10(i) dy B1 integrating → = x² − 5x (+c) dx = 0 when x = 3 M1 Uses the point to find c after ∫ = 0. c=6 A1 x³ 5x² B1 FT Integration again FT if a numerical constant term is integrating again → y = − + 6x ( +d ) present. 3 2 use of (3, 6) M1 Uses the point to find d after ∫ = 0. d = 1½ A1 6 10(ii) dy B1 = x² − 5x + 6 = 0 → x = 2 dx 1 10(iii) d²y B1 www x = 3, =1 and/or +ve Minimum. dx ² d²y x = 2, = −1 and/or −ve Maximum dx ² dy B1 www May use shape of ‘ + x 3 ’ curve or change in sign of dx SC: x = 3 , minimum, x = 2 , maximum, B1 2
Question
y 3 y= 1 + 4x P 2, 1 x O Q 3 The diagram shows part of the curve y = and a point P 2, 1 lying on the curve. The 1 + 4x normal to the curve at P intersects the x-axis at Q. (i) Show that the x-coordinate of Q is 16 9 . [5] (ii) Find, showing all necessary working, the area of the shaded region. [6] 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
11(i) 3 B1 3 × −½ × (1 + 4x ) 2 − dy 3 B1 Must have ‘× 4’ = 3 × −½ × (1 + 4x ) 2 × 4 − dx 2 9 M1 Use of m1.m₂ = − 1 If x = 2, m = − , Perpendicular gradient = 9 2 9 M1 Correct use of line eqn (could use y=0 here) Equation of normal is y − 1 = ( x − 2) 2 16 A1 AG Put y = 0 or on the line before → 9 5 11(ii) 2 3 3 1 + 4x B1 B1 Correct without ‘÷4’. For 2nd B1, ÷4’. Area under the curve = ∫ 1 + 4x 0 dx = 1 ÷4 2 Use of limits 0 to 2 → 4½ − 1½ M1 Use of correct limits in an integral. 3 A1 2 1 2 9 M1 Any correct method. Area of the triangle = ½ × 1 × 9 = 9 or attempt to find ∫ x − 8 dx 16/9 2 1 8 A1 Shaded area = 3 − =2 9 9 6
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Mathematics 9709/11 for May/June 2019 is out of 75 marks, with 1 hour 45 minutes to answer it.
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