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Cambridge AS & A Level Mathematics (9709), Paper 1, October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w19_qp_129709_w19_ms_129709/12/O/N/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 coefficient of x in the expansion of 4 + ax 1 + 2 is 3. Find the value of the constant a. [4] 2
Mark scheme
1 6x x 2 B1 B1 OE In or from a correct expansion. Can be implied by correct , 15 × equation. 2 4 × (4 + ax) → 3a + 15 = 3 M1 2 terms in x2 equated to 3 or 3x2. Condone x2 on one side only. a = −4 A1 CAO 4
Question
The point M is the mid-point of the line joining the points 3, 7 and −1, 1. Find the equation of the x y line through M which is parallel to the line + = 1. [4] 3 2 dy k
Mark scheme
2 Attempt to find the midpoint M M1 (1, 4) A1 Use a gradient of ±⅔ and their M to find the equation of the line. M1 Equation is y – 4 = −⅔(x – 1) A1 AEF Alternative method for question 2 Attempt to find the midpoint M M1 (1, 4) A1 Replace 1 in the given equation by c and substitute their M M1 Equation is y – 4 = −⅔(x – 1) A1 AEF 4
Question
A curve is such that = , where k is a constant. The points P 1, −1 and Q 4, 4 lie on the dx x curve. Find the equation of the curve. [4]
Mark scheme
3 − +1 1 B1 OE kx 2 k x (y =) = (+c) 1 1 − +1 2 2 Substitutes both points into an integrated expression with a ‘+c’ and M1 Expect to see –1 = 2k + c and 4 = 4k + c solve as far as a value for one variable. k = 2½ and c = −6 A1 WWW y = 5 x −6 A1 OE From correct values of both k & c and correct integral. 4
Question
A r cm O 21 rad T B The diagram shows a circle with centre O and radius r cm. Points A and B lie on the circle and angle AOB = 21 radians. The tangents to the circle at A and B meet at T . (i) Express the perimeter of the shaded region in terms of r and 1. [3] (ii) In the case where r = 5 and 1 = 1.2, find the area of the shaded region. [4]
Mark scheme
4(i) Arc length AB = 2rθ B1 AT BT B1 r 2 Tan θ = or → AT or BT = r tan θ r sin θ Accept or − r 2 or NOT (90 – θ) r r cosθ π sin − θ 2 P = 2rθ + 2r tan θ B1FT OE, FT for their arc length + 2 × their AT 3 4(ii) Area ∆AOT = ½ × 5 × 5 tan 1.2 or B1 Area AOBT = 2 × ½ × 5 × 5 tan 1.2 Sector area = ½ × 25 × 2.4 (or 1.2) *M1 Use of ½r2θ with θ = 1.2 or 2.4. Shaded area = 2 triangles – sector DM1 Subtraction of sector, using 2.4 where appropriate, from 2 triangles Area = 34.3 (cm2) A1 AWRT Alternative method for question 4(ii) Area of ∆ ABT = ½ × (5 × tan 1.2)2 × sin(π – 2.4) (= 55.86) B1 Segment area = ½ × 25 × (2.4 – sin 2.4) (= 21.56) *M1 Use of ½r2 (θ – sin θ) with θ = 1.2 or 2.4 Shaded area = triangle – segment DM1 Subtraction of segment from ∆ ABT, using 2.4 where appropriate. Area = 34.3 (cm2) A1 AWRT 4
Question
15 cm h cm The diagram shows a solid cone which has a slant height of 15 cm and a vertical height of h cm. (i) Show that the volume, V cm3, of the cone is given by V = 13 0 225h − h3 . [2] [The volume of a cone of radius r and vertical height h is 31 0r2 h.] (ii) Given that h can vary, find the value of h for which V has a stationary value. Determine, showing all necessary working, the nature of this stationary value. [5]
Mark scheme
5(i) Use of Pythagoras → r2 = 152 − h2 M1 V = ⅓π(225 − h2) × h → ⅓π(225h – h3) A1 AG WWW e.g. sight of r = 15 – h gets A0. 2 5(ii) dv π B1 = (225 – 3h ) 2 dh 3 dv M1 Differentiates, sets their differential to 0 and attempts to solve Their =0 at least as far as h2 ≠ 0. dh (ℎ =) √75, 5√3 or AWRT 8.66 A1 Ignore − 75 OE and ISW for both A marks d2h π M1 Differentiates for a second time and considers the sign of the = (–6h) (→ −ve) second differential or any other valid complete method. dh 2 3 → Maximum A1FT Correct conclusion from correct 2nd differential, value for h not required, or any other valid complete method. FT for their h, if used, as long as it is positive. π SC Omission of π or throughout can score B0M1A1M1A0 3 5 −1 *
Question
(a) Given that x > 0, find the two smallest values of x, in radians, for which 3 tan 2x + 1 = 1. Show all necessary working. [4] (b) The function f : x → 3 cos2 x − 2 sin2 x is defined for 0 ≤ x ≤ 0. (i) Express f x in the form a cos2 x + b, where a and b are constants. [1] (ii) Find the range of f. [2]
Mark scheme
6(a) (2x + 1) = tan (⅓) (= 0.322 or 18.4 OR −0.339 rad or 8.7°) M1 Correct order of operations. Allow degrees. Either their 0.322 + π or 2π DM1 Must be in radians π Or their −0.339 + or π 2 x = 1.23 or x = 2.80 A1 AWRT for either correct answer, accept 0.39π or 0.89π A1 For the second answer with no other answers between 0 and 2.8 SC1 For both 1.2 and 2.8 4 6(b)(i) 5 cos2 x – 2 B1 Allow a = 5, b = −2 1 6(b)(ii) −2 B1FT FT for sight of their b 3 B1FT FT for sight of their a + b 2 JJJG
Question
G Q F D E P C B k j O i A The diagram shows a three-dimensional shape OABCDEFG. The base OABC and the upper surface DEFG are identical horizontal rectangles. The parallelograms OAED and CBFG both lie in vertical planes. Points P and Q are the mid-points of OD and GF respectively. 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, C and D are given by OA = 6i, OC = 8j and OD = 2i + 10k. −−→ −−→ (i) Express each of the vectors PB and PQ in terms of i, j and k. [4] (ii) Determine whether P is nearer to Q or to B. [2] (iii) Use a scalar product to find angle BPQ. [3]
Mark scheme
7(i) ( ) PB = 5i + 8j – 5k B2,1,0 B2 all correct, B1 for two correct components. JJJG ( PQ ) = 4i + 8j + 5k B2,1,0 B2 all correct, B1 for two correct components. Accept column vectors. SC B1 for each vector if all components multiplied by –1. 4 7(ii) (Length of PB =) (5 + 8 + 5 ) = ( 114 ≈ 10.7) 2 2 2 M1 Evaluation of both lengths. Other valid complete comparisons can be accepted. (Length of PQ =) ( 4 + 8 + 5 ) = ( 105 ≈ 10.2) 2 2 2 P is nearer to Q. A1 WWW 2 JJJG JJJG JJJG JJJG 7(iii) ( ) PB.PQ = 20 + 64 – 25 M1 Use of x1x2 + y1y2 + z1z2 on their PB and PQ (Their 114)(their 105) cos BPQ = (their 59) M1 All elements present and in correct places. BPQ = 57.4(°) or 1.00 (rad) A1 AWRT Calculating the obtuse angle and then subtracting gets A0. 3
Question
(a) Over a 21-day period an athlete prepares for a marathon by increasing the distance she runs each day by 1.2 km. On the first day she runs 13 km. (i) Find the distance she runs on the last day of the 21-day period. [1] (ii) Find the total distance she runs in the 21-day period. [2] (b) The first, second and third terms of a geometric progression are x, x − 3 and x − 5 respectively. (i) Find the value of x. [2] (ii) Find the fourth term of the progression. [2] (iii) Find the sum to infinity of the progression. [2]
Mark scheme
8(a)(i) 21st term = 13 + 20 × 1.2 = 37 (km) B1 1 8(a)(ii) S21= ½×21 × (26 + 20 × 1.2) or ½ × 21 × (13 + their 37) M1 A correct sum formula used with correct values for a, d and n. 525 (km) A1 2 8(b)(i) x−3 x−5 M1 Any valid method to obtain an equation in one variable. = oe (or use of a, ar and ar2) x x−3 (a = or x =) 9 A1 2 8(b)(ii) x −3 x−5 x−5 M1 Any valid method to find r and the fourth term with their a & r. r= or or = ⅔. Fourth term = 9 × (⅔)3 x x −3 x 2⅔ or 2.67 A1 OE, AWRT 2 8(b)(iii) a 9 M1 Correct formula and using their ‘r’ and ‘a’, with r <1, to S∞ = = 1 − r 1 − 23 obtain a numerical answer. 27 or 27.0 A1 AWRT 2
Question
Functions f and g are defined by f x = 2x2 + 8x + 1 for x ∈ >, g x = 2x − k for x ∈ >, where k is a constant. (i) Find the value of k for which the line y = g x is a tangent to the curve y = f x. [3] (ii) In the case where k = −9, find the set of values of x for which f x < g x. [3] (iii) In the case where k = −1, find g−1 f x and solve the equation g−1 f x = 0. [3] (iv) Express f x in the form 2 x + a2 + b, where a and b are constants, and hence state the least value of f x. [3]
Mark scheme
9(i) f(x) = g(x) → 2x + 6x + 1 + k (= 0) 2 *M1 Forms a quadratic with all terms on same side. 2 Use of b = 4ac DM1 Uses the discriminant = 0. (k =) 3½ A1 OE, WWW Alternative method for question 9(i) 4 x + 8 = 2 (→ x = –1½) *M1 Differentiating, equating gradients and solving to give x = Substitutes their x value into either 2x² + 6x + 1 + k = 0 OR into the DM1 Substituting appropriately for their x and proceeding to find a −13 value of k. curve to find y = then both values into the line. 2 (k =) 3½ A1 OE, WWW 3 9(ii) 2x² + 6x – 8 (< 0) M1 Forms a quadratic with all terms on same side – 4 and 1 A1 −4<x<1 A1 CAO 3 9(iii) x −1 B1 Needs to be in terms of x. (g−1(x)) = 2 2 x² + 8x + 1 − 1 M1 Substitutes f into g−1 and attempts to solve it = 0 as far as x = (g−1f(x)) = = 0 → (2x² + 8x = 0) → x = 2 0, –4 A1 CAO 3 2 9(iv) 2(x + 2) − 7 B1 B1 or a = +2, b = −7 (Least value of f(x) or y =) −7 or ⩾ −7 B1FT FT for their b from a correct form of the expression. 3
Question
y 4 y= 1− 2x + 12 B x O A 4 The diagram shows part of the curve y = 1 − . The curve intersects the x-axis at A. The 2x + 12 normal to the curve at A intersects the y-axis at B. dy (i) Obtain expressions for and Ó y dx. [4] dx (ii) Find the coordinates of B. [4] (iii) Find, showing all necessary working, the area of the shaded region. [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
10(i) dy B2,1,0 OE. Full marks for 3 correct components. Withhold one mark = [ 0] + (2 x + 1)−3 × [+ 16] for each error or omission. dx ∫ydx = [ x ] + (2 x + 1) −1 × [ +2] (+c) B2,1,0 OE. Full marks for 3 correct components. Withhold one mark for each error or omission. 4 10(ii) At A, x = ½. B1 Ignore extra answer x = −1.5 dy *M1 dy = 2 → Gradient of normal ( = −½ ) With their positive value of x at A and their , uses dx dx m₁m₂ = −1 Equation of normal: DM1 Use of their x at A and their normal gradient. y − 0 = −½ ( x − ½ ) or y − 0 = −½ (0 – ½) or 0 = −½×½ + c B (0, ¼) A1 4 10(iii) 1 *M1 ∫ y dx SOI with 0 and their positive x coordinate of A. 2 4 2 ( ∫ ( 2 x + 1) 1− dx ) 0 [½ + 1] – [0 + 2] = (−½) DM1 Substitutes both 0 and their ½ into their ∫ydx and subtracts. 1 B1 Area of triangle above x-axis = ½ × ½ × ¼ = 16 9 A1 OE (including AWRT 0.563) Total area of shaded region = 16 Alternative method for question 10(iii) 0 1 1 *M1 − 1 ∫ 1 − ( dy ) ∫ x dy SOI. Where x is of the form k 1 − y ) 2 + c with 0 and 2 −3 (1 − y ) 2 their negative y intercept of curve. DM1 Substitutes both 0 and their –3 into their ∫xdy and subtracts. [ −2] − −4 + = (½) 3 2 1 B1 Area of triangle above x-axis = ½ × ½ × ¼ = 16 9 A1 OE (including AWRT 0.563) Total area of shaded region = 16 Alternative method for question 10(iii) 1 *M1 ∫(their normal curve) with 0 and their positive x coordinate 2 1 1 of A. ∫ 0 − 2 x + − y dx 4 Curve [½ + 1] – [0 + 2] = (−½) DM1 Substitutes both 0 and their ½ into their ∫ydx and subtracts. 1 B1 Substitutes both 0 and ½ into the correct integral and subtracts. 2 1 1 − x 2 x −1 1 1 ∫ − x + dx = + = + – [ 0] = 0 2 4 4 4 16 8 16 9 A1 OE (including AWRT 0.563) Total area of shaded region = 16 4
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Mathematics 9709/12 for October/November 2019 is out of 75 marks, with 1 hour 45 minutes to answer it.
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