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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_139709_w19_ms_139709/13/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
(i) Expand 1 + y6 in ascending powers of y as far as the term in y2 . [1] 6 (ii) In the expansion of 1 + px − 2x2 the coefficient of x2 is 48. Find the value of the positive constant p. [3]
Mark scheme
1(i) 1 + 6y + 15y B1 CAO 1 1(ii) ( ) ( ) M1 ( ) ( ) 2 2 1 + 6 px − 2 x 2 + 15 px − 2 x 2 SOI. Allow 6C1 ×15 px − 2 x 2 , 6C2 ×14 px − 2 x 2 (15 p − 12 ) ( x ) = 48 ( x ) 2 2 2 A1 1 term from each bracket and equate to 48 p=2 A1 SC: A1 p = 4 from 15p ‒ 12 = 48 3
Question
The function g is defined by g x = x2 − 6x + 7 for x > 4. By first completing the square, find an expression for g−1 x and state the domain of g−1 . [5]
Mark scheme
2 *B1 DB1 dependent on 3 in 1st bracket ( y = ) ( x − 3)2 [ −2] DB1 x − 3 = ( ± ) y + 2 or y − 3 = ( ± ) x + 2 M1 Correct order of operations (g ( x )) = 3 + x + 2 −1 A1 Must be in terms of x Domain (of g −1 ) is (x) > ‒1 B1 Allow (‒1, ∞). Do not allow y > ‒1 or g(x) > ‒1 or g −1 ( x ) > ‒1 5
Question
The equation of a curve is y = x3 + x2 − 8x + 7. The curve has no stationary points in the interval a < x < b. Find the least possible value of a and the greatest possible value of b. [4]
Mark scheme
3 dy B1 = 3x2 + 2x ‒ 8 dx Set to zero (SOI) and solve M1 (Min) a = ‒2, (Max) b = 4/3. – in terms of a and b. A1 Accept a . − 2, b - 4 3 A1 SC: A1 for a > −2, b < 43 or for −2 < x < 43 4
Question
C A r O r B The diagram shows a semicircle ACB with centre O and radius r. Arc OC is part of a circle with centre A. (i) Express angle CAO in radians in terms of 0. [1] (ii) Find the area of the shaded region in terms of r, 0 and ï3, simplifying your answer. [4]
Mark scheme
4(i) π B1 Angle CAO = 3 1 4(ii) 1 2 π M1 SOI (Sector AOC) = r × their 2 3 1 3 1 M1 π ( r )( 2r ) sin their or ( 2r )( r ) or ( r )( r ) 3 1 π For M1M1, their must be of the form kπ where (∆ ABC) = 2 3 2 2 2 3 0<k<½ 1 3 1 A1 All correct ( r )( 2r ) sin or ( 2r )( r ) or ( r )( r ) 3 1 π (∆ ABC) = 2 3 2 2 2 3 1 2π A1 r 2 − r 2 2 3 4 2 3
Question
x cm 4x cm 2x cm The dimensions of a cuboid are x cm, 2x cm and 4x cm, as shown in the diagram. (i) Show that the surface area S cm2 and the volume V cm3 are connected by the relation 2 S = 7V 3 . [3] (ii) When the volume of the cuboid is 1000 cm3 the surface area is increasing at 2 cm2 s−1 . Find the rate of increase of the volume at this instant. [4]
Mark scheme
5(i) S = 28x , V = 8x B1B1 SOI 2 B1 AG, WWW 7V 3 = 7 × 4 x 2 = S 3 5(ii) dS 14V − 13 14 *M1 Attempt to differentiate = = SOI when V = 1000 A1 dS − 13 dV 3 30 For M mark to be of form kV dV dV dS dV dS 1 DM1 = × OE used with dt = 2 and dt d t dS their 14 30 30 A1 OE or 4.29 7 Alternative method for question 5(ii) S2 3 dV 3 1 1 30 *M1 Attempt to differentiate V= → = ×S × 2 = SOI when S = 700 A1 dV 1 7 7 dS 2 7 7 14 For M mark to be of form kS 2 dS dV dS dV dS 1 DM1 = × OE used with dt = 2 and dt d t dS their 14 30 30 A1 OE or 4.29 7 5(ii) Alternative method for question 5(ii) dV dS dV dx *M1 Attempt to find either or and together with either dt dx dx dS or x dV dS dV 3 x dx 1 A1 = 24x2 or = 56 x and = , = or x = 5 (A1) dx dx dS 7 dt 140 dV DM1 Correct method for dt 30 A1 OE or 4.29 7 4
Question
A line has equation y = 3kx − 2k and a curve has equation y = x2 − kx + 2, where k is a constant. (i) Find the set of values of k for which the line and curve meet at two distinct points. [4] (ii) For each of two particular values of k, the line is a tangent to the curve. Show that these two tangents meet on the x-axis. [3]
Mark scheme
6(i) 3kx − 2k = x − kx + 2 → x − 4kx + 2k + 2 ( = 0 ) 2 2 B1 kx terms combined correctly-implied by correct b 2 − 4ac Attempt to find b 2 − 4ac M1 Form a quadratic equation in k 1 and − 12 A1 SOI k > 1, k < − 12 A1 Allow x > 1, x < −1/ 2 4 6(ii) 3 M1 Use of their k values (twice) in y = 3kx − 2k y = 3x − 2, y = − x +1 2 3 M1 Equate their tangent equations OR substitute y = 0 into both 3x − 2 = − x + 1 OR y + 2 = 2 − 2 y lines 2 2 A1 2 x= , → y = 0 in one or both lines Substitute x = in one or both lines 3 3 3
Question
(i) Show that the equation 3 cos4 1 + 4 sin2 1 − 3 = 0 can be expressed as 3x2 − 4x + 1 = 0, where x = cos2 1. [2] (ii) Hence solve the equation 3 cos4 1 + 4 sin2 1 − 3 = 0 for 0Å ≤ 1 ≤ 180Å. [5] 1
Mark scheme
7(i) ( ) 3cos 4θ + 4 1 − cos 2θ − 3 ( = 0 ) M1 Use s = 1 − c 2 2 3 x 2 + 4 (1 − x ) − 3 ( = 0 ) → 3 x 2 − 4 x + 1( = 0 ) A1 AG 2 7(ii) Attempt to solve for x M1 Expect x = 1, 1/3 cos θ = ( ± )1, ( ± ) 0.5774 A1 1 Accept ( ± ) SOI 3 (θ = ) 0º, 180º, 54.7º, 125.3º A3,2,1,0 A2,1,0 if more than 4 solutions in range 5
Question
A function f is defined for x > 12 and is such that f ′ x = 3 2x − 1 2 − 6. (i) Find the set of values of x for which f is decreasing. [4] (ii) It is now given that f 1 = −3. Find f x. [4]
Mark scheme
8(i) M1 SOI ( 2x − 1) < 2 or 3( 2 x − 1) < 6 1 1 2 2 2x −1 < 4 A1 SOI 1 5 A1 A1 Allow 2 separate statements <x< 2 2 4 8(ii) 3 B1 B1 f(x) = [3 ( 2 x − 1) ÷ ÷ ( 2) ] [ − 6 x] (+c) 3/ 2 2 Subsitute x = 1, y = ‒3 into an integrated expression. M1 Dependent on c being present (c = 2) A1 f(x) = ( 2 x − 1) − 6 x + 2 3 2 4
Question
The first, second and third terms of a geometric progression are 3k, 5k − 6 and 6k − 4, respectively. (i) Show that k satisfies the equation 7k2 − 48k + 36 = 0. [2] (ii) Find, showing all necessary working, the exact values of the common ratio corresponding to each of the possible values of k. [4] (iii) One of these ratios gives a progression which is convergent. Find the sum to infinity. [2]
Mark scheme
9(i) 5k − 6 6k − 4 M1 OR any valid relationship = → ( 5k − 6 )2 = 3k ( 6k − 4 ) 3k 5k − 6 25k 2 − 60k + 36 = 18k 2 − 12k → 7 k 2 − 48k + 36 A1 AG 2 9(ii) 6 B1B1 6 k= ,6 Allow 0.857(1) for 7 7 6 2 B1 Must be exact When k = ,r= − 7 3 4 B1 When k = 6, r = 3 4 9(iii) a 2 6 M1 Provided 0 < |their ‒2/3| < 1 Use of S∞ = with r = their − and a = 3 × their 1− r 3 7 18 2 54 A1 FT if 0.857(1) has been used in part (ii). ÷ 1 + = or 1.54 7 3 35 2
Question
Relative to an origin O, the position vectors of the points A, B and X are given by ` a ` a ` a OA = −4 , OB = 2 and OX = −2 . 2 11 5 −−→ (i) Find AX and show that AXB is a straight line. [3] ` a −−→ 1 The position vector of a point C is given by OC = −8 . 3 (ii) Show that CX is perpendicular to AX . [3] (iii) Find the area of triangle ABC. [3]
Mark scheme
10(i) 6 18 12 −12 B1B1 AX = 2 , and one of AB = 6 , XB = 4 , BX = −4 3 9 6 −6 3 B1 WWW State AB = 3AX ( or XB = 2AX or AB = XB etc) hence straight line A conclusion (i.e. a straight line) is required. 2 OR AX.AB AX.BX = 1 (→ θ = 0) or = –1 (→ θ = 180) AX AB AX BX hence straight line 3 10(ii) −3 B1 CX = 6 2 CX.AX = −18 + 12 + 6 M1 = 0 (hence CX is perpendicular to AX) A1 3 10(iii) CX = 32 + 62 + 22 , AB = 182 + 62 + 92 M1 Both attempted 1 M1A1 Accept answers which round to 73.5 Area ∆ ABC = × their 21 × their 7 = 73 12 2 3
Question
y A 2, 3 B y = x − 1−2 + 2 x O 1 3 The diagram shows part of the curve y = x − 1−2 + 2, and the lines x = 1 and x = 3. The point A on the curve has coordinates 2, 3. The normal to the curve at A crosses the line x = 1 at B. (i) Show that the normal AB has equation y = 12 x + 2. [3] (ii) Find, showing all necessary working, the volume of revolution obtained when the shaded region is rotated through 360Å about the x-axis. [8] 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) dy B1 = −2 ( x − 1) −3 dx 1 M1 m must come from differentiation When x = 2, m = ‒2 → gradient of normal = − m A1 1 Equation of normal is y − 3 = ½ ( x − 2 ) → y = ½ x + 2 AG Through (2, 3) with gradient − . Simplify to AG m 3 11(ii) ( π) ∫ y12 ( dx ) , ( π ) ∫ y 2 ( d x )2 *M1 Attempt to integrate y for at least one of the functions 2 ( π ) ∫ ( 12 x + 2 ) 2 or ( x + 2x + 4) 1 4 2 (2 ) A1A1 A1 for 1 x + 2 2 depends on an attempt to integrate this ( π ) ∫ ( ( x − 1)−4 + 4 ( x − 1)−2 + 4 ) form later ( π ) 23 ( 12 x + 2 ) 3 or 1 x3 + x 2 + 4 x A1A1 Must have at least 2 terms correct for each integral 12 ( x − 1)−3 4 ( x − 1) −1 (π) + + 4x −3 −1 125 2 1 −1 −1 DM1 Apply limits to at least 1 integrated expansion (π) 18 − or +4+8− + 1 + 4 − 2 + 12 − − 4 + 8 12 3 12 24 3 Attempt to add 2 volume integrals (or 1 volume integral + frustum) DM1 { π 7 12 7 +6 7 24 } 7 111 A1 2 1 −1 −1 13 π or π or 13.9π or 43.6 + 4+8− +1+ 4 − 2 + 12 − − 4 + 8 8 8 3 12 24 3 8
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Mathematics 9709/13 for October/November 2019 is out of 75 marks, with 1 hour 45 minutes to answer it.
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