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Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics, February/March 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_m19_qp_129709_m19_ms_129709/12/F/M/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 x3 in the expansion of 1 − px5 is −2160. Find the value of the constant p. [3]
Mark scheme
1 5C3 ( − )( px ) soi 3 B1 Can be part of expansion. Condone omission of ‒ sign ( −1)10 p 3 = −2160 then ÷ and cube root M1 Condone omission of ‒ sign. p=6 A1 3
Question
A curve with equation y = f x passes through the points 0, 2 and 3, −1. It is given that f ′ x = kx2 − 2x, where k is a constant. Find the value of k. [5]
Mark scheme
2 y = 1 3 kx 3 − x 2 ( + c ) M1A1 Attempt integration for M mark Sub (0, 2) DM1 Dep on c present. Expect c = 2 Sub (3, ‒1) → −1 = 9k − 9 + their c DM1 k = 2/3 A1 5
Question
C 7 8 Y X A B D In the diagram, CXD is a semicircle of radius 7 cm with centre A and diameter CD. The straight line YABX is perpendicular to CD, and the arc CYD is part of a circle with centre B and radius 8 cm. Find the total area of the region enclosed by the two arcs. [6]
Mark scheme
3 7 −17 B1 Accept 61.0°, 66° or 122° Angle CBA = sin −1 = 1.0654 or CBD = cos −1 = 2.13 8 32 Sector BCYD = ½ × 82 × 2 × their1.0654 ( rad ) soi M1 Expect 68.1(9). Angle must be in radians (or their 61/360 × 2 × 82) or sector CBY = ½ × 82 × their1.0654 ( rad ) Or sector DBY ∆BCD = 7 × 82 − 7 2 or ½ × 82 × sin ( 2 × their1.0654 ) soi M1 Expect 27.1(1). Award M1 for ABC or ABD Semi-circle CXD = ½π × 7 2 = 76.9 ( 7 ) M1 M1M1 for segment area formula used correctly Total area = their68.19 ‒ their27.11 + their76.97 = 118.0–118.1 M1A1 Cannot gain M1 without attempt to find angle CBA or CBD 6
Question
A curve has equation y = 2x − 1−1 + 2x. dy d2 y (i) Find and 2 . [3] dx dx (ii) Find the x-coordinates of the stationary points and, showing all necessary working, determine the nature of each stationary point. [4]
Mark scheme
4(i) dy / dx = −2 ( 2 x − 1) + 2 B2,1,0 Unsimplified form ok (–1 for each error in ‘–2’, −2 ‘ ( 2 x − 1) ’ and ‘2’) −2 d 2 y / dx 2 = 8 ( 2 x − 1) B1 Unsimplified form ok −3 3 4(ii) Set dy / dx to zero and attempt to solve – at least one correct step M1 x = 0, 1 A1 Expect ( 2 x − 1)2 = 1 When x = 0, d 2 y / dx 2 = −8 (or < 0) . Hence MAX B1 When x = 1, d 2 y / dx 2 = 8 (or > 0) . Hence MIN B1 Both final marks dependent on correct x and correct d 2 y / dx 2 and no errors May use change of sign of dy / dx but not at x = 1 / 2 4
Question
Two vectors, u and v, are such that ` a ` a q 8 u= 2 and v= q−1 , 6 q2 − 7 where q is a constant. (i) Find the values of q for which u is perpendicular to v. [3] (ii) Find the angle between u and v when q = 0. [4]
Mark scheme
5(i) u.v = 8q + 2q − 2 + 6q 2 − 42 B1 May be unsimplified 6q 2 + 10q − 44 = 0 oe M1 Simplify, set to zero and attempt to solve q = 2, ‒11/3 A1 Both required. Accept ‒3.67 3 5(ii) 0 8 M1 Correct method for scalar product u = 2 v = −1 u.v = ‒2 ‒ 42 6 −7 |u| × |v| = 22 + 62 × 82 + 12 + 7 2 M1 Prod of mods. At least one methodically correct. −44 −44 −4 M1 All linked correctly and inverse cos used correctly cos θ = = = 40 × 114 4 285 √ 11 θ = 130.7° or 2.28(05) rads A1 No other angles between 0° and 180° 4
Question
(i) The first and second terms of a geometric progression are p and 2p respectively, where p is a positive constant. The sum of the first n terms is greater than 1000p. Show that 2n > 1001. [2] (ii) In another case, p and 2p are the first and second terms respectively of an arithmetic progression. The nth term is 336 and the sum of the first n terms is 7224. Write down two equations in n and p and hence find the values of n and p. [5]
Mark scheme
6(i) Sn = ( p 2n − 1 ) soi M1 2 −1 ( ) p 2n − 1 > 1000 p → 2 n > 1001 AG A1 2 6(ii) p + ( n − 1) p = 336 B1 Expect np = 336 n B1 n 2 p + ( n − 1) p = 7224 Expect ( p + np ) = 7224 2 2 Eliminate n or p to an equation in one variable M1 Expect e.g. 168(1 + n) = 7224 or 1 + 336/p =43 etc n = 42, p = 8 A1A1 5
Question
(a) Solve the equation 3 sin2 21 + 8 cos 21 = 0 for 0Å ≤ 1 ≤ 180Å. [5] (b) y y = a + tan bx ï3 x − 16 0 O The diagram shows part of the graph of y = a + tan bx, where x is measured in radians and a and b are constants. The curve intersects the x-axis at − 16 0 , 0 and the y-axis at 0, ï3. Find the values of a and b. [3]
Mark scheme
7(a) ( ) 3 1 − cos 2 2θ + 8cos 2θ = 0 → 3cos 2 2θ − 8cos 2θ − 3 ( = 0 ) M1 Use s 2 = 1 − c 2 and simplify to 3-term quadratic in 2θ 1 A1 Ignore other solution cos 2θ = − soi 3 2θ = 109.(47)º or 250.(53)º A1 One solution is sufficient, may be implied by either of the next solns θ = 54.7º or 125.3º A1A1ft Ft for 180º ‒ other solution Use of double angles leads to 3c 4 − 7c 2 + 2 = 0 ⇒ c = ±1 / √ 3 for M1A1A1 then A1A1 for each angle Similar marking if 3sin 2 2θ = −8cos2θ is squared leading to 9sin 4 2θ + 64sin 2 2θ − 64 = 0 5 7(b) √ 3 = a + tan0 → a = √ 3 B1 b = 8 or –4 (or –10, 14 etc) scores M1A0 0 = tan(−bπ / 6) + √ 3 taken as far as tan −1 , angle units consistent ( ) M1 A0 if tan −1 − 3 is not exact; (b=2 no working scores B2) b=2 A1 3
Question
(i) Express x2 − 4x + 7 in the form x + a2 + b. [2] The function f is defined by f x = x2 − 4x + 7 for x < k, where k is a constant. (ii) State the largest value of k for which f is a decreasing function. [1] The value of k is now given to be 1. (iii) Find an expression for f −1 x and state the domain of f −1 . [3] 2 (iv) The function g is defined by g x = for x > 1. Find an expression for gf x and state the x−1 range of gf. [4]
Mark scheme
8(i) ( x − 2 )2 + [3] B1 DB1 2nd B1 dependent on ±2 in 1st bracket 2 8(ii) Largest k is 2 Accept k - 2 B1 Must be in terms of k 1 8(iii) y = ( x − 2) + 3 ⇒ x − 2 = ( ± ) y − 3 M1 2 ⇒ f −1 ( x ) = 2 − x − 3 for x > 4 A1B1 3 8(iv) 2 2 B1 Either form gf(x) = = x − 4 x + 7 − 1 ( x − 2 )2 + 2 2 Since f ( x ) > 4 ⇒ gf ( x ) < 2 / 3 (or since x < 1etc ) M1A1 2/3 in answer implies M1 www range of gf(x) is 0 < gf(x)( < 2/3) B1 Accept 0 < y < 2/3, (0, 2/3) but 0 < x < 2/3 is SCM1A1B0 4
Question
y y= x3 + x2 P x O 3 The diagram shows part of the curve with equation y = x3 + x2 . The shaded region is bounded by the curve, the x-axis and the line x = 3. (i) Find, showing all necessary working, the volume obtained when the shaded region is rotated through 360Å about the x-axis. [4] (ii) P is the point on the curve with x-coordinate 3. Find the y-coordinate of the point where the normal to the curve at P crosses the y-axis. [6]
Mark scheme
9(i) ( ) V = ( π ) ∫ x 3 + x 2 ( dx ) M1 Attempt ∫ y 2 dx x 4 x3 3 A1 (π ) + 4 3 0 DM1 May be implied by a correct answer (π ) ( −0 ) 81 +9 4 117π A1 Accept 91.9 oe If additional areas rotated about x-axis, maximum of 4 M1A0DM1A0 4 9(ii) B2,1,0 Omission of 3 x 2 + 2 x is one error dy 1 3 ( ) ( ) −1/2 = x + x2 × 3x 2 + 2 x dx 2 (At x = 3,) y = 6 B1 1 1 11 DB1ft Ft on their dy / dx providing differentiation attempted At x = 3, m = × × 33 = soi 2 6 4 4 DM1 Equation through (3, their 6) and with gradient ‒1/their Equation of normal is y − 6 = − ( x − 3) m 11 1 A1 When x = 0, y = 7 oe 11 6
Question
y 1 y = 4x 2 x O 1 The diagram shows the curve with equation y = 4x 2 . (i) The straight line with equation y = x + 3 intersects the curve at points A and B. Find the length of AB. [6] (ii) The tangent to the curve at a point T is parallel to AB. Find the coordinates of T . [3] (iii) Find the coordinates of the point of intersection of the normal to the curve at T with the line AB. [3] 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) 4 x1/ 2 = x + 3 → M1 Eliminate y from the 2 equations and then: ( x1/ 2 ) 2 − 4 x1/2 + 3 ( = 0 ) OR 16 x = x 2 + 6 x + 9 Either treat as quad in x1/ 2 OR square both sides and RHS is 3-term x1/ 2 = 1or 3 x 2 − 10 x + 9 ( = 0 ) A1 If in 1st method x1/ 2 becomes x, allow only M1 unless subsequently squared x = 1 or 9 A1 y = 4 or12 A1ft Ft from their x values If the 2 solutions are found by trial substitution B1 for the first coordinate and B3 for the second coordinate AB 2 = ( 9 − 1) + (12 − 4 ) M1 2 2 AB = 128 or 8 2 oe or 11.3 A1 6 10(ii) dy/dx = 2 x −1/ 2 B1 2 x −1/ 2 = 1 M1 Set their derivative = their gradient of AB and attempt to solve (4, 8) A1 Alternative method without calculus: MAB = 1, tangent is y = mx + c where m = 1 and meets y = 4x1/2 when 4x1/2 = x + c. This is a quadratic with b2 = 4ac, so 16 – 4 × 1 × = 0 so c = 4 B1 Solving 4x1/2 = x + 4 gives x = 4 and y = 8 M1A1 3 10(iii) Equation of normal is y − 8 = −1( x − 4 ) M1 Equation through their T and with gradient ‒1/their gradient of AB. Expect y = − x + 12 , Eliminate y (or x) → − x + 12 = x + 3 or y − 3 = 12 − y M1 May use their equation of AB (4½, 7½) A1 3
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Mathematics 9709/12 for February/March 2019 is out of 75 marks, with 1 hour 45 minutes to answer it.
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