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Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w22_qp_119709_w22_ms_119709/11/O/N/22
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
Solve the equation 3x + 2 = . [3] x−1 dy −4
Mark scheme
1 (3x + 2)( x −1) = 2 3x2 − x − 4 [= 0] M1 OE Multiply by denominator and obtain a quadratic. ( 3x − 4 )( x + 1) [ = 0] M1 Solve by factorising, formula or completing the square. 4 A1 Allow 1.33 [x =] −1, If M1 M0, SC B1 possible for two correct answers. 3 3
Question
The equation of a curve is such that = 12 12 x − 1 . It is given that the curve passes through the dx point P 6, 4. (a) Find the equation of the tangent to the curve at P. [2] (b) Find the equation of the curve. [4] 1
Mark scheme
2(a) −4 M1 1 3 dy 3 12 × 6 − 1 = 12 ( 2 ) = Substitute x = 6 into −4 SOI by gradient used. 2 4 dx 4 3 A1 3 1 3 y−4= ( x − 6) OE e.g. y = x − or evaluates c in y = x + c 4 4 2 4 1 3 OR evaluates c = − using (6, 4) and gradient . ISW 2 4 2 2(b) 1 −3 B2, 1, 0 12 x − 1 1 −3 1 [ y =] 2 ÷ = −8 x − 1 2 −3 2 1 −3 M1 Must have +c . 12 × × 6 − 1 Substitute y = 4, x = 6 and solve for c in an integrated 4= 2 +c 4 = −8 × 2−3 + c c = [5] 1 expression. May be unsimplified. ×−3 2 −3 A1 OE Must see ‘ y = ’ or ‘ f ( x ) = ’ in the working. [ y =] − 8 x − 1 1 +5 2 4
Question
A curve has equation y = ax 2 − 2x, where x > 0 and a is a constant. The curve has a stationary point at the point P, which has x-coordinate 9. Find the y-coordinate of P. [5] A @ 2 5 4 The coefficient of x in the expansion of 1 + x + 1 + px6 is 70. 2 p Find the possible values of the constant p. [6]
Mark scheme
3 dy 1 − 2 1 B2, 1, 0 = ax −2 dx 2 1 1 a M1 dy 0 = a ( 9 ) 2 − 2 − 2 = 0 a = [12] − Substitute x = 9 and = 0 into their derivative and 2 6 dx solve a linear equation for a. [ a = ]12 A1 1 A1 FT FT on their a. y = their a × ( 9 ) 2 − 18 = 18 5 .4 5 2 B1 Accept with x 2 present. Must evaluate 5C2 2 2 10 × 22 40 Coefficient of x in 1 + x is 10 = 2 = p p p2 p2 Coefficient of x 2 in (1 + px ) is 15 ( p ) = 15 p 2 B1 Accept with x 2 present. Must evaluate 6C2 6 2 40 *M1 Forming an equation in p with their coefficients, the 2 + 15 p 2 = 70 p given 70, no x terms and no extra terms. 15 p4 − 70 p2 + 40 [= 0] or 3 p4 − 14 p2 + 8 [= 0] DM1 Forming a 3-term equation in p (or another variable) with all terms on one side and their coefficients. 70 ± 702 − 4 (15)( 40 ) DM1 Attempt to solve 3-term quartic (or quadratic in another [5]( p2 − 4)( 3 p2 − 2) [= 0] or or variable) by factorisation, formula or completing the 30 square. 14 ± 14 − 4 ( 3)(8) 2 6 2 A1 6 p = ±2 , ± OE e.g. ± or AWRT ±0.816 3 3 If *M1 DM1 DM0, allow SC B1 for 4 correct values. 6
Question
8 cm A B O The diagram shows a sector OAB of a circle with centre O. The length of the arc AB is 8 cm. It is given that the perimeter of the sector is 20 cm. (a) Find the perimeter of the shaded segment. [4] (b) Find the area of the shaded segment. [2]
Mark scheme
5(a) [ 2r + 8 = 20 ] r = 6 B1 8 *M1 4 Angle AOB = Expect OE ( 76.4° ) . their 6 3 M0 Assume triangle is equilateral. 2 4 DM1 For 6 read their 6 . AB = 2 × 6sin their or 62 + 62 − 2 × 62 cos their 3 3 6 4 or AB = × sin their π 2 3 sin − their 2 3 Perimeter = [ 7.42 + 8 = ] 15.4 A1 AWRT 4 5(b) 1 4 1 4 M1 Sector area – whole triangle area. Area = × 62 × their − × 62 × sin their 2 3 2 3 For 6 read their 6 . 1 4 1 2 2 or Area = × 62 × their − 2 × 6 sin their 6cos their Sector area – 2(half triangle area). 2 3 2 3 3 = [ 24 −17.49 = ] 6.51 A1 AWRT 2
Question
(a) Show that the equation 1 1 + =1 sin 1 + cos 1 sin 1 − cos 1 may be expressed in the form a sin2 1 + b sin 1 + c = 0, where a, b and c are constants to be found. [3] 1 1 (b) Hence solve the equation + = 1 for 0Å ≤ 1 ≤ 360Å. [3] sin 1 + cos 1 sin 1 − cos 1
Mark scheme
6(a) sin θ − cosθ + sin θ + cosθ sin θ − cosθ + sin θ + cosθ *M1 Use common denominator and equate to 1. = =1 ( sin θ + cosθ )(sinθ − cosθ ) sin 2θ − cos2θ ( 2sin θ [= sin 2θ − cos2θ ] = sin 2θ − 1 − sin 2θ ) DM1 Multiply by common denominator and replace cos2 θ by 1 – sin2 θ. 2sin 2θ − 2sin θ − 1 = 0 A1 OE In the given form. 3 6(b) 2± ( −2 )2 − 4 ( 2 )( −1) 2 ± 4 + 8 1 ± 3 M1 Use formula or complete the square to solve a quadratic [sin θ =] = = equation of the correct form. 4 4 2 201.5° or 338.5° A1 A1 FT AWRT; A1 for either solution correct. A1 FT for 540 ‒ (first value). If M0, allow SC B1 B1FT similarly. 3
Question
A tool for putting fence posts into the ground is called a ‘post-rammer’. The distances in millimetres that the post sinks into the ground on each impact of the post-rammer follow a geometric progression. The first three impacts cause the post to sink into the ground by 50 mm, 40 mm and 32 mm respectively. (a) Verify that the 9th impact is the first in which the post sinks less than 10 mm into the ground. [3] (b) Find, to the nearest millimetre, the total depth of the post in the ground after 20 impacts. [2] (c) Find the greatest total depth in the ground which could theoretically be achieved. [2] 3 p
Mark scheme
7(a) r = 0.8 B1 SOI 50 × ( their 0.8) = 10.5 M1 Evaluate 8th or 9th term in GP. 7 50 × ( their 0.8) = 8.39 . Hence 9th impact required A1 AG Two terms correct + conclusion (mention of 9th 8 impact or u9 somewhere in the solution). Statement that one is <10 (and the other >10) is insufficient unless it mentions 9th impact or u9. Alternative method for final two marks: Logarithm method 50 × ( their 0.8) < 10 ( their 0.8) < 0.5 M1 n n n log ( their 0.8 ) < log 0.5 log 0.5 n> [ n > ] 7.2 log ( their 0.8 ) n = 8 hence 9th impact required A1 AG Need conclusion that mentions 9th impact or u9. 3 7(b) ( 50 1 − ( their 0.8 ) 20 ) M1 OE Use of formula with their r SOI. 1 − their 0.8 = 247 A1 Must be to the nearest mm (not 247.1). 2 7(c) 50 M1 Use of sum to infinity formula with their r SOI. 1 − their 0.8 Substituting a value of n into the sum formula M0. = 250 A1 2
Question
The function f is defined by f x = 2 − for x > , where p is a constant. 4x − p 4 (a) Find f ′ x and hence determine whether f is an increasing function, a decreasing function or neither. [3] p b (b) Express f −1 x in the form − , where a, b, c and d are integers. [4] a cx − d (c) Hence state the value of p for which f −1 x f x. [1]
Mark scheme
8(a) 12 B2, 1, 0 f ′ ( x ) = −3 ( −1)( 4 )( 4 x − p ) = −2 ( 4 x − p ) 2 > 0 Hence increasing function B1FT Correct conclusion from their f ′ ( x ) . 3 8(b) 3 M1 OE Form horizontal equation. Sign errors only, no y =2− ( y − 2 )( 4 x − p ) = −3 or 4 xy − py = 8 x − 2 p − 3 missing terms. 4x − p 3 May go directly to 4 y = p − OE M1 M1 x−2 3 M1 OE Factorise out [ 4 ] x or [ 4] y . 4 xy − 8 x = py − 2 p − 3 4 x ( y − 2 ) = p ( y − 2 ) − 3 or 4 x = − +p x−2 3 M1 OE Make x (or y ) the subject. p ( y − 2) − 3 − +p p 3 x−2 x= = − 4 ( y − 2 ) 4 4 y − 8 x or 4 p 3 A1 OE in correct form (must be in terms of x). f −1 ( x ) = − 4 4x −8 4 8(c) [ p =]8 B1 1
Question
Functions f and g are both defined for x ∈ > and are given by f x = x2 − 4x + 9, g x = 2x2 + 4x + 12. (a) Express f x in the form x − a2 + b. [1] (b) Express g x in the form 2 x + c2 + d . [2] (c) Express g x in the form kf x + h, where k and h are integers. [1] (d) Describe fully the two transformations that have been combined to transform the graph of y = f x to the graph of y = g x. [4]
Mark scheme
9(a) ( x − 2 )2 + 5 B1 1 9(b) 2 (( x + 1) + 5) 2 B2, 1, 0 2 9(c) g ( x ) = 2f ( x + 3) or k = 2, h = 3 B1 In correct form. B0 if contradiction. 1 9(d) −3 B2, 1, 0 FT FT on their x + 3 or h = 3 . {Translation} 0 {Stretch} {y direction, factor 2} B2, 1, 0 FT FT on their 2 or k = 2 . 4 1 ( )
Question
y !Å B 4, 5 1 y = 2x 2 + 1 A 0, 1 y = 12 x2 − x + 1 x O 1 Curves with equations y = 2x 2 + 1 and y = 12 x2 − x + 1 intersect at A 0, 1 and B 4, 5, as shown in the diagram. (a) Find the area of the region between the two curves. [5] The acute angle between the two tangents at B is denoted by !Å, and the scales on the axes are the same. (b) Find !. [5]
Mark scheme
10(a) 1 *M1 ± 2 x1/2 + 1 − x 2 − x + 1 dx [ = ± 2 x1/2 − x 2 + xdx] 2 2 4 x3/2 x3 x 2 4 x3/2 x3 x 2 B2, 1, 0 OE Coefficients may be unsimplified. ± + x − − + x or ± − + 3 6 2 3 6 2 32 32 44 20 DM1 ± (F(4) – F(0)) using their integral(s). ± − + 8 or ± − 0 − + 0 3 3 3 3 =8 A1 Depends on all previous marks. If *M1 B2 DM0 and limits stated, SC B1 for +8 5 10(b) dy − 1 dy M1 A1 Attempt at differentiating one function. Upper curve: = x 2 . Lower curve: = x −1 A1 if both correct. dx dx 1 M1 Evaluate two gradients using x = 4 . At x = 4: gradient of upper curve = , gradient of lower curve = 3 2 1 M1 Use inverse tan to find angles then subtract. α = tan −1 3 − tan −1 [= 71.57 − 26.57 ] OR find equations of both tangents then Pythagoras using 2 a point on each e.g. on axes. OR cosine rule using intercepts or proportion. [α = ] 45° A1 AWRT 5
Question
y A 0, 10 B C D x O x2 + y2 = 20 The diagram shows the circle with equation x2 + y2 = 20. Tangents touching the circle at points B and C pass through the point A 0, 10. (a) By letting the equation of a tangent be y = mx + 10, find the two possible values of m. [4] (b) Find the coordinates of B and C. [3] The point D is where the circle crosses the positive x-axis. (c) Find angle BDC in degrees. [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
11(a) y − 10 2 *M1 Substitute equation of line into equation of circle. x + ( mx + 10) = 20 or y + 2 = 20 or mx + 10 = 20 − x 2 2 2 m ( ) x 2 1 + m2 + 20mx + 80 [= 0] or A1 Collect terms into a 3 term quadratic. y ( m + 1) − 20 y + (100 − 20m ) [ = 0] 2 2 2 ( 20m )2 − 4 (1 + m2 ) × 80[= 0 80m2 − 320 = 0 [80]( m2 − 4) = 0] DM1 Use b2 − 4ac [= 0] . ( )( ) or ( −20) − 4 m2 + 1 100 − 20m2 [= 0 2 [80](m4 − 4m2 ) = 0] m = ±2 A1 Two values for m . 4 11(b) Method 1: Use of quadratic (1 + 2 ) x ± 20 ( 2) x + 80 = 0 5x ± 40x + 80 = 0 2 2 2 M1 Sub their m into their quadratic in x or y or restart with their tangent equation and equation of circle. or y ( 2 + 1) − 20 y + (100 − 20 ( 2 ) ) [ = 0 [ 5 ( y − 4 y + 4 ) = 0 2 2 2 2 [5]( x ± 4)2 = 0 x = ±4 or y = 2 A1 Correct solutions or one correct pair (x, y). ( −4, 2 ) , ( 4, 2 ) A1 Two correct points with x and y paired correctly. Method 2: Using equation of normal 1 1 M1 Equate tangent and normal and solve for x . 2 x + 10 = − x or −2 x + 10 = x 2 2 x = ±4 A1 Two correct x -values or one correct pair (x, y). ( −4, 2 ) , ( 4, 2 ) A1 Two correct points with x and y paired correctly. 3 11(c) Method 1: Using angle at circumference 20 80 80 *M1 Use a trig function in triangle AOB. cos BOA = or sin BOA = or tan BOA = [ = 2] 10 10 20 BOA = 63.4° BOC = 126.8° or126.9° DM1 Strategy involving doubling [ BDC =] 63.4° A1 AWRT Metho 2: Using cosine rule ( 20 + 4) + 2 , CD = ( 20 − 4) + 2 2 2 *M1 Calculate two lengths in triangle BCD. BC = 8 , BD = 2 2 64 = 80 − 16 5 cosBDC DM1 Use cosine rule with their lengths 5 A1 AWRT cosBDC = [ BDC =] 63.4° 5 Method 3: Subtract angles from 90 ° Calculate one angle at D [ = 13.28] *M1 ODB or angle between CD and the vertical from D Calculate a second angle at D [ = 13.28] and subtract both from 90° DM1 [ BDC =] 63.4° A1 AWRT 3
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Mathematics 9709/11 for October/November 2022 is out of 75 marks, with 1 hour 50 minutes to answer it.
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