Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 1 · Variant 1

2 1 Solve the equation 3x + 2 = x −1. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 ( 3x + 2 )( x − 1) = 2  3x 2 −−x 4  = 0 M1 OE Multiply by denominator and obtain a quadratic. ( 3 x − 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

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dy 12 The equation of a curve is such that = 12 −1 −4. It is given that the curve passes through the 2x 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] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: −42(a) M1 d y 3  1   −4 3  SOI by gradient used. Substitute x = 6 into −6 1 = 12 ( 2 ) = 12     4 d x  2   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)  −3  B2, 1, 0  1    12  x − 1   −3   1   2    1 x − 1  y =    = −8     −3  2   2         −3 M1 Must have +c .  1  12    6 − 1  Substitute y = 4, x = 6 and solve for c in an integrated  2  −3 4 = + c   4 = −8 2 + c   c = 5 expression. May be unsimplified.   1 − 3 2 −3 A1 OE Must see ‘ y = ’ or ‘ f ( x ) = ’ in the working.  1  x − 1 + 5  y =  − 8    2  4

More questions on Integration

1 3 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] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 dy 1 − 12 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 = 1 2 A1  1  A1 FT FT on their a. 2 − 18 = 9 )  y = their a  (  18   5 .4 2  2  5  2  2 10  2 2  40  B1 Accept with x 2 present. Must evaluate 5C2 Coefficient of x in  1 + x  is 10   = 2  = 2   p   p  p  p  Coefficient of x 2 in (1 + px ) 6 is 15 ( p ) 2  = 15 p 2  B1 Accept with x 2 present. Must evaluate 6C2   40 2 *M1 Forming an equation in p with their coefficients, the + 15 p = 70 p 2 given 70, no x terms and no extra terms. 15 p 4 − 70 p 2 + 40  = 0 or 3 p 4 − 14 p 2 + 8  = 0 DM1 Forming a 3-term equation in p (or another variable) with all terms on one side and their coefficients. 2 DM1 Attempt to solve 3-term quartic (or quadratic in another 70  70 − 4 (15 )( 40 ) 2 2 5 p − 4 3 p − 2 = 0  or or variable) by factorisation, formula or completing the ( )( )  30 square. 14  14 2 − 4 ( 3 )( 8 ) 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

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5 8 cm A B O The diagram shows a sector OAB of a circle with centre O. The length of the arc AB is 8cm. It is given that the perimeter of the sector is 20cm. (a) Find the perimeter of the shaded segment. 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(b) Find the area of the shaded segment. 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Mark scheme: 5(a)  2 r + 8 = 20   r = 6 B1 8 *M1 4 Angle AOB = Expect OE ( 76.4.) their 6 3 M0 Assume triangle is equilateral. 2 2 2 2 4 DM1 For 6 read their 6 . AB = 2  6sin their or 6 + 6 −2 6 cos their 3 3 6 4 or AB =  sin their   2  3 sin  − their   2 3  Perimeter =  7.42 + 8 =  15.4 A1 AWRT 4 5(b) 1 2 4 1 2 4 M1 Sector area – whole triangle area. Area =  6  their −  6  sin their 2 3 2 3 For 6 read their 6 . 1 2 4 1  2  2  or Area =  6  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

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6 (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. 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[3] sin 1 + cos 1 sin 1 −cos 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) sin− cos+ sin+ cos  sin− cos+ sin+ cos *M1 Use common denominator and equate to 1. = = 1 ( sin+ cos)( sin− cos)  sin 2− cos 2  DM1 Multiply by common denominator and replace cos2  by 2sin [ = sin 2− cos 2] = sin 2− 1 − sin 2 ( ) 1 – sin2 . 2sin 2− 2sin−=1 0 A1 OE In the given form. 3 6(b) 2 M1 Use formula or complete the square to solve a quadratic 2  ( −2 ) − 4 ( 2 )( −1)  2  4 + 8 1  3  [sin = ]  = =  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

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7 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 50mm, 40mm and 32mm respectively. (a) Verify that the 9th impact is the first in which the post sinks less than 10mm into the ground. 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(b) Find, to the nearest millimetre, the total depth of the post in the ground after 20 impacts. 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(c) Find the greatest total depth in the ground which could theoretically be achieved. 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Mark scheme: 7(a) r = 0.8 B1 SOI 7 M1 Evaluate 8th or 9th term in GP. 50  ( their 0.8 ) = 10.5 8 A1 AG Two terms correct + conclusion (mention of 9th 50  ( their 0.8 ) = 8.39 . Hence 9th impact required 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 n n M1 50  ( their 0.8 )  10  ( their 0.8 )  0.5 n log ( their 0.8 )  log0.5 log0.5 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) 20 M1 OE Use of formula with their r SOI. 50 1 − ( their 0.8 ) ( ) 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. Substituting a value of n into the sum formula M0. 1 −their 0.8 = 250 A1 2

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10 y !Å B 4, 5 1 2 + 1 y = 2x A 0, 1 1 y = 2x2 −x + 1 x O 1 Curves with equations y = 2x 2 + 1 and y = 12x2 −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. 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The acute angle between the two tangents at B is denoted by !Å, and the scales on the axes are the same. (b) Find !. 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Mark scheme: 10(a) 1/2  1 2  1/2 1 2 *M1  2 x + 1 − x − x + 1 dx [ =  2 x − x + xdx ] ( )    2  2  4 x 3/2  x 3 x 2    4 x 3/2 x 3 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) d y − 12 dy M1 A1 Attempt at differentiating one function. Upper curve: = x . Lower curve: = x − 1 A1 if both correct. d x dx 1 M1 Evaluate two gradients using x = 4 . At x = 4: gradient of upper curve = , gradient of lower curve = 3 2 −1 −1 1 M1 Use inverse tan to find angles then subtract. = tan 3 − tan  = 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. =  45  A1 AWRT 5

More questions on Integration

11 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. 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(b) Find the coordinates of B and C. 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The point D is where the circle crosses the positive x-axis. (c) Find angle BDC in degrees. 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Mark scheme: 11(a) 2 *M1 Substitute equation of line into equation of circle. 2 2 2  y − 10  2 = 20 or mx + 10 = 20 − x x + ( mx + 10 ) = 20 or y +    m  2 2 A1 Collect terms into a 3 term quadratic. x 1 + m + 20mx + 80  = 0 or ( ) y 2 m 2 + 1 − 20 y + 100 − 20 m 2 = 0  ( ) ( ) 1 + m 2 m 2 − 4 = 0] DM1 Use b2 − 4ac  = 0 . ( 20 m ) 2 − 4 ( )  80[ = 0  80m 2 − 320 = 0  80 ( ) m 2 + 1 100 − 20m 2 [ = 0  80 ( m4 − 4m2 ) = 0] or ( −20 ) 2 − 4 ( )( ) m = 2 A1 Two values for m . 4 11(b) Method 1: Use of quadratic M1 Sub their m into their quadratic in x or y or restart with 1 + 2 2 = 0  5 x 2  40 x + 80 = 0  x 2  20 ( 2 ) x + 80  ( )   their tangent equation and equation of circle. 2 2 + 1 − 20 y + 100 − 20 2 2 y 2 − 4 y + 4 = 0  or y 2 = 0  5  ( ) ( ) ( ) )  (  (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 2 2 *M1 Calculate two lengths in triangle BCD. BC = 8 , BD = 20 + 4 + 2 2 , CD = 20 − 4 + 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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