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Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s22_qp_139709_s22_ms_139709/13/M/J/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
The coefficient of x in the expansion of p + x is 144. 3 p Find the possible values of the constant p. [4]
Mark scheme
1 1 3 B1 OE 4C1× p × x soi Can be seen in an expansion. p3 4 B1 OE = 144 Correct with correct power of p and only one p term. p2 1 B1 B1 2 p=± OE ± etc. Allow ±0.167 for B1 B1. 6 12 1 SC B1 for ± B1 only, 36 4
Question
y 1 0 1 0 π 2π 3π 4π −1 −2 −3 y = p sin q1 + r −4 −5 The diagram shows part of the curve with equation y = p sin q1 + r, where p, q and r are constants. (a) State the value of p. [1] (b) State the value of q. [1] (c) State the value of r. [1]
Mark scheme
2(a) [p =] 3 B1 1 2(b) 1 B1 [q =] 2 1 2(c) [r =] ‒2 B1 1
Question
An arithmetic progression has first term 4 and common difference d . The sum of the first n terms of the progression is 5863. 11726 (a) Show that n − 1d = − 8. [1] n (b) Given that the nth term is 139, find the values of n and d , giving the value of d as a fraction. [4] @ A −1
Mark scheme
3(a) n B1 Must show a useful intermediate step. 8 + ( n − 1) d = 5863 leading to n 8 + ( n − 1) d = 11726 2 WWW AG. 11726 leading to ( n − 1) d = −8 n 1 3(b) 11726 *M1 OE Use of correct un formula with expression from (a) or 4 + ( n − 1) d = 139 leading to − 8 = 135 Sn formula to eliminate d. n 11726 A1 n= = 82 143 11726 DM1 Substitute their n into a correct un or Sn formula 81d = −8 82 5 A1 138 d= Accept OE fraction only 3 81 If M0 DM0 scored them SC B1 B1 for correct n and d values only. 4
Question
(a) The curve with equation y = x2 + 2x − 5 is translated by . 3 Find the equation of the translated curve, giving your answer in the form y = ax2 + bx + c. [3] (b) The curve with equation y = x2 + 2x − 5 is transformed to a curve with equation y = 4x2 + 4x − 5. Describe fully the single transformation that has been applied. [2] 2
Mark scheme
4(a) ( x + 1) + 2 ( x + 1) − 5 + 3 , or ( x + 1 + 1) + −6 + 3 2 2 M1 M1 −1 0 M1 for dealing with and M1 for dealing with . 0 3 [ y =] x2 + 4 x + 1 A1 Answer only given full marks. 3 4(b) {Stretch}{x direction or horizontally or y-axis invariant}{ factor ½} B2, 1, 0 Additional transformation B0. 2
Question
(a) Solve the equation 6 y + − 7 = 0. [4] y 2 (b) Hence solve the equation 6 tan x + − 7 = 0 for 0Å ≤ x ≤ 360Å. [3] tan x
Mark scheme
5(a) 6 y + 2 − 7 y1/2 [= 0] *M1 OE Rearrange to a 3-term quadratic. 1 1 DM1 Or use of formula or completing the square. 2 y 2 − 1 3 y 2 − 2 [= 0] or e.g. ( 2u − 1)( 3u − 2 ) [= 0] 1 2 A1 Answers only SC B1 if DM1 not scored. [ y1/2 =] , 2 3 1 4 A1 Answers only SC B1 if DM1 not scored. [ y =] , 4 9 4 5(b) Use of tan x = their y values M1 Must have at least 2 values of y from part (a). x = 14[.0], 24[.0], A1 FT for 180 + angle (twice). x = 194[.0], 204[.0] A1 FT AWRT 3
Question
The function f is defined by f x = 2x2 − 16x + 23 for x < 3. (a) Express f x in the form 2 x + a2 + b. [2] (b) Find the range of f. [1] (c) Find an expression for f −1 x. [3] The function g is defined by g x = 2x + 4 for x < −1. (d) Find and simplify an expression for fg x. [2]
Mark scheme
6(a) 2 ( x − 4) −9 2 B1 B1 OE When a and b stated give priority to marking algebraic expression. 2 6(b) y > −7 B1 Allow f(x) > –7 or (–7, ∞) Don’t allow x > –7. 1 6(c) y+9 M1 2 operations correct. Allow a sign error. ( x − 4 )2 = 2 y+9 M1 2 operations correct. Allow a sign error. x = 4 [±] 2 x+9 A1 FT x −b [f −1 ( x ) =] 4 − OE FT on their answer to (a) i.e. – a − . 2 2 3 6(d) fg(x) = f(2x + 4) = 2 ( 2 x + 4 − 4 ) − 9 M1 Allow 2 ( 2 x + 4 )2 − 16 ( 2 x + 4 ) + 23 . 2 8 x 2 − 9 only A1 2
Question
y B 0, 2 O P x C x − 22 + y + 42 = 20 The diagram shows the circle with equation x − 22 + y + 42 = 20 and with centre C. The point B has coordinates 0, 2 and the line segment BC intersects the circle at P. (a) Find the equation of BC. [2] (b) Hence find the coordinates of P, giving your answer in exact form. [5]
Mark scheme
7(a) Equation of BC is y =2−3 x B2, 1, 0 OE forms y + 4 = −3 ( x − 2 ) or y − 2 = −3 ( x − 0 ) . 2 7(b) ( x − 2 ) + ( 2 − 3x + 4 ) = 20 *M1 OE 2 2 Sub line equation into equation of circle to eliminate y. 10(x – 2)2 = 20 or [10](x2 – 4x + 2)[= 0] A1 OE Accept (10x2 – 40x + 20). 4 [ ± ] 16 − 8 DM1 Correctly solving their quadratic. x − 2 = [ ± ] 2 or x = 2 x =2− 2 A1 OE only solution. Answer only SC B1 If DM1 not scored. y =3 2 −4 A1 OE only solution. Answer only SC B1 If DM1 not scored. 5
Question
y 1 1 y = x 2 + 4x− 2 A 1, 5 B 16, 5 x O −1 The diagram shows the curve with equation y = x 2 + 4x 2 . The line y = 5 intersects the curve at the 1 points A 1, 5 and B 16, 5. (a) Find the equation of the tangent to the curve at the point A. [4] (b) Calculate the area of the shaded region. [4]
Mark scheme
8(a) dy −1/2 B1 B1 Allow unsimplified versions. dx = ½ x − 2 x −3/2 dy 1 3 M1 Substitute x = 1 into a differentiated y. At x = 1, = − 2 =− dx 2 2 3 A1 3 13 Equation of tangent is y − 5 = − ( x − 1) WWW Or y = − x + . 2 2 2 4 8(b) x 3/ 2 B1 OE Integrate to find area under curve, allow unsimplified + 8 x1/ 2 versions. 3/ 2 128 2 M1 Apply limits 1 → 16 to an integrated expression. 3 + 32 − 3 + 8 Area under line = 15 × 5 = 75 B1 16 Or by 5dx . 1 Required area = 75 ‒ 66 = 9 A1 4
Question
D B 1.8 rad 6 cm 6 cm C A The diagram shows triangle ABC with AB = BC = 6 cm and angle ABC = 1.8 radians. The arc CD is part of a circle with centre A and ABD is a straight line. (a) Find the perimeter of the shaded region. [5] (b) Find the area of the shaded region. [3]
Mark scheme
9(a) AC M1 OE Correct working in degrees is acceptable throughout. 6sin 0.9 = or AC 2 = 62 + 6 2 − 2 × 6 × 6cos1.8 2 AC = 9.40 A1 SOI Accept 9.39 – 9.41, may be used but not seen for A1. Angle CAB = ½(π ‒ 1.8) M1 SOI Expect 0.6708 (or 0.671). Arc CD = their 9.40 × their 0.6708 M1 Expect 6.306 (or 6.31), do not accept 6 for their AC or 1.8 for CAB. [Perimeter = 6 + 3.40 + 6.306 =] 15.7 A1 Accept 15.69 – 15.72. 5 9(b) Sector ADC ‒ ΔABC = ½ × their 9.402 × their 0.6708 – ½ × 62 × sin 1.8 M1 M1 Accept correct use of their answers from part (a). [29.64 ‒ 17.53 =] 12.1 A1 AWRT 3
Question
The function f is defined by f x = 4x + 2−2 for x > − 12 . ∞ (a) Find Ó f x dx. [4] 1 A point is moving along the curve y = f x in such a way that, as it passes through the point A, its y-coordinate is decreasing at the rate of k units per second and its x-coordinate is increasing at the rate of k units per second. (b) Find the coordinates of A. [6]
Mark scheme
10(a) ( 4 x + 2 )−1 B1 B1 OE If more than one function of x present then B0 B0. −1 1 ÷4 or eg − ( x + 0.5 ) 16 −1 or −1 (16 x + 8) 0 − (−1 / 24) M1 Apply limits to an integral, ∞ must be used correctly. 1/24 A1 Allow 0.0417 AWRT. 4 10(b) dy dx = −2 ( 4 x + 2 ) −3 ×4 B1 B1 Allow unsimplified forms. dy B1 SOI Recognise = ‒1 dx −8 M1 Must be numerical. their = their − 1 dy ( 4x + 2)3 Must be some attempt to solve their equation and ≠0. dx (0, ¼) A1 A1 Accept x = 0, y = ¼. y = ¼ must be from x = 0 not x = –1. 6
Question
The point P lies on the line with equation y = mx + c, where m and c are positive constants. A curve m has equation y = − . There is a single point P on the curve such that the straight line is a tangent to x the curve at P. (a) Find the coordinates of P, giving the y-coordinate in terms of m. [6] The normal to the curve at P intersects the curve again at the point Q. (b) Find the coordinates of Q in terms of m. [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
11(a) m M1 All x terms in the numerator. OE e.g. mx 2 + cx = − m . mx + c = − mx 2 + cx + m = 0 x b 2 − 4ac = 0 c 2 − 4m 2 = 0 M1 OE b2 – 4ac = 0 is implied by c2 – 4m2 = 0. c = [ ± ] 2m A1 SOI. Allow ± at this stage. mx 2 [ ± ] 2mx + m = 0 x 2 [ ± ] 2 x + 1 = 0 M1 Sub c = +2m Ignore substitution of -2m. ( x + 1) = 0 x = −1 only A1 2 y = m only or (‒1, m) only A1 Alternative method to question 11(a) dy m M1 As this is a method mark a sign error is allowed. = dx x 2 m M1 A1 dy 2 = m x2 = 1 Equating their and m and attempt to solve. x dx x = ±1 or x = −1 A1 If x = −1 and y = m are the only answers offered here award the final M1 A1. Selecting x = –1 as the only answer and attempt to find y M1 y = m or (‒1, m) A1 6 11(b) −1 *M1 −1 Equation of normal is y − m = ( x + 1) Through their P with gradient , OE m m −1 m2 − 1 e.g. y = x+ . m m 1 Allow use of the gradient of the curve as − with m ( their x )2 their P. Coordinates of P must be in terms of m only. −x 1 −m DM1 OE Equating their normal equation to the equation of the − +m= m m x ( ) x 2 + x 1 − m 2 − m 2 [ = 0] curve and removing x from the denominator. ( x + 1) ( x − m 2 ) [ = 0] x = m 2 A1 or x= 2 m 2 − 1 ± 1 − 2m 2 + m 4 + 4m 2 m − 1 ± m + 1 = ( = m2 2 ) 2 2 −m −1 A1 −1 y= = or m2 , , ignore the coordinates of P. m2 m m 4
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Mathematics 9709/13 for May/June 2022 is out of 75 marks, with 1 hour 50 minutes to answer it.
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