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Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s21_qp_139709_s21_ms_139709/13/M/J/21
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
A curve with equation y = f x is such that f ′ x = 6x2 − . It is given that the curve passes through x2 the point 2, 7. Find f x. [3] 3
Mark scheme
1 8 B1 Allow any correct form f ( x ) = 2 x + x [ +c ] 3 7 = 16 + 4 + c M1 Substitute f(2) = 7 into an integral. c must be present. Expect c = ‒13 8 A1 Allow y = , f ( x ) or y can appear earlier in f ( x ) = 2 x3 + − 13 x answer 3
Question
The function f is defined by f x = 13 2x − 1 2 − 2x for 21 < x < a. It is given that f is a decreasing function. Find the maximum possible value of the constant a. [4]
Mark scheme
2 B2, 1, 0 Expect ( 2 x − 1)1/ 2 − 2 f ( x ) = −1 (( 2 x − 1)1/2 ) × × 2 × 1 3 3 ( −2 ) 2 ( 2 x − 1)1/ 2 − 2 0 → 2 x − 1 4 or 2 x − 1 < 4 M1 SOI. Rearranging and then squaring, must have power of ½ not present Allow ‘=0’at this stage but do not allow ‘ ≥ 0’ or ‘ ˃ 0’ If ‘–2’ missed then must see ⩽ or < for the M1 Value [of a] is 2½ or a = 2½ A1 5 WWW, OE e.g. , 2.5 2 Do not allow from ‘=0’ unless some reference to negative gradient. 4
Question
A line with equation y = mx − 6 is a tangent to the curve with equation y = x2 − 4x + 3. Find the possible values of the constant m, and the corresponding coordinates of the points at which the line touches the curve. [6]
Mark scheme
3 x 2 − 4 x + 3 = mx − 6 leading to x 2 − x ( 4 + m ) + 9 *M1 Equating and gathering terms. May be implied on the next line. b 2 − 4ac leading to ( 4 + m ) − 4 × 9 2 DM1 SOI. Use of the discriminant with their a, b and c 4 + m = ±6 or ( m − 2 )( m + 10 ) = 0 leading to m = 2 or −10 A1 Must come from b 2 − 4ac = 0 SOI Substitute both their m values into their equation in line 1 DM1 m = 2 leading to x =3 ; m = ‒10 leading to x = ‒3 A1 (3, 0), (‒3, 24) A1 Accept 'when x = 3, y = 0; when x = ‒3, y = 24' If final A0A0 scored, SC B1 for one point correct WWW Alternative method for Question 3 dy *M1 = 2x − 4 → 2x − 4 = m dx x2 − 4 x + 3 = ( 2 x − 4) x − 6 DM1 x 2 − 4 x + 3 = 2 x 2 − 4 x − 6 → 9 = x 2 → x = ±3 A1 y = 0, 24 or (3, 0), (‒3, 24) A1 Substitute both their x values into their equation in line 1 DM1 Or substitute both their ( x, y ) into y = mx − 6 When x = 3, m = 2; when x = ‒3, m = ‒10 A1 If A0, DM1, A0 scored, SC B1 for one point correct WWW 6
Question
(a) Show that the equation tan x + sin x = k, tan x − sin x where k is a constant, may be expressed as 1 + cos x = k. [2] 1 − cos x (b) Hence express cos x in terms of k. [2] tan x + sin x (c) Hence solve the equation = 4 for −π < x < π. [2] tan x − sin x
Mark scheme
4(a) tanx + sinx sinx + sinxcosx M1 Multiply numerator and denominator by cos x , [ = k ] leading to [= k ] or divide numerator and denominator by tan x or tanx − sinx sinx − sinxcosx 1 sin x +1 tanx + tanxcosx or cos x [=k] or [=k] 1 tanx − tanxcosx −1 cos x 1 A1 AG, WWW sinx (1 + cosx ) cos x + 1 cos x tanx (1 + cosx ) 1 + cosx or . or leading to [= k ] sinx (1 − cosx ) 1 − 1 cos x tanx (1 − cosx ) 1 − cosx cos x 2 4(b) k − k cos x = 1 + cos x leading to k − 1 = k cos x + cos x M1 Gather like terms on LHS and RHS k −1 A1 WWW, OE k − 1 = ( k + 1) cos x leading to cos x = k +1 2 4(c) Obtaining cos x from their (b) or (a) M1 3 Expect cos x = 5 ±0.927 (only solutions in the given range) A1 AWRT. Accept ±0.295π 2
Question
A D 4 cm B C The diagram shows a triangle ABC, in which angle ABC = 90Å and AB = 4 cm. The sector ABD is part of a circle with centre A. The area of the sector is 10 cm2. (a) Find angle BAD in radians. [2] (b) Find the perimeter of the shaded region. [4]
Mark scheme
5(a) ½ × 42 × angle BAD = 10 M1 Use of sector area formula Angle BAD = 1.25 A1 OE. Accept 0.398π, 71.6o for SC B1 only 2 5(b) Arc BD = 4 × their1.25 M1 Use of arc length formula. Expect 5. BC = 4 tan ( their1.25) M1 Expect 12.0(4). May use ACB=0.321 or 18.4o 4 M1 Expect 12.69 ‒ 4 = 8.69. May use ACB. 42 + ( their BC ) − 4 2 CD = − 4 or cos ( their1.25 ) Perimeter = 5 + 12.0(4) + 8.69 = 25.7 (cm) A1 AWRT 4
Question
Functions f and g are both defined for x ∈ > and are given by f x = x2 − 2x + 5, g x = x2 + 4x + 13. (a) By first expressing each of f x and g x in completed square form, express g x in the form f x + p + q, where p and q are constants. [4] (b) Describe fully the transformation which transforms the graph of y = f x to the graph of y = g x. [2]
Mark scheme
6(a) f ( x ) = ( x − 1) + 4 2 B1 g ( x ) = ( x + 2) + 9 2 B1 g ( x ) = f ( x + 3) + 5 B1 B1 B1 for each correct element. Accept p = 3, q = 5 4 6(b) Translation or Shift B1 −3 B1 FT If given as 2 single translations both must be or acceptable explanation −3 0 5 described correctly e.g. & 0 5 FT from their f ( x + p ) + q or their f ( x) → g ( x) 1 −2 Do not accept or 4 9 2
Question
(a) Write down the first four terms of the expansion, in ascending powers of x, of a − x6 . [2] @ A 2 (b) Given that the coefficient of x2 in the expansion of 1 + a − x6 is −20, find in exact form ax the possible values of the constant a. [5]
Mark scheme
7(a) ( a − x )6 = a 6 − 6a5 x + 15a 4 x 2 − 20a3 x3 + … B2, 1, 0 Allow extra terms. Terms may be listed. Allow a 6 x0 . 2 7(b) 2 M1 Attempting to find 2 terms in x 2 ( ) ( 1 + …15a x − 20a x + … leading to x 15a − 40a ax 4 2 3 3 2 4 2 ) 15a 4 − 40a 2 = −20 leading to 15a 4 − 40a 2 + 20 [ = 0] A1 Terms on one side of the equation ( 5a − 10 )( 3a − 2 ) [ = 0] 2 2 M1 OE. M1 for attempted factorisation or solving for a2 or u (=a2) using e.g. formula or completing the square 2 B1 B1 OE exact form only a = ± 2, ± 3 2 If B0B0 scored then SC B1 for 2, WWW 3 or ±1.41,±0,816 WWW 5
Question
Functions f and g are defined as follows: f : x → x2 − 1 for x < 0, 1 g:x→ for x < − 12 . 2x + 1 (a) Solve the equation fg x = 3. [4] (b) Find an expression for fg−1 x. [3]
Mark scheme
8(a) fg ( x ) = 1 / ( 2 x + 1) − 1 2 B1 SOI 1 / ( 2 x + 1) − 1 = 3 leading to 4 ( 2 x + 1) = 1 2 2 M1 Setting fg ( x ) = 3 and reaching a stage before 1 2 x + 1 = ±½ or reaching a 3 term quadratic in or = [ ± ] 2 or 16 x 2 + 16 x + 3 = 0 x ( 2 x + 1) 2 x + 1 = ±½ or 2 x + 1 = −½ or ( 4 x + 1)( 4 x + 3) [ = 0] A1 Or formula or completing square on quadratic 3 A1 x=− only 4 Alternative method for Question 8(a) x2 − 1 = 3 M1 g ( x ) = −2 A1 1 M1 = −2 ( 2 x + 1) 3 A1 x=− only 4 4 8(b) 1 1 1 *M1 Obtain 2 x + 1 or 2 y + 1 as the subject − 1 leading to ( 2 x + 1) = leading to 2 x + 1 = [ ± ] 2 y= ( 2 x + 1) 2 y +1 y +1 1 1 DM1 Make x ( or y ) the subject x = [±] − 2 y +1 2 1 1 A1 x +1 1 −x 1 1 − − OE e.g. – – , – + + 2 x +1 2 2x + 2 2 4x + 4 4 2 3
Question
(a) A geometric progression is such that the second term is equal to 24% of the sum to infinity. Find the possible values of the common ratio. [3] (b) An arithmetic progression P has first term a and common difference d . An arithmetic progression Q has first term 2 a + 1 and common difference d + 1. It is given that 5th term of P 1 Sum of first 5 terms of P 2 = and = . 12th term of Q 3 Sum of first 5 terms of Q 3 Find the value of a and the value of d . [6]
Mark scheme
9(a) 24 a M1 Form an equation using a numerical form of the ar = × percentage and correct formula for u2 and S∞ 100 1 − r 100r 2 − 100r + 24 [ = 0] A1 OE. All 3 terms on one side of an equation. 2 3 A1 Dependent on factors or formula seen from their ( 20r − 8)( 5r − 3) [ = 0] → r = , quadratic. 5 5 3 9(b) { } 3 × {( a + 4 d )} = ( 2 ( a + 1) + 11( d + 1) ) *M1 SOI Attempt to cross multiply with contents of at least one { } correct Simplifies to a + d = 13 A1 5 5 *M1 SOI Attempt to cross multiply with contents of { } 2 × 3{( 2a + 4d )} = 2 × 2 ( 4 ( a + 1) + 4 ( d + 1) ) at least one { } correct Simplifies to −a + 2d = 8 A1 Solve 2 linear equations simultaneously DM1 Elimination or substitution expected d = 7, a = 6 A1 SC B1 for a=6, d=7 without complete working 6
Question
Points A −2, 3, B 3, 0 and C 6, 5 lie on the circumference of a circle with centre D. (a) Show that angle ABC = 90Å. [2] (b) Hence state the coordinates of D. [1] (c) Find an equation of the circle. [2] The point E lies on the circumference of the circle such that BE is a diameter. (d) Find an equation of the tangent to the circle at E. [5]
Mark scheme
10(a) 3 5 M1 Attempting to find required gradients, sides or Gradient of AB = − , gradient of BC = or lengths of all 3 sides or vectors vectors 5 3 ρρ mab mbc = −1 or Pythagoras or AB.BC =0 or cos ABC = 0 from cosine rule A1 WWW 2 10(b) Centre = mid-point of AC = (2,4) B1 1 10(c) ( x − their x c ) + ( y − their yc ) = r 2 or ( their xc − x ) + ( their yc − y ) = r 2 2 2 2 2 M1 Use of circle equation with their centre ( x − 2 )2 + ( y − 4 )2 = 17 A1 Accept x 2 − 4 x + y 2 − 8 y + 3 = 0 OE 2 10(d) x+3 y+0 −1 M1 Use of mid-point formula, vectors, steps on a , = ( 2, 4 ) or BE = 2BD = 2 4 diagram 2 2 Or Equation of BE is y = −4 ( x − 3) or y − 4 = −4 ( x − 2 ) leading to y = −4 x + 12 May be seen to find x coordinate at E Substitute equation of BE into circle and form a 3-term quadratic. 3 −2 1 A1 E = (1, 8) ( x, y ) = (1,8) or OE = + = Accept without working for both marks SC B2 0 8 8 1 B1 Or gradient of BE = -4 Gradient of BD, m, = ‒4 or gradient AC = = gradient of tangent 4 Equation of tangent is y − 8 = ¼ ( x − 1) OE M1 A1 For M1, equation through their E or (1, 8) (not, −1 A, B or C) and with gradient their − 4 5
Question
y 1 1 y = x 2 + k2 x− 2 x O 9 2 4k2 4k 1 −1 The diagram shows part of the curve with equation y = x 2 + k2 x 2 , where k is a positive constant. (a) Find the coordinates of the minimum point of the curve, giving your answer in terms of k. [4] The tangent at the point on the curve where x = 4k2 intersects the y-axis at P. (b) Find the y-coordinate of P in terms of k. [4] The shaded region is bounded by the curve, the x-axis and the lines x = 94 k2 and x = 4k2 . (c) Find the area of the shaded region in terms of k. [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) dy 1 −1/2 1 2 −3/2 B1 B1 Allow any correct unsimplified form = x − k x dx 2 2 1 −1/2 1 2 −3/2 1 1 M1 OE. Set to zero and one correct algebraic step x − k x = 0 leading to x −1/2 = k 2 x −3/2 towards the solutions. 2 2 2 2 dy must only have 2 terms. dx ( k , 2k ) 2 A1 4 11(b) dy 1 1 3 B1 OE When x = 4k2, = − = dx 4k 16k 16k 1 5k B1 k y = 2k + k 2 × = OE. Accept 2k + 2k 2 2 5k 3 5k 3 M1 Use of line equation with their gradient and Equation of tangent is y − = 2 16k ( ) x − 4k 2 or y = mx + c → = 2 16k 4k 2 + c( ) ( 4k 2 , their y ) , 5k 3k 7k 7k A1 OE When x = 0, y = − = or from y = mx + c, c = 2 4 4 4 4 11(c) 3 B1 Any unsimplified form 1 2 − 1 2x 2 2 2 1 x + k x dx = 2 2 + 2k x 3 16k 3 9k 3 M1 9 2 + 4k 3 − + 3k 3 Apply limits k → 4k 2 to an integration of y. 4 3 4 M0 if volume attempted. 49k 3 A1 OE. Accept 4.08 k 3 12 3
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Mathematics 9709/13 for May/June 2021 is out of 75 marks, with 1 hour 50 minutes to answer it.
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