Free until 14 Nov · 80% off for the first 100 · Pricing
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w20_qp_119709_w20_ms_119709/11/O/N/20
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
Find the set of values of m for which the line with equation y = mx − 3 and the curve with equation y = 2x2 + 5 do not meet. [3] dy 1
Mark scheme
1 2 x + 5 = mx − 3 → 2 x − mx + 8 ( = 0 ) 2 2 B1 Form 3-term quadratic m 2 − 64 M1 Find b 2 − 4 ac . −8 < m < 8 A1 Accept (-8, 8) and equality included 3
Question
The equation of a curve is such that = + x. It is given that the curve passes through the dx x − 32 point 2, 7. Find the equation of the curve. [4]
Mark scheme
2 B1 B1 ( y = ) − ( x − 3) + −1 1 2 x ( +c ) 2 M1 Substitute x = 2, y = 7 into an integrated expansion (c present). 7=1+2+c Expect c = 4 −1 1 A1 OE y = − ( x − 3) + x 2 +4 2 4
Question
Air is being pumped into a balloon in the shape of a sphere so that its volume is increasing at a constant rate of 50 cm3 s−1 . Find the rate at which the radius of the balloon is increasing when the radius is 10 cm. [3]
Mark scheme
3 B1 dr (Derivative =) 4πr 2 (→ 400π) SOI Award this mark for dV 50 M1 Can be in terms of r their derivative 1 A1 AWRT or 0.0398 8π 3
Question
y 5 4 3 2 1 y = cos 1 1 0 π 2π 3π 4π −1 In the diagram, the lower curve has equation y = cos 1. The upper curve shows the result of applying a combination of transformations to y = cos 1. Find, in terms of a cosine function, the equation of the upper curve. [3] 0 a 16
Mark scheme
4 B1 B1 ( y = ) [3] + [ 2] cos θ 1 B1 2 3
Question
In the expansion of 2x2 + , the coefficients of x6 and x3 are equal. x (a) Find the value of the non-zero constant a. [4] 0 a 16 (b) Find the coefficient of x6 in the expansion of 1 − x3 2x2 + . [1] x
Mark scheme
5(a) a 2 3 a 3 B1 B1 SOI Can be seen in an expansion ( ) ( ) 4 6C 2 × 2 x 2 × , 6C 3 × 2 x 2 × ( x ) ( x ) 15 × 2 4 × a 2 = 20 × 23 × a 3 M1 SOI Terms must be from a correct series 15 × 24 3 A1 OE a= = 20 × 23 2 4 5(b) 0 B1 1
Question
The equation of a curve is y = 2 + 25 − x2 . Find the coordinates of the point on the curve at which the gradient is 34 . [5] sin 1 sin 1
Mark scheme
6 dy 1 −1/2 B1 B1 = dx 2 ( 25 − x 2 ) × [ −2 x ] −x 4 x2 16 M1 4 = → = Set = and square both sides ( 25 − x ) 2 1/2 3 25 − x 2 9 3 ( ) 16 25 − x 2 = 9 x 2 → 25 x 2 = 400 → x = ( ± ) 4 A1 When x = ‒ 4, y = 5 → (‒ 4, 5) A1 5
Question
(a) Show that − 2 tan2 1. [3] 1 − sin 1 1 + sin 1 sin 1 sin 1 (b) Hence solve the equation − = 8, for 0Å < 1 < 180Å. [3] 1 − sin 1 1 + sin 1
Mark scheme
7(a) sin θ sin θ sin θ (1 + sin θ ) − sin θ (1 − sin θ ) *M1 − = Put over a single common denominator 1 − sin θ 1 + sin θ 1 − sin 2θ 2sin 2θ DM1 Replace 1 − sin 2 θ by cos 2 θ and simplify numerator cos 2θ 2tan 2θ A1 AG 3 7(b) 2 tan 2 θ = 8 → tanθ = ( ± ) 2 B1 SOI B1 FT on 180 – 1st solution (with justification) (θ = ) 63.4°, 116.6° B1 FT 3
Question
A geometric progression has first term a, common ratio r and sum to infinity S. A second geometric progression has first term a, common ratio R and sum to infinity 2S. (a) Show that r = 2R − 1. [3] It is now given that the 3rd term of the first progression is equal to the 2nd term of the second progression. (b) Express S in terms of a. [4]
Mark scheme
8(a) a a B1 S= , 2S = SOI at least one correct 1− r 1− R 2a a M1 = SOI 1− r 1− R 2 − 2R = 1 − r → r = 2R − 1 A1 AG 3 8(b) ar 2 = aR → ( a )( 2 R − 1) = R ( a ) 2 *M1 4 R 2 − 5 R + 1 ( = 0 ) → ( 4 R − 1)( R − 1) ( = 0 ) DM1 Allow use of formula or completing square. 1 A1 R= Allow R = 1 in addition 4 2a A1 S= 3 Alternative method for question 8(b) ar 2 = aR → ( a ) r 2 = ½ ( r + 1)( a ) *M1 Eliminating 1 variable DM1 Allow use of formula or completing square. Must solve a 2r 2 − r − 1 ( = 0 ) → ( 2r + 1)( r − 1) ( = 0 ) quadratic. 1 A1 r=− Allow r = 1 in addition 2 2a A1 S= 3 4
Question
C A B D The diagram shows a circle with centre A passing through the point B. A second circle has centre B and passes through A. The tangent at B to the first circle intersects the second circle at C and D. The coordinates of A are −1, 4 and the coordinates of B are 3, 2. (a) Find the equation of the tangent CBD. [2] (b) Find an equation of the circle with centre B. [3] (c) Find, by calculation, the x-coordinates of C and D. [3]
Mark scheme
9(a) 4−2 1 B1 mAB = =− −1 − 3 2 B1 FT 1 Equation of tangent is y − 2 = 2 ( x − 3) (3, 2) with their gradient − mAB 2 9(b) AB 2 = 42 + 2 2 = 20 or r 2 = 20 or r = 20 or AB = 20 B1 Equation of circle centre B is ( x − 3) + ( y − 2 ) = 20 2 2 M1 A1 FT their 20 for M1 3 9(c) ( x − 3)2 + ( 2 x − 6 )2 = their 20 M1 Substitute their y − 2 = 2 x − 6 into their circle, centre B 2 5 x 2 − 30 x + 25 = 0 or 5 ( x − 3) = 20 A1 ( 5)( x − 5)( x − 1) or x − 3 = ±2 x = 5, 1 A1 3
Question
C 1 rad F E r O A B D The diagram shows a sector CAB which is part of a circle with centre C. A circle with centre O and radius r lies within the sector and touches it at D, E and F, where COD is a straight line and angle ACD is 1 radians. (a) Find CD in terms of r and sin 1. [3] It is now given that r = 4 and 1 = 16 π. (b) Find the perimeter of sector CAB in terms of π. [3] (c) Find the area of the shaded region in terms of π and 3. [4]
Mark scheme
10(a) r r M1 A1 sin θ = → OC = OC sin θ r A1 CD = r + sin θ 3 10(b) 4 B1 SOI Radius of arc AB = 4 + = 4 + 8 = 12 π sin 6 2π 1 π M1 Expect 4π, must use their CD, not 4 (Arc AB =) their 12 × or AB = their 12 × 6 2 6 Perimeter = 24 + 4π A1 3 10(c) 1 π M1 Area FOC = × 4 × their OC × sin 2 3 8 3 A1 1 2π 2 16π B1 Area sector FOE = × ×4 = 2 3 3 16π A1 Shaded area = 16 3 − 3 Alternative method for question 10(c) M1 48 or 4 3 FC = ( their OC )2 − 42 1 A1 Area FOC = × 4 × 4 3 = 8 3 2 1 π 2 8π B1 Area of half sector FOE = × ×4 = 2 3 3 16π A1 Shaded area = 16 3 − 3 4
Question
The functions f and g are defined by f x = x2 + 3 for x > 0, g x = 2x + 1 for x > − 12 . (a) Find an expression for fg x. [1] (b) Find an expression for fg−1 x and state the domain of fg−1 . [4] (c) Solve the equation fg x − 3 = gf x. [4]
Mark scheme
11(a) fg(x) = ( 2 x + 1) + 3 2 B1 OE 1 11(b) y = ( 2 x + 1) + 3 → 2 x + 1 = ( ± ) y − 3 2 M1 1st two operations. Allow one sign error or x/y interchanged 1 M1 OE 2nd two operations. Allow one sign error or x/y interchanged x = (±) 2 ( y − 3 −1) A1 B1 Allow (3, ∞) ( fg ( x ) = ) 12 ( x − 3 −1) for (x) > 3 −1 4 11(c) ( gf(x) = 2 x 2 + 3 + 1 ) B1 SOI ( 2 x + 1)2 + 3 − 3 = 2 ( x 2 + 3) + 1 → 2 x 2 + 4 x − 6 ( = 0 ) *M1 Express as 3-term quadratic ( 2 )( x + 3)( x − 1) ( = 0 ) DM1 Or quadratic formula or completing the square x =1 A1 4
Question
y B A 4, 0 O x 1 y = 4x 2 − 2x y=3−x C 1 The diagram shows a curve with equation y = 4x 2 − 2x for x ≥ 0, and a straight line with equation y = 3 − x. The curve crosses the x-axis at A 4, 0 and crosses the straight line at B and C. (a) Find, by calculation, the x-coordinates of B and C. [4] (b) Show that B is a stationary point on the curve. [2] (c) Find the area of the shaded region. [6] 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
12(a) 1 1 *M1 4x 2 − 2x = 3 − x → x − 4x 2 + 3 ( = 0) 3-term quadratic. Can be expressed as e.g. u 2 − 4u + 3 (=0) 1 1 DM1 x 2 − 1 x 2 − 3 ( = 0 ) or ( u − 1)( u − 3)( = 0 ) Or quadratic formula or completing square 1 A1 x 2 = 1, 3 SOI x = 1, 9 A1 Alternative method for question 12(a) 1 2 *M1 1 4 x 2 = ( 3 + x ) 2 Isolate x 2 16 x = 9 + 6 x + x 2 → x 2 − 10 x + 9 ( = 0 ) A1 3-term quadratic DM1 Or formula or completing square on a quadratic obtained by a ( x − 1)( x − 9 ) ( = 0 ) correct method x = 1, 9 A1 4 12(b) dy *B1 = 2 x1/2 − 2 dx dy DB1 or 2 x1/2 − 2 = 0 when x =1 hence B is a stationary point dx 2 12(c) M1 9 1 1 2 Area of correct triangle = 2 (9 ‒ 3) × 6 or ( 3 − x )( dx ) = 3x − 2 x → −18 3 3 B1 B1 1 4x 2 (4 x 2 − 2 x) ( dx ) = − x2 3 2 M1 ( 72 − 81) − 64 − 16 Apply limits 4 → their 9 to an integrated expression 3 −14 13 A1 OE Shaded region = 18 − 14 13 = 3 23 A1 OE 6
Checked your answers against these? Let AI mark the whole paper, on your own working.
Mark my 9709/11Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Question papers and mark schemes © Cambridge University Press & Assessment. CIE Buddy is independent and not connected to Cambridge.
On this page, free: the mark scheme for AS & A Level Mathematics 9709/11, October/November 2020 (9709_w20_ms_11), and the question paper (9709_w20_qp_11), to read or download, with each question's marking points written out underneath.
Mathematics 9709/11 for October/November 2020 is out of 75 marks, with 1 hour 50 minutes to answer it.
Yes. Upload your answers to 9709/11/O/N/20 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.