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Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics, February/March 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_m20_qp_129709_m20_ms_129709/12/F/M/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
The function f is defined by f x = + x2 for x < −1. 3x + 2 Determine whether f is an increasing function, a decreasing function or neither. [3]
Mark scheme
1 f ′ ( x ) = − ( 3 x + 2 ) × [3] + [ 2 x ] −2 B2, 1, 0 < 0 hence decreasing B1 Dependent on at least B1 for f ′ ( x ) and must include < 0 or ‘(always) neg’ 3
Question
The graph of y = f x is transformed to the graph of y = 1 + f 12 x . Describe fully the two single transformations which have been combined to give the resulting transformation. [4]
Mark scheme
2 [Stretch] [factor 2, x direction (or y-axis invariant)] *B1 DB1 [Translation or Shift] [1 unit in y direction] or B1B1 Accept transformations in either order. Allow (0, 1) for the vector 0 [Translation/Shift] 1 4
Question
y 5 y = x2 + 1 1 x O The diagram shows part of the curve with equation y = x2 + 1. The shaded region enclosed by the curve, the y-axis and the line y = 5 is rotated through 360Å about the y-axis. Find the volume obtained. [4]
Mark scheme
3 ( π ) ( y − 1) dy *M1 SOI Attempt to integrate x2 or ( y − 1) y2 A1 ( π) − y 2 25 1 DM1 Apply limits 1 → 5 to an integrated expression ( π ) − 5 − − 1 2 2 8π or AWRT 25.1 A1 4
Question
A curve has equation y = x2 − 2x − 3. A point is moving along the curve in such a way that at P the y-coordinate is increasing at 4 units per second and the x-coordinate is increasing at 6 units per second. Find the x-coordinate of P. [4]
Mark scheme
4 dy B1 = 2x − 2 dx dy 4 B1 OE, SOI = dx 6 4 M1 dy their ( 2 x − 2 ) = their LHS and RHS must be their expression and value 6 dx 4 A1 x= oe 3 4
Question
Solve the equation tan 1 + 3 sin 1 + 2 =2 tan 1 − 3 sin 1 + 1 for 0Å ≤ 1 ≤ 90Å. [5] @ A 1 a 5
Mark scheme
5 2 tan θ − 6sin θ + 2 = tan θ + 3sin θ + 2 → tan θ − 9sin θ ( = 0 ) M1 Multiply by denominator and simplify sin θ − 9sin θ cos θ ( = 0 ) M1 Multiply by cosθ 1 M1 Factorise and attempt to solve at least one of the factors = 0 sin θ (1 − 9cosθ ) ( = 0 ) → sin θ = 0, cosθ = 9 θ = 0 or 83.6° (only answers in the given range) A1A1 5
Question
The coefficient of in the expansion of 2x + 2 is 720. x x (a) Find the possible values of the constant a. [3] 1 (b) Hence find the coefficient of in the expansion. [2] x7
Mark scheme
6(a) a 2 B1 SOI 3 5C2 2 ( x ) 2 Can include correct x's ( ) x x3 1 B1 SOI 10 × 8 × a 2 4 = 720 Can include correct x's x x a = ±3 B1 3 6(b) their a 4 B1 SOI 5C4 ( ) 2 2 x Their a can be just one of their values (e.g. just 3). x ( ) Can gain mark from within an expansion but must use their value of a 810 identified B1 Allow with x −7 2
Question
A D 0.8 rad O B C 6 cm The diagram shows a sector AOB which is part of a circle with centre O and radius 6 cm and with angle AOB = 0.8 radians. The point C on OB is such that AC is perpendicular to OB. The arc CD is part of a circle with centre O, where D lies on OA. Find the area of the shaded region. [6]
Mark scheme
7 OC = 6cos0.8 = 4.18(0) M1A1 SOI 1 *M1 OE Area sector OCD = ( their 4.18)2 × 0.8 2 1 M1 OE ΔOCA = × 6 × their 4.18 × sin 0.8 2 Required area = their ΔOCA ‒ their sectorOCD DM1 SOI. If not seen their areas of sector and triangle must be seen 2.01 A1 CWO. Allow or better e.g. 2.0064 6
Question
A woman’s basic salary for her first year with a particular company is $30 000 and at the end of the year she also gets a bonus of $600. (a) For her first year, express her bonus as a percentage of her basic salary. [1] At the end of each complete year, the woman’s basic salary will increase by 3% and her bonus will increase by $100. (b) Express the bonus she will be paid at the end of her 24th year as a percentage of the basic salary paid during that year. [5]
Mark scheme
8(a) 2% B1 1 8(b) Bonus = 600 + 23 × 100 = 2900 B1 Salary = 30000 × 1.0323 M1 Allow 30000 × 1.0324 (60984) = 59207.60 A1 Allow answers of 3significant figure accuracy or better their 2900 M1 SOI their 59200 4.9(0)% A1 5
Question
(a) Express 2x2 + 12x + 11 in the form 2 x + a2 + b, where a and b are constants. [2] The function f is defined by f x = 2x2 + 12x + 11 for x ≤ −4. (b) Find an expression for f −1 x and state the domain of f −1 . [3] The function g is defined by g x = 2x − 3 for x ≤ k. (c) For the case where k = −1, solve the equation fg x = 193. [2] (d) State the largest value of k possible for the composition fg to be defined. [1] dy 1
Mark scheme
9(a) 2 ( x + 3 ) 2 [ −7 ] B1B1 Stating a = 3, b = −7 gets B1B1 2 9(b) y+7 M1 First 2 operations correct. y = 2 ( x + 3) − 7 → 2 ( x + 3) = y + 7 → ( x + 3 ) = 2 2 2 2 Condone sign error or with x/y interchange y+7 y+7 x+7 A1FT FT on their a and b. Allow y = ... x + 3 = (±) → x = (±) − 3 → f −1 ( x ) = − −3 2 2 2 Domain: x ⩾ ‒5 or ⩾ -5 or [–5, ∞) B1 Do not accept y = …, f ( x ) = …, f −1 ( x ) = … 3 9(c) fg(x) = 8 x 2 − 7 B1FT SOI. FT on their –7 from part (a) 8 x 2 − 7 = 193 → x 2 = 25 → x = −5 only B1 Alternative method for question 9(c) g(x) = f −1 (193) → 2 x − 3 = − 100 − 3 M1 FT on their f −1 ( x ) x = ‒5 only A1 2 9(d) 1 B1 1 1 (Largest k is) − Accept − or k ⩽ − 2 2 2 1
Question
The gradient of a curve at the point x, y is given by = 2 x + 3 2 − x. The curve has a stationary dx point at a, 14, where a is a positive constant. (a) Find the value of a. [3] (b) Determine the nature of the stationary point. [3] (c) Find the equation of the curve. [4]
Mark scheme
10(a) 1 M1 SOI. 2 ( a + 3) 2 − a = 0 dy Set = 0 when x = a. Can be implied by an answer in terms of a dx 4 ( a + 3) = a 2 → a 2 − 4a − 12 = 0 M1 Take a to RHS and square. Form 3-term quadratic ( a − 6 )( a + 2 ) → a = 6 A1 Must show factors, or formula or completing square. Ignore a = ‒2 SC If a is never used maximum of M1A1 for x = 6 ,with visible solution 3 10(b) d2 y − 1 B1 ( ) 2 2 = x + 3 −1 dx d2 y 1 2 M1A1 A mark only if completely correct Sub their a → 2 = − 1 = − (or < 0) → MAX 2 dx 3 3 If the second differential is not − correct conclusion must be 3 drawn to award the M1 3 10(c) 3 2 ( x + 3) 2 B1B1 1 ( y =) 3 − x 2 ( +c ) 2 2 4 3 M1 Substitute into an integrated expression. c must be present. Sub x = their a and y = 14 → 14 = ( 9 ) 2 − 18 + c Expect c = ‒4 3 4 3 1 A1 Allow f ( x ) = …. y= ( x + 3) 2 − x 2 − 4 3 2 4
Question
(a) Solve the equation 3 tan2 x − 5 tan x − 2 = 0 for 0Å ≤ x ≤ 180Å. [4] (b) Find the set of values of k for which the equation 3 tan2 x − 5 tan x + k = 0 has no solutions. [2] (c) For the equation 3 tan2 x − 5 tan x + k = 0, state the value of k for which there are three solutions in the interval 0Å ≤ x ≤ 180Å, and find these solutions. [3]
Mark scheme
11(a) (tan x − 2)(3tan x + 1) ( = 0 ) . or formula or completing square M1 Allow reversal of signs in the factors. Must see a method 1 A1 tan x = 2 or - 3 x = 63.4° ( only value in range ) or161.6°(only value in range ) B1FT B1FT 4 11(b) Apply b 2 − 4ac < 0 M1 SOI. Expect 25 − 4 ( 3)( k ) < 0, tan x must not be in coefficients 25 A1 25 k> Allow b 2 − 4ac = 0 leading to correct k > for M1A1 12 12 2 11(c) k=0 M1 SOI 5 A1 tan x = 0 or 3 x = 0° or 180° or 59.0° A1 All three required 3
Question
A diameter of a circle C1 has end-points at −3, −5 and 7, 3. (a) Find an equation of the circle C1 . [3] y C2 R C1 x O S @ A 8 The circle C1 is translated by to give circle C2 , as shown in the diagram. 4 (b) Find an equation of the circle C2 . [2] The two circles intersect at points R and S. (c) Show that the equation of the line RS is y = −2x + 13. [4] (d) Hence show that the x-coordinates of R and S satisfy the equation 5x2 − 60x + 159 = 0. [2] 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) Centre = (2, ‒1) B1 2 2 M1 1 r 2 = 2 − ( −3) + − 1 − ( −5) [ 2 − 7] + [ −1 − 3] 2 2 or OE OR ( −3 − 7 )2 + ( −5 − 3)2 OE 2 ( x − 2 ) + ( y + 1) = 41 2 2 A1 Must not involve surd form SCB3 ( x + 3)( x − 7 ) + ( y + 5 )( y − 3) = 0 3 12(b) 8 B1FT SOI Centre = their (2, ‒1) + = (10, 3) FT on their (2, ‒1) 4 ( x − 10) + ( y − 3) = their 41 2 2 B1FT FT on their 41 even if in surd form SCB2 ( x − 5 )( x − 15 ) + ( y + 1)( y − 7 ) = 0 2 12(c) 4 B1 Gradient m of line joining centres = OE 8 Attempt to find mid-point of line. M1 Expect (6, 1) Equation of RS is y − 1 = −2 ( x − 6 ) M1 −1 Through their (6, 1) with gradient m y = −2 x + 13 A1 AG Alternative method for question 12(c) ( x − 2) + ( y + 1) − 41 = ( x − 10 ) + ( y − 3) − 41 OE 2 2 2 2 M1 x 2 − 4 x + 4 + y 2 + 2 y + 1 = x 2 − 20 x + 100 + y 2 − 6 y + 9 OE A1 Condone 1 error or errors caused by 1 error in the first line 16 x + 8 y = 104 A1 y = −2 x + 13 A1 AG 4 12(d) ( x − 10) + ( −2 x + 13 − 3) = 41 2 2 M1 Or eliminate y between C1 and C2 x 2 − 20 x + 100 + 4 x 2 − 40 x + 100 = 41 → 5 x 2 − 60 x + 159 = 0 A1 AG 2
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Mathematics 9709/12 for February/March 2020 is out of 75 marks, with 1 hour 50 minutes to answer it.
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