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Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, February/March 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_m22_qp_129709_m22_ms_129709/12/F/M/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
A curve with equation y = f x is such that f ′ x = 2x 3 − x 3 . It is given that f 8 = 5. Find f x. [4]
Mark scheme
1 2 4 B1 B1 2 4 2x 3 x 3 and may be seen as sums of 1 and a fraction. f ( x ) = 2 − 4 [ +c ] 3 3 3 3 5 = 12 ‒ 12 + c M1 Substituting (8,5) into an integral. 2 4 A1 Fractions in the denominators scores A0. f ( x ) = 3 x 3 − 34 x 3 + 5 4
Question
A curve has equation y = x2 + 2cx + 4 and a straight line has equation y = 4x + c, where c is a constant. Find the set of values of c for which the curve and line intersect at two distinct points. [5]
Mark scheme
2 x 2 + 2cx + 4 = 4 x + c leading to x 2 + 2cx − 4 x + 4 − c [ = 0 ] *M1 Equate ys and move terms to one side of equation. b 2 − 4 ac = ( 2c − 4 ) − 4 ( 4 − c ) 2 DM1 Use of discriminant with their correct coefficients. 4c 2 − 16c + 16 − 16 + 4c = 4c 2 − 12c A1 b 2 − 4ac > 0 leading to ( 4 ) c ( c − 3) > 0 M1 Correctly apply ‘> 0’ considering both regions. c < 0, c > 3 A1 Must be in terms of c. SC B1 instead of M1A1 for c ⩽ 0, c ⩾ 3 5
Question
Find the term independent of x in each of the following expansions. @ A 2 6 (a) 3x + 2 [3] x @ A 2 6 (b) 3x + 2 1 − x3 [3] x
Mark scheme
3(a) 4 2 2 B1 Can be seen within an expansion. 6 C2 × ( 3x ) 2 x 15 × 34 × 22 B1 Identified. Powers must be correct. 4860 B1 Without any power of x 3 3(b) Their 4860 and one other relevant term M1 Using their 4860 and an attempt to find a term in x‒3 3 2 3 A1 Must be identified. Other term = 6C 3 ( 3 x ) 2 or 6C 3 × 33 × 23 or 4320 If M0 scored then SC B1 for 4320 as the only answer. x [4860 ‒ 4320 =] 540 A1 3
Question
The first term of a geometric progression and the first term of an arithmetic progression are both equal to a. The third term of the geometric progression is equal to the second term of the arithmetic progression. The fifth term of the geometric progression is equal to the sixth term of the arithmetic progression. Given that the terms are all positive and not all equal, find the sum of the first twenty terms of the arithmetic progression in terms of a. [6]
Mark scheme
4 ar 2 = a + d B1 ar 4 = a + 5d B1 a 2 r 4 = a ( a + 5d ) leading to a 2 + 5ad = ( a + d ) 2 *M1 Eliminating r or complete elimination of a and d. 3ad − d 2 = 0 leading to d = 3a OR [ r = 2 leading to ] d = 3a A1 20 DM1 Use of formula with their d in terms of a. S20 = [ 2a + 19 × 3a ] 2 590a A1 6
Question
(a) Express 2x2 − 8x + 14 in the form 2 x − a2 + b. [2] The functions f and g are defined by f x = x2 for x ∈ >, g x = 2x2 − 8x + 14 for x ∈ >. (b) Describe fully a sequence of transformations that maps the graph of y = f x onto the graph of y = g x, making clear the order in which the transformations are applied. [4]
Mark scheme
5(a) 2[{ ( x − 2 ) } {+3}] 2 B1 B1 B1 for a = 2, B1 for b = 3. 2 ( x − 2 ) +6 gains B1B0 2 2 5(b) {2} B2,1,0 B2 for fully correct, B1 with two elements correct. {} {Translation} OR {Stretch} {y direction} {factor 2} indicates different elements. {3} {2} B2,1,0 B2 for fully correct, B1 with two elements correct. {} {Stretch} {y direction} {factor 2} OR {Translation} indicates different elements. {6} 4
Question
y y = 3x − 20 x + 12 + y − 22 = 85 A C x O B The circle with equation x + 12 + y − 22 = 85 and the straight line with equation y = 3x − 20 are shown in the diagram. The line intersects the circle at A and B, and the centre of the circle is at C. (a) Find, by calculation, the coordinates of A and B. [4] (b) Find an equation of the circle which has its centre at C and for which the line with equation y = 3x − 20 is a tangent to the circle. [4] sin 1 + 2 cos 1 sin 1 − 2 cos 1 4
Mark scheme
6(a) ( x + 1)2 + ( 3 x − 22 )2 = 85 M1 OE. Substitute equation of line into equation of circle. 10 x 2 − 130 x + 400 [ = 0 ] A1 Correct 3-term quadratic [10] ( x − 8)( x − 5) leading to x = 8 or 5 A1 Dependent on factors or formula or completing of square seen. (8, 4), (5, ‒5) A1 If M1A1A0A0 scored, then SC B1 for correct final answer only. 4 6(b) Mid-point of AB = (6 12 ,- 12 ) M1 Any valid method Use of C = (‒1, 2) B1 SOI r 2 = ( −1 − 6 12 ) + ( 2 + 12 ) 2 2 M1 Attempt to find r2. Expect r 2 = 62 1 . 2 Equation of circle is ( x + 1) + ( y − 2 ) = 62 12 2 2 A1 OE. 4
Question
(a) Show that − 2 . [4] cos 1 − 2 sin 1 cos 1 + 2 sin 1 5 cos 1 − 4 sin 1 + 2 cos 1 sin 1 − 2 cos 1 (b) Hence solve the equation − = 5 for 0Å < 1 < 180Å. [3] cos 1 − 2 sin 1 cos 1 + 2 sin 1
Mark scheme
7(a) ( sin θ + 2cosθ )( cosθ + 2sin θ ) − ( sin θ − 2cosθ )( cosθ − 2sin θ ) *M1 Obtain an expression with a common denominator ( cosθ − 2sin θ )( cosθ + 2sin θ ) ( 5sin θ cosθ + 2cos 2 θ + 2sin 2 θ − 5sin θ cos θ − 2sin 2 θ − 2cos 2 θ ) A1 cos 2 θ − 4sin 2 θ = ( 4 cos 2 θ + sin 2 θ ) 2 2 cos θ − 4sin θ 4 DM1 Use cos 2 θ + sin 2 θ = 1 twice ( cos 2 θ − 4 1 − cos 2 θ ) 4 A1 AG 2 5cos θ − 4 4 7(b) 4 M1 Make cos θ the subject 2 = 5 leading to 25cos 2 θ = 24 5cos θ − 4 24 leading to cosθ = = ( ± ) 0.9798 25 θ = 11.5° or 168.5° A1 A1 FT FT on 180º ‒ 1st solution 3
Question
y A B x O x − 22 + y2 = 8 The diagram shows the circle with equation x − 22 + y2 = 8. The chord AB of the circle intersects the positive y-axis at A and is parallel to the x-axis. (a) Find, by calculation, the coordinates of A and B. [3] (b) Find the volume of revolution when the shaded segment, bounded by the circle and the chord AB, is rotated through 360Å about the x-axis. [5]
Mark scheme
8(a) ( −2 )2 + y 2 = 8 leading to y = 2 leading to A = ( 0, 2 ) B1 Substitute y = their 2 into circle leading to ( x − 2 ) + 4 = 8 2 M1 Expect x = 4. B = (4, 2) A1 3 (8 − ( x − 2) )dx 8(b) Attempt to find [ π ] 2 *M1 ( x − 2 )3 x3 A1 [ ] π 8 x − or [ π ] 8 x − − 2 x 2 + 4 x 3 3 DM1 Apply limits 0 → their 4. [ π ] 32 − 16 64 or [ π ] 32 − − 32 + 16 3 3 Volume of cylinder = π × 22 × 4 = 16π B1 FT OR from π 22 dx with their limits from (a). FT on their A and B éê Volume of revolution = 26 2 π -16π =ùú 10 2 π A1 Accept 33.5 ë 3 û 3 5
Question
Functions f, g and h are defined as follows: 1 f : x → x − 4x 2 + 1 for x ≥ 0, 2 g : x → mx + n for x ≥ −2, where m and n are constants, 1 h : x → x2 − 2 for x ≥ 0. (a) Solve the equation f x = 0, giving your solutions in the form x = a + b c, where a, b and c are integers. [4] (b) Given that f x gh x, find the values of m and n. [4]
Mark scheme
9(a) 1 4 ± 16 − 4 M1 A1 OE. Answer must come from formula or completing square. x 2 = = 2± 3 If M0A0 scored then SC B1 for 2 ± 3 only. 2 [ x =]( 2 ± 3 ) M1 Attempt to square their 2 ± 3 2 7+4 3 , 7−4 3 A1 Accept 7 ± 4 3 or a = 7, b = ±4, c = 3 SC B1 instead of second M1A1 for correct final answer only. Alternative method for question 9(a) 1 2 *M1 A1 OE -4 x 2 + 1 = 0 leading to ( x + 1) = 16 x leading to x 2 - 14 x + 1 = 0 14 ± 196 - 4 DM1 Attempt to solve for x x= 2 7+4 3 , 7−4 3 A1 SC B1 instead of second M1A1 for correct final answer only. 4 9(b) 1 2 M1 SOI [gh ( x ) =] m x 2 − 2 + n 1 1 A1 SOI gh ( x ) = m x − 4 x 2 + 4 + n ≡ x − 4 x 2 + 1 m = 1, n = −3 A1 A1 WWW 4
Question
D E A 5 C 8 B The diagram shows a circle with centre A of radius 5 cm and a circle with centre B of radius 8 cm. The circles touch at the point C so that ACB is a straight line. The tangent at the point D on the smaller circle intersects the larger circle at E and passes through B. (a) Find the perimeter of the shaded region. [5] (b) Find the area of the shaded region. [3]
Mark scheme
10(a) 12 5 12 M1 5 12 5 tan A = or cos A = or sin A = OR tanB = or cos B = or sin B = 5 13 13 12 13 13 A = 1.176 B = 0.3948 A1 Allow 1.18 or 67.4°, Allow 0.395 or 22.6°. π May be implied by − 1.176 2 DE = 4 B1 If trigonometry used accept AWRT 4.00 Arcs = 5 × their1.176 and 8 × their 0.3948 M1 Or corresponding calculations in degrees. [Perimeter = 5.880 + 3.158 + 4 =] 13.0 A1 Accept 13. If DE is outside the given range this mark cannot be awarded. 5 10(b) Area of triangle = 12 × 5 × their12 [ = 30] B1 FT Area of sectors = 12 × 52 × their 1.176 + 12 × 82 × their 0.3948 M1 Or corresponding calculations in degrees [Area = 30 ‒ 14.70 ‒12.63 =] 2.67 A1 Allow 2.66 to 2.67 3
Question
It is given that a curve has equation y = k 3x − k−1 + 3x, where k is a constant. (a) Find, in terms of k, the values of x at which there is a stationary point. [4] The function f has a stationary value at x = a and is defined by f x = 4 3x − 4−1 + 3x for x ≥ 32 . (b) Find the value of a and determine the nature of the stationary value. [3] (c) The function g is defined by g x = − 3x + 1−1 + 3x for x ≥ 0. Determine, making your reasoning clear, whether g is an increasing function, a decreasing function or neither. [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
11(a) dy dx { = − k ( 3x − k ) −2 } {×3} {+3} B2, 1, 0 −3k M1 dy + 3 = 0 leading to ( 3)( 3 x − k ) = ( 3) k 2 Set = 0 and remove the denominator ( 3x − k ) 2 dx leading to 3 x − k = [ ± ] k k± k A1 OE x= 3 4 11(b) 4± 4 B1 Substitute x = a when k = 4. Allow x = 2. a= leading to a = 2 3 f '' ( x ) = f ' −12 ( 3x − 4 ) + 3 = 72 ( 3x − 4 ) −2 −3 B1 Allow 18k ( 3 x − k )−3 > 0 (or 9) when x = 2 → minimum B1 FT 3 FT on their x = 2, providing their x and f ′′ ( x ) is correct 2 3 11(c) 3 M1 Condone one error. Substitute k = ‒1 leading to g′ ( x ) = +3 ( 3x + 1) 2 g′ ( x ) > 0 or g′ ( x ) always positive, hence g is an increasing function A1 WWW. A0 if the conclusion depends on substitution of values into g′ ( x ) . Alternative method for question 11(c) k± k M1 Allow the statement ‘no turning points’ for increasing or x= when k = ‒1 has no solutions, so g is increasing or decreasing 3 decreasing Show g’(x) is positive for any value of x, hence g is an increasing A1 Or show g(b) > g(a) for b > a → g, hence g is an increasing function function 2
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Mathematics 9709/12 for February/March 2022 is out of 75 marks, with 1 hour 50 minutes to answer it.
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