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Cambridge AS & A Level Mathematics (9709), Paper 1, October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w19_qp_119709_w19_ms_119709/11/O/N/19
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 term independent of x in the expansion of 2x + 2 . [3] 4x
Mark scheme
1 1 B1 SOI 6C2 × ( 2 x ) × 4 (4 x 2 ) 2 SC: Condone errors in (4−1 )2 evaluation or interpretation for B1 only 1 B1 Identified as required term. 15 × 24 × 42 15 B1 3
Question
An increasing function, f, is defined for x > n, where n is an integer. It is given that f ′ x = x2 − 6x + 8. Find the least possible value of n. [3]
Mark scheme
2 Attempt to solve f ′ ( x ) = 0 or f ′ ( x ) > 0 or f ′ ( x ). 0 M1 SOI (x – 2)(x – 4) A1 2 and 4 seen (Least possible value of n is) 4 A1 Accept n = 4 or n . 4 3
Question
The line y = ax + b is a tangent to the curve y = 2x3 − 5x2 − 3x + c at the point 2, 6. Find the values of the constants a, b and c. [5]
Mark scheme
3 dy B1 = 6 x 2 − 10 x − 3 dx dy M1 At x = 2, = 24 − 20 − 3 = 1 → a = 1 A1 dx 6= 2+b → b = 4 B1FT Substitute x = 2, y = 6 in y = ( their a ) x + b 6 = 16 − 20 − 6 + c → c = 16 B1 Substitute x =2, y = 6 into equation of curve 5
Question
A runner who is training for a long-distance race plans to run increasing distances each day for 21 days. She will run x km on day 1, and on each subsequent day she will increase the distance by 10% of the previous day’s distance. On day 21 she will run 20 km. (i) Find the distance she must run on day 1 in order to achieve this. Give your answer in km correct to 1 decimal place. [3] (ii) Find the total distance she runs over the 21 days. [2] 1
Mark scheme
4(i) Identifies common ratio as 1.1 B1 Use of x (1.1) = 20 M1 SOI 20 A1 Accept 2.97 20 x= = 3.0 (1.1)20 3 4(ii) (1.1)21 − 1 M1 Correct formula used for M mark. their 3.0 × → 192 A1 Allow 2.97 used from (i) 1.1 − 1 Accept 190 from x = 2.97… 2
Question
(i) Given that 4 tan x + 3 cos x + = 0, show, without using a calculator, that sin x = − 23 . [3] cos x (ii) Hence, showing all necessary working, solve the equation 1 4 tan 2x − 20Å + 3 cos 2x − 20Å + =0 cos 2x − 20Å for 0Å ≤ x ≤ 180Å. [4]
Mark scheme
5(i) 1 M1 Multiply by cos x or common denominator of cos x 4 tan x + 3cos x + = 0 → 4sin x + 3cos 2 x + 1 = 0 cos x ( ) 4sin x + 3 1 − sin 2 x + 1 = 0 → 3sin 2 x − 4sin x − 4 = 0 M1 Use cos2 x = 1 − sin 2 x and simplify to 3-term quadratic in sin x 2 A1 AG sin x = − 3 3 5(ii) 2x ‒20º = 221.8º, 318.2º M1A1 Attempt to solve sin ( 2 x − 20 ) = −2 / 3 ( M1) . At least 1 correct (A1) x = 120.9º, 169.1º A1 FT for 290º ‒ other solution. A1FT SC A1 both answers in radians 4
Question
A straight line has gradient m and passes through the point 0, −2. Find the two values of m for which the line is a tangent to the curve y = x2 − 2x + 7 and, for each value of m, find the coordinates of the point where the line touches the curve. [7]
Mark scheme
6 Equation of line is y = mx − 2 B1 OR x 2 − 2 x + 7 = mx − 2 → x 2 − x ( 2 + m ) + 9 = 0 M1 Apply b 2 − 4ac ( = 0 ) → ( 2 + m ) − 4 × 9 ( = 0 ) *M1 2 m = 4 or −8 A1 2 m = 4→ x ‒ 6x + 9 = 0 → x = 3 DM1 m = ‒8→ x2 + 6x + 9 = 0 → x = ‒3 (3, 10), (‒3, 22) A1A1 Alternative method for question 6 dy B1 = 2x ‒ 2 dx 2x − 2 = m M1 x2 − 2 x + 7 = ( 2 x − 2) x − 2 = 2 x2 − 2 x − 2 M1 x 2 − 9 = 0 → x = ±3 A1 (3, 10), (‒3, 22) A1A1 When x = 3, m = 4; when x = ‒3, m = ‒8 A1 7
Question
Functions f and g are defined by 3 f:x→ for x > 0, 2x + 1 1 g : x → + 2 for x > 0. x (i) Find the range of f and the range of g. [3] ax (ii) Find an expression for fg x, giving your answer in the form , where a, b and c are integers. bx + c [2] (iii) Find an expression for fg−1 x, giving your answer in the same form as for part (ii). [3]
Mark scheme
7(i) Range of f is 0 < f(x) < 3 B1B1 OE. Range cannot be defined using x Range of g is g(x) > 2 B1 OE 3 7(ii) B1B1 Second B mark implies first B mark ( fg ( x ) = ) 2 1 +32 + 1 = 2 +3x5 x (x ) 2 7(iii) 3x M1 Correct order of operations y= → 2 y + 5 xy = 3 x → 3 x − 5 xy = 2 y 2 + 5x 2y M1 Correct order of operations x (3 − 5 y ) = 2 y → x= 3 − 5y (( fg ) ( x )) = 3 −2 x5x −1 A1 3
Question
C B 6 cm 3 8 0 rad O A The diagram shows a sector OAC of a circle with centre O. Tangents AB and CB to the circle meet at B. The arc AC is of length 6 cm and angle AOC = 38 0 radians. (i) Find the length of OA correct to 4 significant figures. [2] (ii) Find the perimeter of the shaded region. [2] (iii) Find the area of the shaded region. [4] dy 1
Mark scheme
8(i) 3 M1 OA × π = 6 8 16 A1 OA = = 5.093 ( 0 ) π 8(ii) 3 M1 AB = their 5.0930 × tan π 16 Perimeter = 2 × 3.4030 + 6 = 12.8 A1 8(iii) Area OABC = ( 2 × ½ ) × their 5.0930 × their 3.4030 M1 3 M1 Area sector = ½ × ( their 5.0930 ) × π 2 8 Shaded area = their17.331 − their15.279 = 2.05 M1A1
Question
A curve for which = 5x − 1 2 − 2 passes through the point 2, 3. dx (i) Find the equation of the curve. [4] d2 y (ii) Find . [2] dx2 (iii) Find the coordinates of the stationary point on the curve and, showing all necessary working, determine the nature of this stationary point. [4]
Mark scheme
9(i) y = [( 5 x − 1) ÷ 32 ÷5 ] [ − 2 x ] B1 1/ 2 B1 27 M1 Substitute x = 2, y = 3 3= −4+c (3 / 2) × 5 3 A1 18 17 2 ( 5 x − 1) 2 17 c=7− = → y= − 2x + 5 5 15 5 9(ii) d 2 y / dx 2 = ½ ( 5 x − 1) [×5] −1/2 B1 B1 9(iii) ( 5 x − 1)1/ 2 − 2 = 0 → 5 x − 1 = 4 M1A1 dy Set = 0 and attempt solution (M1) x =1 dx 16 17 37 A1 37 y= −2+ = Or 2.47 or 1, 25 5 15 15 d2 y 5 1 5 A1 OE = × = (> 0) hence minimum dx x 2 2 4
Question
D C A B Relative to an origin O, the position vectors of the points A, B, C and D, shown in the diagram, are given by ` a ` a ` a ` a OA = 3 , OB = −3 , OC = −2 and OD = 2 . −4 5 5 −1 (i) Show that AB is perpendicular to BC. [3] (ii) Show that ABCD is a trapezium. [3] (iii) Find the area of ABCD, giving your answer correct to 2 decimal places. [3]
Mark scheme
10(i) 2 −1 3 4 2 2 B1B1 Condone reversal of labels AB = −3 − 3 = −6 , BC = −2 − −3 = 1 5 −4 9 5 5 0 AB.BC = 6 ‒ 6 → = 0 (hence perpendicular) B1 AG 10(ii) 4 2 2 B1 −2 DC = −2 − 2 = −4 Or: CD = 4 5 −1 6 −6 AB = kDC M1 OE 3 Expect k = 2 Or: DC.BC = 4 ‒ 4 = 0 hence BC is also perpendicular to DC Or: AB.DC = 1 or AB.CD = –1, angle between lines is 0 or 180 AB is parallel to DC, hence ABCD is a trapezium A1 10(iii) |AB| = 9 + 36 + 81 = 126 = 11.22 M1 Method for finding at least 2 magnitudes |DC| = 4 + 16 + 36 = 56 = 7.483 |BC| = 4 + 1 + 0 = 5 = 2.236 Area = 12 ( theirAB + theirDC ) × theirBC = 20.92 M1A1 OE
Question
y 3 y = x2 + 4x + 3 x −2 O y = −1 −1 The diagram shows a shaded region bounded by the y-axis, the line y = −1 and the part of the curve y = x2 + 4x + 3 for which x ≥ −2. (i) Express y = x2 + 4x + 3 in the form y = x + a2 + b, where a and b are constants. Hence, for x ≥ −2, express x in terms of y. [4] (ii) Hence, showing all necessary working, find the volume obtained when the shaded region is rotated through 360Å about the y-axis. [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
11(i) ( y =) ( x + 2) − 1 2 B1 2nd B1 dependent on 2 in bracket DB1 x + 2 = ( ± )( y + 1) M1 1/ 2 x = −2 + ( y + 1) A1 1/ 2 11(ii) x 2 = 4 + ( y + 1) − / + 4 ( y + 1) 2 1 *M1A1 SOI. Attempt to find x 2 .The last term can be ‒ or + at this stage 3 A2,1,0 y 2 4 ( y + 1) 2 ( π ) ∫ x ( dy ) = ( π ) 5 y + − 3 2 2 2 9 64 1 DM1 Apply y limits ( π ) 15 + − − −5 + 2 3 2 8π A1 or 8.38 3
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Mathematics 9709/11 for October/November 2019 is out of 75 marks, with 1 hour 45 minutes to answer it.
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