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Cambridge AS & A Level Mathematics (9709), Paper 2, 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_229709_w19_ms_229709/22/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
The polynomial f x is defined by f x = x4 − 3x3 + 5x2 − 6x + 11. Find the quotient and remainder when f x is divided by x2 + 2. [3]
Mark scheme
1 Divide at least as far as the x term in the quotient ( )( ) M1 Allow use of x 2 + 2 x 2 + ax + b + cx + d Obtain at least x 2 − 3x A1 Obtain x 2 − 3x + 3 and remainder 5 A1 3
Question
(i) Solve the equation 4x + 5 = x − 7. [3] (ii) Hence, using logarithms, solve the equation 2y+2 + 5 = 2y − 7, giving the answer correct to 3 significant figures. [2]
Mark scheme
2(i) State or imply non-modular equation (4 x + 5) 2 = ( x − 7) 2 or pair of B1 different linear equations Attempt solution of 3-term quadratic equation or pair of linear M1 equations Obtain 52 and −4 A1 SC For x = −4 only, from correct work, allow B1 3 2(ii) Apply logarithms and use power law for 2 y = k where k > 0 M1 from (i) Obtain –1.32 only A1 AWRT 2
Question
ln y 0.22, 3.96 1.32, 2.43 ln x O The variables x and y satisfy the equation y = kxa , where k and a are constants. The graph of ln y against ln x is a straight line passing through the points 0.22, 3.96 and 1.32, 2.43, as shown in the diagram. Find the values of k and a correct to 3 significant figures. [5]
Mark scheme
3 ln y = ln k + a ln x B1 SOI Equate gradient of line to a M1 Obtain a = −1.39 A1 OE Substitute appropriate values into a correct equation to find ln k M1 Obtain ln k = 4.266... and k = 71.2 A1 SC1 for gradient = −1.39 and no other relevant working Alternative method for question 3 ln y = ln k + a ln x B1 SOI 3.96 = ln k + 0.22a M1 For one correct equation 2.43 = ln k + 1.32a M1 For a second correct equation and attempt to solve to find one unknown Obtain a = −1.39 A1 OE Obtain ln k = 4.266... and k = 71.2 A1 SC1 for gradient = −1.39 and no other relevant working Alternative method for question 3 e3.96 = k × 0.22a and e2.43 = k ×1.32a B1 Apply a correct method to obtain a M1 Obtain a = −1.39 A1 OE Substitute appropriate values into a correct equation to find k M1 Obtain k = 71.2 A1 AWRT 5
Question
The sequence x1 , x2 , x3 , à defined by x1 = 1, xn+1 = xn ln 2xn converges to the value !. (i) Use the iterative formula to find the value of ! correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] (ii) State an equation satisfied by ! and hence determine the exact value of !. [2] −1x
Mark scheme
4(i) Use iteration correctly at least once M1 Must see correct attempt at x3 Obtain final answer 1.359 A1 Show sufficient iterations to 6 sf to justify answer or show sign A1 Answer required to exactly 4 sf change in interval [1.3585, 1.3595] Must see to at least x5 3 4(ii) Form correct equation in x (or α ) B1 x x= OE ln 2 x Obtain 12 e B1 2
Question
Find the exact coordinates of the stationary point of the curve with equation y = e 2 2x + 5. [5] 18 (a) Show that Ô dx = ln 27. 3
Mark scheme
5 Differentiate using the product rule *M1 du Must have u and v correct in a correct formula with =2 dx 1 dv − x and = me 2 dx −1 x −1 x Obtain correct 2e 2 − 12 e 2 (2 x + 5) A1 OE Equate first derivative to zero and solve for x DM1 Solution must come from linear terms Obtain x = − 12 only A1 1 Obtain 4e 4 or exact equivalent only A1 5
Question
[4] 2x 2 10 (b) Find the exact value of Ó 6 4 sin2 32 x dx. Show all necessary working. [5] 0
Mark scheme
6(a) Obtain 3 2 ln x or 3 2 ln(2 x) or 3 2 ln ( kx ) B1 Use subtraction law of logarithms correctly, showing sufficient M1 216 detail ln 216 − ln 8 = ln 8 Use power law of logarithms correctly M1 n ln ( kx ) = ln ( kx ) n Confirm ln 27 with sufficient working and no incorrect working A1 AG 4 6(b) Use appropriate identity to express integrand in form k1 + k2 cos3x *M1 3 k1 ≠ 0 . Allow 2 × x for 3x 2 Obtain correct 2 − 2cos3x A1 Integrate to obtain form k3 x + k4 sin 3x DM1 Obtain correct 2 x − 23 sin 3x A1 Use limits to obtain 13 π − 32 or exact equivalent A1 5
Question
The parametric equations of a curve are x = 3 sin 21, y = 1 + 2 tan 21, for 0 ≤ 1 < 14 0. (i) Find the exact gradient of the curve at the point for which 1 = 16 0. [4] (ii) Find the value of 1 at the point where the gradient of the curve is 2, giving the value correct to 3 significant figures. [4]
Mark scheme
7(i) Obtain dx dθ = 6cos 2θ B1 Obtain ddθy = 4sec 2 2θ B1 Divide ddθy by ddθx with θ equated to 16 π M1 Obtain 163 or exact equivalent A1 Allow FT on A1 if dx = 3cos 2θ and dy = 2sec 2 2θ dθ dθ 4 7(ii) Equate expression for ddyx to 2 with only one trigonometry ratio used *M1 Either cos 2θ or sec2θ Obtain cos3 2θ = 13 or sec3 = 3 A1 Attempt correct steps to find a value of θ from cos3 2θ = m , DM1 0 < m <1 Obtain θ = 0.402 and no others within the range A1 AWRT SC: Allow FT if ddθx = 3cos 2θ and ddθy = 2sec 2 2θ 4
Question
(i) Express 0.5 cos 1 − 1.2 sin 1 in the form R cos 1 + !, where R > 0 and 0Å < ! < 90Å, giving the value of ! correct to 2 decimal places. [3] (ii) Hence solve the equation 0.5 cos 1 − 1.2 sin 1 = 0.8 for 0Å < 1 < 360Å. [4] (iii) Determine the greatest and least possible values of 3 − cos 1 + 2.4 sin 12 as 1 varies. [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
8(i) 10 B1 Not 1.69 State R = 1.3 or 3 Use appropriate trigonometry to find α M1 AWRT ±1.18 rads, AWRT ±0.391 rads, AWRT ±67.4o , AWRT ±22.6o Obtain 67.38 with no errors seen A1 AWRT 3 8(ii) Carry out correct method to find one value of θ between 0 and 360 M1 Obtain 240.6 (or 344.6) A1 Carry out correct method to find second value of θ between 0 M1 Must be using either degrees throughout or radians throughout and 360 for M marks Obtain 344.6 (or 240.6) A1 4 8(iii) Recognise expression as [3 − 2 R cos(θ + α )]2 M1 Obtain [3 − 2 × (−1.3)]2 and hence 31.36 or 31.4 A1 Obtain [3 − 2 × 1.3]2 and hence 0.16 A1 3
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Mathematics 9709/22 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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