Mathematics 9709/21 May/June 2019: question paper and mark scheme

Mark scheme

1 Use logarithm subtraction property to produce logarithm of quotient M1 Factorise at least as far as x( x 2 − 4) and x ( x − 2) or use correct algebraic B1 Allow B1 either before or after application of log property long division to obtain a quotient of x + 2 and a remainder of 0 from Allow B1 for equivalent using factorisation then use of addition rule correct working ( x + 2) ( x2 − 2 x ) Allow B1 for ( x − 2x) 2 Obtain final answer ln( x + 2) using correct process A1 With no errors seen 3 Question Answer Guidance

Mark scheme

2(i) State or imply non-modular inequality (3x − 5) 2 < ( x + 3) 2 or B1 SC: Allow B1 for x < 4 from only one linear inequality corresponding equation or pair of different linear equations/inequalities Attempt solution of 3-term quadratic equation/inequality or of two M1 For M1, must get as far as 2 critical values different linear equations/inequalities Obtain critical values 12 and 4 A1 State answer 12 < x < 4 or equivalent A1 If given as 2 separate statements, condone omission of ‘and’ or ∩ but penalise inclusion of ‘or’ or ∪ 4 2(ii) Attempt to find n (not necessarily an integer so far) from M1 0/2 for trial and improvement 30.1n = or < their positive upper value from part (i) or 30.1n+1 = or < 3 × their positive upper value from part (i) Conclude 12 A1 2

Mark scheme

3 Use product rule to differentiate x 2 ln y M1 Allow M1 for 2 x ln y + x 2 y −1 oe 1 dy A1 Obtain 2 x ln y + x 2 × × y dx dy B1 dy Obtain … +2 + 5 =0 B1 for +2 + 5 = 0 , maybe implied by later work dx dx Substitute x = 3 and y = 1 to find value of their dx dy *M1 dy Dependent on at least one dx present dy A1 Obtain dx = − 14 2 Attempt equation of line through (3, 1) with gradient of normal DM1 Allow one sign error Obtain y = 7 x − 20 or equivalent unsimplified A1 FT on their perpendicular gradient 7

Mark scheme

4(a) Use identity tan 2 3 x = sec 2 3 x − 1 B1 Integrate to obtain form k1 tan 3 x + k 2 x M1 Obtain correct 13 tan 3x − x + c A1 3 4(b) Express integrand as e 2 x + 4e − x B1 Integrate to obtain form k3e 2 x + k4 e − x M1 Obtain correct 12 e 2 x − 4e − x A1 Use limits to obtain 12 e 2 − 4e −1 + 72 or similarly simplified equivalent A1 4

Mark scheme

5(i) Substitute x = 2 and equate to zero M1 Allow synthetic division for each– must result in an equation from each division Substitute x = −1 and equate to 27 M1 Allow unsimplified Obtain 4a + 2b = −24 and a − b = 48 or equivalents A1 Allow one error in each equation Solve a relevant pair of simultaneous linear equations M1 Dependent at least one M mark Obtain a = 12, b = −36 A1 5 5(ii) Divide by x − 2 at least as far as the x term to obtain M1 For synthetic division need to see 5 and their a + 10 in the 5 x 2 + ( their a + 10 ) x... bottom line Obtain 5 x 2 + 22 x + 8 A1 Obtain ( x − 2)(5 x + 2)( x + 4) A1 If solved using a calculator and then forming factors, must be correct for full marks 3

Mark scheme

6(i) Use quotient rule (or product rule) to differentiate M1 Penalise missing brackets by withholding the A mark unless recovered later dy 3 x 2 (2 − 5 x) − (−5)(8 + x3 ) A1 Obtain = or equivalent dx (2 − 5 x) 2 State or imply curve crosses x-axis when x = −2 B1 Substitute −2 to obtain 1 A1 4 6(ii) Equate numerator of first derivative to zero and rearrange as far as M1 kx 3 = ... or equivalent Confirm given result x = 0.6 x + 4 x −1 AG A1 Condone in this part error(s) in denominator of derivative 2 6(iii) Use iterative process correctly at least once M1 Obtain final answer 1.81 A1 Show sufficient iterations to 5 sf to justify answer or show a sign change A1 in the interval [1.805, 1.815] 3

Mark scheme

7(i) 1 B1 State or imply cosec 2θ = 2sin θ cosθ Attempt to express left-hand side in terms of sin θ and cos θ only M1 Simplify to confirm cosec 2θ AG A1 3 7(ii) Use identity to express left-hand side in terms of sin 30 or cosec30 M1 2 A1 Obtain or 2cosec30 and confirm 4 AG sin 30 2 7(iii) φ φ *M1 Allow sign errors Solve quadratic equation of the form kcosec2 + cosec − 12 = 0 or 2 2 φ φ A1 12sin 2 − sin − k = 0 correctly for cosec 12 φ or sin 12 φ to find two 2 2 values of sin 2 φ or cosec 12 φ 1 Obtain sin 12 φ = − 14 , 13 Use correct process to find at least one correct value of φ from DM1 Allow for any rounded or truncated value sin 12 φ = ± 14 , ± 13 Obtain any two of –331.0, –29.0, 38.9, 321.1 A1 Allow greater accuracy Obtain all four values and no others between –360 and 360 A1 Allow greater accuracy 5