Mathematics 9709/23 May/June 2019: question paper and mark scheme

Mark scheme

1 Substitute –1 into p( x ) and equate to zero M1 Allow algebraic long division or the use of an identity with the remainder, in terms of m and k, equated to zero Obtain −4 + (k + 1) + m + 3k = 0 or equivalent A1 Obtain m = 3 − 4 k A1 3

Mark scheme

2(i) State or imply non-modular equation (4 + 2 x) 2 = (3 − 5 x)2 or pair B1 of linear equations Attempt solution of 3-term quadratic eqn or pair of linear equations M1 Obtain − 17 , 73 A1 1 SC B1 for x = − from one linear equation 7 3 2(ii) Attempt correct process to solve e3 y = k where k > 0 from (i) M1 Obtain 0.282 and no others A1 2

Mark scheme

3 Use quotient rule to find first derivative or equivalent *M1 dy 3ln x − 3 x × 1x A1 Condone lack of brackets in denominator unless specifically Obtain = or equivalent simplified to 2 ln x dx (ln x ) 2 Equate first derivative to zero and attempt value of x from ln x = k DM1 Must get as far as x = oe Obtain x = e A1 Allow e1 Obtain y = 3e A1 Allow 3 e1 SC1: If 3ln x − 3x × 1x = 0 seen with no reference to dy , then allow M1 A1 then following marks dx SC2: If denominator incorrect and numerator correct/reversed/added then max marks M0A0M1A1A1 SC3: If numerator reversed then max marks M1A0M1A1A1 5

Mark scheme

4(a) Use identity 2 cos 2 x = 1 + cos 2 x B1 1 B1 Integrate to obtain form x + sin 2 x 2 Integrate to obtain −2 cos 2 x B1 Apply limits correctly, retaining exactness M1 Dependent on at least one B mark Obtain 4 + 12 π or similarly simplified equivalent A1 5 4(b) Use y values ln 3, ln 6, ln 9 or decimal equivalents B1 Allow awrt 1.05, 1.34, 1.48, the correct level of accuracy may be implied by a correct answer Use correct formula, or equivalent, with h = 3 , and three y values M1 Obtain 12 × 3 ( ln 3 + 2 ln 6 + ln 9 ) and hence 7.81 A1 Allow greater accuracy 3

Mark scheme

5(i) Carry out division to obtain quotient of form x 2 + k M1 Obtain quotient x 2 − 4 A1 Allow use of an identity Obtain remainder 4 A1 SC: If only the remainder theorem is used to obtain 4 then B1 3 5(ii) Integrate to obtain at least k1 x3 and k 2 ln(2 x + 1) terms using the *M1 result from (i) Obtain correct 13 x 3 − 4 x + 2 ln(2 x + 1) A1 Apply limits and note or imply that constant k3 can be written DM1 ln e k3 Apply appropriate logarithm properties correctly M1 Obtain ln(49e−3 ) A1 5

Mark scheme

6(i) Equate 4t 2 e − t to 1, rearrange to t 2 = ... and hence t = ... M1 1 −t Allow M1 for t = e 4 1t Confirm t = 12 e 2 with necessary detail needed as answer is given A1 2 6(ii) Use iterative process correctly at least once M1 Obtain final answer t = 0.715 A1 Show sufficient iterations to 5 sf to justify answer or show a sign A1 SC: M1A1 from iterations to 4sf resulting in 0.71 change in the interval [0.7145, 0.7155] 3 6(iii) Obtain ddxt = 3 + 12e −2t B1 dy M1 Use product rule to find dt Obtain 8te − t − 4t 2 e − t A1 dy M1 Divide correctly to obtain dx Substitute value from part (ii) to obtain 0.31 A1 Allow greater accuracy 5

Mark scheme

7(a)(i) State R = 32 or equivalent or 5.657… B1 Use appropriate trigonometry to find α M1 Obtain α = 45 A1 3 7(a)(ii) Carry out correct process to find one value of θ M1 Obtain 17.1 A1 Ignore other positive values greater than 17.1 2 7(b) 1 B1 Use or imply cot 2 x = tan 2 x ±2 tan x M1 Use identity of form tan 2 x = to obtain equation in tan x 1 ± tan 2 x Obtain 6 tan 2 x + 10 tan x − 4 = 0 or equivalent A1 Attempt solution of 3-term quadratic equation for tan x M1 Obtain tan x = 13 and hence 0.32 A1 Allow greater accuracy Obtain tan x = −2 and hence 2.03 and no others between 0 and π A1 Allow greater accuracy 6