Mathematics 9709/23 May/June 2021: question paper and mark scheme

Mark scheme

1(a) Use at least one relevant logarithm property correctly M1 2+ x A1 Obtain correct equation = 9 or equivalent with no logarithms x 1 A1 Obtain x = 4 3 1(b) Attempt value of y from tan y = 1 ÷ (their (a)) M1 May be implied by an answer in degrees (76.0) Obtain 1.326 A1 AWRT; and no other answers in the range 2

Mark scheme

2 5 B1 Allow AWRT 0.714 Solve 5 x = 5 − 2 x to obtain x = 7 Attempt solution of linear equation where signs of 5x and 2x are M1 the same 5 A1 Allow AWRT –1.67 Obtain x = − 3 Substitute their values correctly M1 Substitution must be seen unless implied by a correct answer. Their values must come from consideration of 5 x = 5 − 2 x Obtain −6 + 4 and hence 10 A1 Alternative method for Question 2 State or imply non-modulus equation 25 x 2 = (5 − 2 x)2 B1 Attempt solution of 3-term quadratic equation M1 5 5 A1 Allow AWRT 0.714 and AWRT -1.67 Obtain − and 3 7 Substitute their values correctly M1 Substitution must be seen unless implied by a correct answer. Their values must come from consideration of 5 x = 5 − 2 x Obtain −6 + 4 and hence 10 A1 5

Mark scheme

3 Expand at least one of sin(2θ + 30) and cos(2θ + 60) correctly B1 Attempt expansions and division by cos 2θ to obtain equation in *M1 tan 2θ only 4 2 2 3 A1 OE Obtain tan 2θ = , , , 0.3849... 6 3 3 3 9 Obtain θ = 10.5 A1 AWRT, e.g. 10.525… A0 for 0.184 radians Use correct process to find second value of θ DM1 180o + their 2θ or 90o + theirθ 2 Allow if using radians correctly Obtain θ = 100.5 A1 FT AWRT; and no other answers within the range. 6

Mark scheme

4(a) Integrate to obtain form ke 2 x +1 M1 k ≠ 12 If k = 6 need to see evidence of integration, e.g. use of square bracket notation Obtain correct 3e 2 x +1 A1 Apply limits correctly to obtain 3e5 − 3e A1 or exact equivalent, indices must be simplified, but allow e1 . A0 for addition of + c 3 4(b) State tan 2 x = sec 2 x − 1 B1 Express 4sin 2 2x in the form k1 + k 2 cos 4 x M1 Where k1k 2 ≠ 0 Obtain correct 2 − 2cos 4 x A1 Integrate to obtain form k3 tan x + k 4 x + k5 sin 4 x M1 Where k3 k 4 k5 ≠ 0 1 A1 Condone absence of ... + c Obtain correct tan x + x − sin 4 x 2 5

Mark scheme

5(a) Carry out division at least as far as x 2 + kx or equivalent … M1 OE, e.g. comparing coefficients with coefficient of x 2 equal to 1 and attempt at a second coefficient. Obtain quotient x 2 + 4 x + 12 A1 Confirm remainder is 7 A1 AG 3 5(b) Include ( x − 2)2 as a factor M1 Must be a product of factors only ( )( SC B1 for x 2 − 4 x + 4 x 2 + 4 x + 12 ) Conclude ( x − 2)2 ( x 2 + 4 x + 12) A1 isw any attempt to factorise the quotient. 2 5(c) Apply logarithms and use power law for e −3 y = k where k > 0 M1 1 1 1 A1 Or exact equivalent Obtain y = − ln 2, ln 3 3 2 6 Must be simplified e.g. not lne or 3 ISW extra solutions but A0 if undefined solutions are included. 2

Mark scheme

6(a) Obtain x = 1 . B1 Allow e 0 . Must come from correct work, e.g. ln x = 0 . Obtain x = e 2 B1 Differentiate to obtain at least one correct term *M1 2ln x 2 A1 ln x ln x 2 Obtain correct first derivative − Allow + − x x x x x 2x Allow − 2 x Substitute at least one of their x-values corresponding to y = 0 to DM1 Allow unsimplified. find gradient Obtain gradient −2 [at A] and gradient 2e −2 at [B] A1 Must be simplified. 6 6(b) Equate first derivative to zero M1 Their derivative must have at least 2 terms. Obtain x = e A1 Allow e1 2

Mark scheme

7(a) 1 B1 AG Equate y to 3 and confirm p = 2sin 2 p 1 7(b) 1 M1 Consider sign of p − or equivalent for 0.5 and 0.6 2sin 2 p Obtain −0.09... and 0.06... or equivalents and justify conclusion A1 AG 2 7(c) Use iteration process correctly at least once M1 Need to see 0.55494… Obtain final answer 0.557 only A1 Allow recovery. Allow if iterations are to 4sf Allow if insufficient iterations seen. Show sufficient iterations to 5 s.f. to justify answer or show sign A1 If not starting at 0.55 then max marks M1A1A0 change in interval [0.5565, 0.5575] 3 7(d) dx B1 Obtain = 4 + 2e2t dt dy M1 Must be of the form p sin 2t + qt cos 2t Use product rule to find dt Obtain 6sin 2t + 12t cos 2t A1 Allow unsimplified. dy dy dx DM1 Must have either B1 or previous M1. Divide to obtain using their and correctly dx dt dt Obtain 0.826 A1 AWRT 5