Cambridge A Level Mathematics 9709 — 2013 May/June Paper 3 · Variant 1

1 Find the quotient and remainder when 2x2 is divided by x 2. [3] +

Mark scheme: 1 Carry out division or equivalent at least as far as two terms of quotient M1 Obtain quotient 2 x − 4 A1 Obtain remainder 8 A1 [3] 1

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4 (i) Solve the equation 4x x . [3] −1 = −3 4y correct to 3 significant figures. [3] −3 (ii) Hence solve the equation 4y+1 −1 =

Mark scheme: 4 (i) Either State or imply non-modular equation (4 x − 1) 2 = ( x − 3) 2 or pair of linear equations 4 x −=1 ± ( x − 3) B1 Solve a three-term quadratic equation or two linear equations M1 2 4 Obtain − and A1 3 5 2 Or Obtain value − from inspection or solving linear equation B1 3 4 Obtain value similarly B2 [3] 5 y 4 (ii) State or imply at least 4 = , following a positive answer from part (i) B1√ 5 Apply logarithms and use log a b = b log a property M1 Obtain –0.161 and no other answer A1 [3]

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5 For each of the following curves, find the gradient at the point where the curve crosses the y-axis: 1 x2 (i) y ; [3] + = 1 e2x + (ii) 2x3 5xy y3 8. [4] + + =

Mark scheme: 5 (i) Use correct quotient rule or equivalent M1 (1 + e 2 x )2 x − (1 + x 2 )2e 2 x Obtain 2 x 2 or equivalent A1 (1 + e ) 1 Substitute x = 0 and obtain − or equivalent A1 [3] 2 d y 3 2 (ii) Differentiate y and obtain 3 y B1 d x d y Differentiate 5xy and obtain 5 y + 5 x B1 d x 2 d y 2 d y Obtain 6 x + 5 y + 5 x + 3 y = 0 B1 d x d x GCE AS/A LEVEL – May/June 2013 9709 31 5 Substitute x = 0, y = 2 to obtain − or equivalent following correct work B1 [4] 6

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6 The points P and Q have position vectors, relative to the origin O, given by OP 7i 7j and OQ j k. −−→ −−→ = + −5k = −5i + + The mid-point of PQ is the point A. The plane is perpendicular to the line PQ and passes through A. (i) Find the equation of , giving your answer in the form ax by d. [4] + + cÏ = (ii) The straight line through P parallel to the x-axis meets at the point B. Find the distance AB, correct to 3 significant figures. [5]

Mark scheme: 6 (i) State or imply A is (1, 4, –2) B1 uuur State or imply QP = 12i + 6j – 6k or equivalent B1 Use QP as normal and A as mid-point to find equation of plane M1 Obtain 12 x + 6 y − 6 z = 48 or equivalent A1 [4] (ii) Either State equation of PB is r = 7i + 7j – 5k + λ i B1 Set up and solve a relevant equation for λ . M1 Obtain λ = −9 and hence B is (–2, 7, –5) A1 Use correct method to find distance between A and B. M1 Obtain 5.20 A1 Or Obtain 12 for result of scalar product of QP and i or equivalent B1 Use correct method involving moduli, scalar product and cosine to find angle APB M1 Obtain 35.26° or equivalent A1 Use relevant trigonometry to find AB M1 Obtain 5.20 A1 [5]

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4 8 (a) Show that 4x ln x dx 56 ln 2 [5] Ó2 = −12. 1 240 (b) Use the substitution u sin 4x to find the exact value of cos34x dx. [5] = Ó 0

Mark scheme: 8 (a) Carry out integration by parts and reach ax 2 ln x + b ∫ 12 x 2 d x M1* Obtain 2 x 2 ln x −∫ 1x . 2 x 2 d x A1 Obtain 2 x 2 ln x − x 2 A1 Use limits, having integrated twice M1 (dep*) Confirm given result 56ln 2 − 12 A1 [5] GCE AS/A LEVEL – May/June 2013 9709 31 (b) State or imply ddux = 4cos4 x B1 Carry out complete substitution except limits M1 Obtain ∫ ( 14 − 14 u 2 ) d u or equivalent A1 Integrate to obtain form k1u + k 2 u 3 with non-zero constants k1 , k 2 M1 Use appropriate limits to obtain 1196 A1 [5]

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