Cambridge A Level Mathematics 9709 — 2016 Oct/Nov Paper 1 · Variant 1

1 8 2 Find the term independent of x in the expansion of 2x + . [4] 2x3

Mark scheme: 6  1  soi B1 May be seen within a number of2 8C6( 2 x )  3   2 x  terms 1 28 × 64 × oe (powers and factorials evaluated) B2,1,0 May be seen within a number of 4 terms 448 B1 Identified as answer [4]

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3 A r C r O ! rad D B In the diagram OCA and ODB are radii of a circle with centre O and radius 2r cm. Angle AOB = ! radians. CD and AB are arcs of circles with centre O and radii r cm and 2r cm respectively. The perimeter of the shaded region ABDC is 4.4r cm. (i) Find the value of !. [2] (ii) It is given that the area of the shaded region is 30 cm2. Find the value of r. [3]

Mark scheme: 3 (i) 2 rα+ rα+ 2 r = 4.4 r M1 At least 3 of the 4 terms required α = 0.8 A1 [2] 2 2 (ii) ½ ( 2 r ) 0.8 − ½( r )0.8 = 30 M1A1 Ft through on their α (3 / 2) r 2 × 0.8 = 30 → r = 5 A1 [3]

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4 C is the mid-point of the line joining A 14, −7 to B −6, 3 . The line through C perpendicular to AB crosses the y-axis at D. (i) Find the equation of the line CD, giving your answer in the form y = mx + c. [4] (ii) Find the distance AD. [2]

Mark scheme: 4 (i) C = (4, −2) B1 m AB = − 1/ 2 → mCD = 2 M1 Use of m1 m2 = − 1 on their m AB Equation of CD is y + 2 = 2 ( x − 4 ) oe M1 Use of their C and mCD in a line equation y = 2 x − 10 A1 [4] 2 2 2 (ii) AD = (14 − 0 ) + ( −−−7 ( 10 ) ) M1 Use their D in a correct method AD = 14.3 or √205 A1 [2] a 1− r 2 ( )

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5 The sum of the 1st and 2nd terms of a geometric progression is 50 and the sum of the 2nd and 3rd terms is 30. Find the sum to infinity. [6]

Mark scheme: ( )5 a (1 + r ) = 50 or = 50 B1 1 −r a 1 − r 3 ( ) ar (1 + r ) = 30 or = 30 + a B1 Or otherwise attempt to solve 1 − r for r Eliminating a or r M1 Any correct method r = 3 / 5 A1 a = 125 / 4 oe A1 S = 625 / 8 oe A1 Ft through on their r and a [6] ( −<1 r < 1) 2 4 2 2 4 ( )

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6 (i) Show that cos4x  1 −2 sin2x + sin4x. [1] (ii) Hence, or otherwise, solve the equation 8 sin4x + cos4x = 2 cos2x for 0Å ≤x ≤360Å. [5]

Mark scheme: 2 1 − sin 2 x = 1 − 2sin 2 x + sin 4 x AG B1 Could be LHS to RHS or vice6 (i) cos 4 x = ( ) [1] versa (ii) 4 2 4 2 4 2 8sin x + 1 − 2sin x + sin x = 2 1 − sin x M1 Substitute for cos x and cos x or ( ) 9sin 4 x = 1 A1 OR sub for sin 4 x → 3cos 2 x = 2 x = 35.3o (or any correct solution) A1 → cos x = ( ± ) 2 / 3 Any correct second solution from 144.7˚, 215.3˚, Allow the first 2 A1 marks for 324.7˚ A1 radians The remaining 2 solutions A1 (0.616, 2.53, 3.76, 5.67) [5]

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7 y y = 2x −1 2 B x O A y2 = 1 −2x The diagram shows parts of the curves y = 2x −1 2 and y2 = 1 −2x, intersecting at points A and B. (i) State the coordinates of A. [1] (ii) Find, showing all necessary working, the area of the shaded region. [6]

Mark scheme: 7 (i) A = (½, 0) B1 Accept x = 0 at y = 0 [1] 3/2 1  (1 − 2 x )  2 d x =  ÷ B1B1 May be seen in a single (ii) ∫ (1 −x2 )  ( − 2 )   3 / 2    expression 2  ( 2 x − 1) 3  1 ÷ x dy , may expand ∫ ( 2 x − 1) d x =  [ 2 ] B1B1 May use ∫ 3   a   [ 0 −−( 1/ 3) ] − [ 0 −−( 1/ 6) ] M1 ( 2 x − 1) 2 1/6 A1 Correct use of their limits [6]

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9 G P F p cm C D B 2 cm k E 10 cm j X O i 5 cm 4 cm A The diagram shows a cuboid OABCDEFG with a horizontal base OABC in which OA = 4 cm and AB = 15 cm. The height OD of the cuboid is 2 cm. The point X on AB is such that AX = 5 cm and the point P on DG is such that DP = p cm, where p is a constant. Unit vectors i, j and k are parallel to OA, OC and OD respectively. (i) Find the possible values of p such that angle OPX = 90Å. [4] −−→ (ii) For the case where p = 9, find the unit vector in the direction of XP. [2] −−→ (iii) A point Q lies on the face CBFG and is such that XQ is parallel to AG. Find XQ. [3]

Mark scheme: 9 (i) XP = −4i + (p – 5)j + 2k B1 Or PX [−4i + (p – 5)j + 2k].(pj + 2k) = 0 M1 Attempt scalar prod with OP/PO and set = 0 p 2 − 5 p + 4 = 0 A1 ( = 0 could be implied) p = 1 or 4 A1 [4] (ii) XP = −4i + 4j + 2k → │XP│ = 16 + 16 + 4 M1 Expect 6 Unit vector = 1/ 6 (−4i + 4j + 2k) oe A1 [2] (iii) AG = −4i + 15j + 2k B1 XQ = λAG soi M1 4 λ = 2/3 → XQ = −8 i + 10j + k A1 3 3 [3] 2 2

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1 9 11 The point P 3, 5 lies on the curve y = − x −1 x −5. (i) Find the x-coordinate of the point where the normal to the curve at P intersects the x-axis. [5] (ii) Find the x-coordinate of each of the stationary points on the curve and determine the nature of each stationary point, justifying your answers. [6]

Mark scheme: dy −2 −211 (i) = − ( x − 1) + 9 ( x − 5 ) M1A1 May be seen in part (ii) dx 1 9 mtangent =− + = 2 B1 4 4 Equation of normal is y − 5 = −½ ( x − 3 ) M1 Through (3, 5) and with m = −1/ mtangent x = 13 A1 [5] (ii) 2 2 dy ( x − 5 ) = 9 ( x − 1) B1 Set = 0 and simplify dx x 2 − x − 2 = 0 M1 Simplify further and attempt x − 5 = ( ± ) 3 ( x − 1) or ( 8 )( ) solution x = − 1 or 2 A1 d 2 y − 3 − 3 = 2 ( x − 1) − 18 ( x − 5 ) B1 If change of sign used, x values 2 d x close to the roots must be used and all must be correct d 2 y 1 When x = − 1, =− < 0 MAX B1 2 d x 6 d 2 y 8 When x = 2, = > 0 MIN B1 2 d x 3 [6]

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