Cambridge IGCSE Mathematics - Additional 0606 — 2024 May/June Paper 2 · Variant 3

1 The point A has coordinates (1, 4) and the point B has coordinates (5, 6). The perpendicular bisector of AB intersects the x-axis at the point C and the y-axis at the point D. Given that O is the origin, find the area of triangle OCD. [5]

Mark scheme: Question Answer Marks Partial Marks 1 y  5 2  x  3  oe 3  5  1 6  4  M1 for midpoint  ,  or  3, 5   2 2  1 M1for m  oe or  2 1 2 121 2 B1 FT for x-intercept (5.5, 0) and or 30.25 oe cao y-intercept (0, 11) soi; 4 FT their perpendicular bisector providing M1 M1 awarded

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2 Given that the equation kx 2 + ( 2k - 1) x + k + 1 = 0 has no real roots, find the set of possible values of k. [4]

Mark scheme: 2 Uses b2 – 4ac correctly: M1 (2k – 1)2 – 4(k)(k + 1) [*0 where * is any inequality sign or =] Simplifies to 8k + 1[*0] A1 1 M1 FT their ak + b where a and b are Critical Value: k = soi constants 8 1 A1 k > mark final answer 8

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3 (a) y 10 9 8 7 6 5 4 3 2 1 x -2 -1 0 1 2 3 4 5 6 -1 -2 -3 Draw the graphs of y = 2x - 5 and y = 4 - x for - 2 G x G 6 . [4] (b) Use your graphs to solve the inequality 4 - x G 2x - 5 . [2]

Mark scheme: 3(a) Accurate, ruled graphs drawn 4 M1 for y  4  x : 10 y 9  shape with vertex at (4, 0) 8 A1 Correct graph with y-intercept at 7 (0, 4) 6 5 M1 for y  2 x  5 : 4  shape with vertex at (2.5, 0) 3 A1 Correct graph with y-intercept at 2 (0, 5) 1 x -2 -1 0 1 2 3 4 5 6 -1 -2 3(b) x  1, x  3 final answer 2 FT their (a) providing at least M1 awarded and a pair of V-shaped graphs attempted B1 for exactly two correct critical values or B1 FT for exactly two correct FT critical values soi, FT their (a) providing at least M1 awarded and a pair of V-shaped graphs attempted

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4 (a) Find and simplify the term independent of x in the expansion of x 2 - 3 . [2] 2x (b) DO NOT USE A CALCULATOR IN THIS PART OF THE QUESTION. 4 4 (i) Use the binomial theorem to show that 1 + 2 2 - 1 - 2 2 = k 2 , where k is an integer ` j ` j to be found. [4] 4 4 1 + 2 2 - 1 - 2 2 (ii) Hence write ` j ` j in the form a + b 2 , where a and b are integers. [2] 1 + 2

Mark scheme: 4 2 4(a) 105 1   isw or 13.125 oe oe B1 for 10C 4 ( x 2 ) 6    3 8  2 x  4(b)(i) 1  4(2 2)  6(2 2) 2  4(2 2) 3  (2 2) 4 M1 soi or 1  4(  2 2)  6(  2 2) 2  4(  2 2) 3 ( 2 2) 4 soi 1  8 2  48  64 2  64 or A1 1  8 2  48  64 2  64 Correct difference stated or clearly M1 dep on sight of correct expansions with implied numerical coefficients 1  8 2  48  64 2  64  (1  8 2  48  64 2  64) 144 2 nfww A1 4(b)(ii)  their k 2 1  2 2 STRICT FT of their integer value of k  = 1  2 1  2  their k 2  2their k oe 1 B1 STRICT FT of their integer value of k simplified to 2(their k) – (their k) 2 their k 2 1  2 mark final answer for  1  2 1  2

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- 1 x 1 .5 (a) The function f is defined by f ( x) = 2 for r r cos x 2 2 (i) Show that f ( x) can be written as a tan 2 x + b , where a and b are integers. [2] (ii) Hence solve the equation f ( x) = 4 . [3] (iii) Hence also find the gradient of the curve y = f ( x) at each of the points where y = 4 . [4] (b) Solve the equation 50 cos 2i = 5 sin i + 47 for 0° G i G 360 ° . [5]

Mark scheme: 5(a)(i) sec 2 x  2tan 2 x oe and 2 M1 for use of a relationship to form a 2 convincing correct statement from correct completion to 3tan x  1 nfww which the answer can be easily 2 2sin 2 x determined e.g. sec x  or cos 2 x 1 2  2tan x cos 2 x 5(a)(ii) tan x 1 soi M1 4  their1 FT tan x  providing their 3 4  their1 > 0 their 3 π A2 A1 for each, ignoring extra solutions  x  π ,  or 0.785[39…] nfww 4 4 and no other solutions 5(a)(iii) f ( x )  6tan x sec 2 x oe M2 FT 2(their 3)tan x sec 2 x M1 for f ( x )  k tan x sec 2 x where k ≠ 2their 3   π     π   A2 A1 for each nfww  f      12;  f      12 nfww   4     4   5(b) Correct use of sin 2  cos 2  1 to form M1 Condone one sign or arithmetic error in rearrangement a 3-term quadratic in sinin solvable form 50sin 2  5sin 3   0  oe Solves or factorises their 3-term quadratic M1 FT their 3-term quadratic in sin in sine.g. (10sin+ 3)(5sin– 1) [= 0] sin= 0.3 sin= 0.2 soi A1 11.5 or 11.53[69...] A2 with no extras in range 168.5 or 168.46[30...] 197.5 or 197.45[76...] A1 for any two correct angles, ignoring 342.5 or 342.54[23...] extras

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6 DO NOT USE A CALCULATOR IN THIS QUESTION. (a) Given that x - 3 and x + 1 are both factors of 2x 3 - 3x 2 - 8x - 3 , solve the equation 2x 3 - 3 x 2 - 8x - 3 = 0 . [2] (b) The polynomial p ( x) = x 3 + ax 2 + bx + c , where a, b and c are constants, has remainder - 5 4 when divided by x - 1. The curve y = p ( x) has stationary points at x = and x = 2 . 3 (i) Find the values of a, b and c. [7] (ii) Hence use the second derivative test to show that the stationary point at x = 2 is a minimum. [2]

Mark scheme: 6(a) (2x + 1)(x – 3)(x + 1) nfww M1 Correct method leading to A1 1 x =  , x = 3, x = 1 2 6(b)(i)  p ( x )   3 x 2  2 ax  b   0  B1 2 B1 OR forms the product (3x – 4)(x – 2) = 0  4   4  3    2 a    b  0  3   3  3 2 2  2 a 2  b  0 B1 OR multiplies out to find 3 x 2  10 x  8  0 Solves to find the value of one unknown M1 FT their linear equations in a and b oe OR compares coefficients to state a value of a or b a = 5, b = 8 A1 [p(1)=] 1 + a + b + c = –5 oe, soi M1 [1 – 5 + 8 + c = 5] A1 c = 9 6(b)(ii)  p( x )   6 x  2(their a ) soi M1 FT their a 6(2) – 10 = 2 > 0 [therefore minimum] A1

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7 C D i rad O A B 5 cm 4 cm In the diagram, AD and BC are arcs of circles with common centre O. ODC and OAB are straight lines with OA = 5 cm and AB = 4 cm . Angle BOC = i radians . The area of the shaded region ABCD is 4rcm 2. (a) Find i. [3] (b) C D i rad O A B 5 cm 4 cm The straight line AC is added to the diagram and the region ACD is now shaded. Find the perimeter of the shaded region ACD. [5]

Mark scheme: 7(a) 1 2 1 2 M2 1 2 1 2  9   5   4π oe, soi M1 for  9  or  5  oe, 2 2 2 2 soi π A1  oe or 0.449 or 0.4487 to 0.4488 7 7(b) 5π 2 π [Arc AD = ] M1 for [Arc AD = ] 5  their FT any 7 7 stated value of from (a) [AC = ] 4.991[27...] rot to 4 or more sf 2 M1 for  π  [AC2 = ] 92 + 52 – 2(9)(5) cos  their   7  π FT their providing 0 <  < 2 11.2 or 11.23[526...] rot to 4 or more sf A1

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= cos 4x - . Given that = at the point , on the curve, find8 A curve is such that 2 r r r d x 4 dx 4 16 4 the equation of the curve. [7]

Mark scheme: 8  π  1  π  B2 B1 for  cos  4 x  4  dx  4 sin  4 x  4    c   π  1 k sin  4 x   where k > 0 or k   4  4 3 1   3π  π  M1 FT their k providing B1 awarded  sin  4      c 4 4   16  4  1  π  1 2 M1 FT for  cos  4 x    x   A  16  4  2  π   1  A  m cos  4 x     their  x    4   2   π  FT their k sin  4 x    their c  4  providing at least B1 M1 awarded 1  π  1 5π 2 M1 FT for y  cos  4 x    x  oe, cao 16  4  2 32 π 1  3π π  1  3π   cos      A   4 16  4 4  2  16  FT  π   1   their m  cos  4 x     their  x  A  4   2  providing previous M1 awarded

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9 y -1 x = 9 y = 4 + 3x - 1 ` j A O B x The diagram shows a sketch of part of the curve y = 4 + ( 3 x - 1 ) -1 and the line x = 9 . The point A has x-coordinate 1. The tangent to the curve at A meets the x-axis at the point B. Find the area of the shaded region. [10] Question 10 is printed on the next page.

Mark scheme: 9 d y  2 M2 d y  2  (3 x  1)  3 M1 for  k (3 x  1) where k > 0 d x d x d y 3 9 A1 d y [When x = 1]  and y = FT their providing M1 has been dx 4 2 d x awarded Equation of tangent: M1 d y FT the value of their at x = 1 9 3 d x y   ( x  1) oe isw 2 4 and their y B(7, 0) oe A1 Area of triangle: M1 FT their 7 and their –0.75x + 5.25 of the 1 9 27 form mx + c if needed     their 7   1 or nfww or 2 2 2 3 21  3 21    49   (7)    8 4  8 4  [Area under curve = F(x) = ] B2 B1 for  1  9 1 d x  k ln(3 x  1) or 1 ln3 x  1 4 x  ln(3 x  1)   oe  3 x  1 3  3 1 Correct and actioned plan e.g. M1 dep on at least previous B1 and correct 27 plan or correct FT area of triangle oe F(9) – F(1) – their 2 1 1 A1 18  ln13 2 3 or 19.4 or 19.35[49...]

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10 A B P O D C The diagram shows a parallelogram OABC. The point D divides the line OC in the ratio 2 : 3. OA = a and OC = c The point P lies on AD such that OP = m OB and AP = nAD , where m and n are scalars. Find two expressions for OP, each in terms of a, c and a scalar, and hence show that P divides both DA and OB in the ratio m : n, where m and n are integers to be found. [7]

Mark scheme:  10 OP   a + c  oe B1  2  B2   2   OP  a    a + c   oe B1 for OP  a   a +  their  c   5    5   Equates components at least once: M1 FT providing at least B1 awarded 2 1  or   5 Equates components: A1 2 1  and   5 5 2 A2 5 2   and A1 for  or  7 7 7 7 DP : PA = 2 : 5 = OP : PB oe

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