Cambridge IGCSE Mathematics - Additional 0606 — 2017 Oct/Nov Paper 2 · Variant 2

2 Solve the equation 0 .5 -0 .5 = x . [5] x + 5 x

Mark scheme: 2 2 x1.5 + 6 x −0.5 = x x 0.5 + 5 x −0.5 M1 Attempt to multiply by x 0.5 + 5 x− 0.5 ( ) or x 0.5 or divide by x 0.5 2 x1.5 + 6 x − 0.5 = x1.5 + 5 x 0.5 or A1 Simplified numerical powers x1.5 − 5 x 0.5 + 6 x− 0.5 = 0 6 2 2 x + 2 x + 6 x or = x or = x x + 5 5 1 + x x 2 − 5 x + 6 = 0 M1 M1dep obtain a three term quadratic. Allow errors in signs and coefficients but not powers ( x − 3 )( x − 2 ) = 0 M1 Solve a three term quadratic x = 3 or 2 only A1

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4 Solve the simultaneous equations log 2 x + 4 = 2 log 2 y , ^ h log 2 7y - x = 4 . ^ h [5]

Mark scheme: 4 x + 4 = y 2 B1 7 y − x = 16 B1 allow 2 4 for1 6 7 y − 16 + 4 = y 2 y 2 − 7 y + 12 → ( y − 3 )( y − 4 )( = 0 ) M1 Attempt to eliminate x or y to 2 obtain a three term quadratic. or x − 17 x + 60 → ( x − 5 )( x − 12 )( = 0 ) Solve a three term quadratic M1 M1dep → y = 3, x = 5 or y = 4 x = 12 A1 Allow for values seen even if correct pairs not clear.

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5 Naomi is going on holiday and intends to read 4 books during her time away. She selects these books from 5 mystery, 3 crime and 2 romance books. Find the number of ways in which she can make her selection in each of the following cases. (i) There are no restrictions. [1] (ii) She selects at least 2 mystery books. [3] (iii) She selects at least 1 book of each type. [3]

Mark scheme: 5(i) 10C 4 = 210 B1 5(ii) 2 Mystery 2 others = 5C2 × 5C2 =100 B3 B1 for one combination, 5 5 unsimplifiied 3 Mystery 1 other = C3 × C1 = 50 B1 for second combination, 4 Mystery = 5C 4 =5 unsimplifiied B1 for third combination, Total 155 unsimplifiied and total Alternative Method All – 0 Mystery – 1 Mystery B1 All minus 0 or 1 or both = 210 − 5C 4 − 5C1 × 5C3 B1 B1dep 1Mystery and 0 mystery unsimplified = 210 − 5 −×5 10 = 155 B1 B1dep final answer 5(iii) 2M1C1R = 5C 2 × 3C1 × 2C1 = 60 B3 B1 for one combination, 5 3 2 unsimplifiied 1M2C1R = C1 × C 2 × C1 = 30 B1 for second combination, = 5C1 × 3C1 × 2C 2 unsimplifiied 1M1C2R B1 for third combination, = 15 unsimplifiied and total Total 105

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6 The volume of a closed cylinder of base radius x cm and height h cm is 500 cm3. (i) Express h in terms of x. [1] 2 1000 2 (ii) Show that the total surface area of the cylinder is given by A = 2r x + cm . [2] x (iii) Given that x can vary, find the stationary value of A and show that this value is a minimum. [5]

Mark scheme: 6(i) 2 500 B1 Ignore units Condone r for x πx h = 500 → h = πx 2 6(ii) A = 2 πx 2 + 2 πxh M1 Correct expression for A and insert for their h. 2 500 2 1000 A1 Answer given = 2 πx + 2 πx × = 2 πx + 2 Condone r for x . πx x 6(iii) Differentiate: at least one power reduced by 1 M1 d A 1000 A1 = 4πx − d x x 2 dA 1000 A1 = 0 → x = 3 isw or ( x = 4.3 ( 0 ) ) dx 4π 2 1000 2 A1 awrt 349 A = 2π( 4.3 ) + = 349 cm 4.3 d 2 A 2000 B1 Correct second differential (need = 4 π + ( > 0 ) or a positive value not be evaluated) and conclusion. 2 3 dx x or ( → min) Examine correct gradient either side of x = 4.3 and conclusion

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7 The gradient of the normal to a curve at the point with coordinates ,x y is given by . 1 - 3x (i) Find the equation of the curve, given that the curve passes through the point (1, −10). [5] (ii) Find, in the form y = mx + c , the equation of the tangent to the curve at the point where x = 4 . [4]

Mark scheme: 7(i) d y 3 x − 1 B1 Gradient = Negative reciprocal. (Gradient or ) = Can be implied. d x x 1 1 B1 ± One correct term − = 3x 2 − x 2 3 1 M1 at least 1 fractional power increased y = 2 x 2 − 2 x 2 ( + C ) by1. − 10 = 2 − 2 + C → C = − 10 A1 one term correct with simplified coefficients 3 1 A1 For C from correct working. y = 2 x 2 − 2 x 2 − 10 7(ii) x = 4 → y = 16 − 4 − 10 = 2 B1 dy 1 B1 → = 6 − = 5.5 dx 2 Eqn with their grad and point (4, ...) M1 y − 2 A1 Must be in the form y = mx + c but Eqn of tangent: = 5.5 → y = 5.5 x − 20 oe x − 4 accept 2 y = 11x − 40

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8 The matrix A is . 4 3 (i) Find 2A -1 . [3] ^ h (ii) Hence solve the simultaneous equations 2y + 4x + 5 = 0 , 6y + 8x + 9 = 0 . [4]

Mark scheme: 8(i)  4 2  B1 2A=    8 6  −1 1  6 − 2  B2  6 − 2  ( 2 A ) =   B1 for   8  − 8 4   − 8 4  1 B1 for 8 8(ii) 4 x + 2 y = −5 B1 8 x + 6 y = − 9  − 5  M1 Allow recovery Pre multiply   by a 2 × 2 matrix.  − 9   x  1  6 − 2  − 5  M1  − 5    =    Pre multiply their   by their  y  8  − 8 4  − 9   − 9  answer to (i)  x  1  − 12   − 1.5  A2 A1 for x value   =   =   A1 for y value oe  y  8  4   0.5  Allow both unsimplified

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9 (i) Find x ln x . [2] dx (ii) Hence find lnx d x . [2] y 2k (iii) Hence, given that k 2 0 , show that ln x d x = k ln 4 k - 1 . [4] y k ^ h

Mark scheme: 9(i) d 1 M1A1 Product rule. One correct term + ( xlnx ) = x × + lnx isw another term. Allow unsimplified. dx x 9(ii) ∫ 1+ln x d x = x ln x M1 Correct use of (i) and must be dealing with 2 terms. soi ∫ ln xdx = xln x − x + ( C ) A1 Correct answer with no working is fine. 9(iii) 2 k M1 Insert limits and subtract correctly ∫k ln x d x = [ 2 k ln2 k − 2 k ] − [ k ln k − k ] using their result from (ii) which = k (2ln2 k − l n k − 1) must contain an ln function = k ln ( 2 k ) 2 − ln k − 1 M1 Uses n ln a = ln a n somewhere oe ( )   4 k 2   M1  a  = k  ln   − 1  Uses lna − lnb = ln   or  k   b      ln a + ln b = ln ab somewhere = k ( ln4 k − 1) A1 Answer given Correct completion.

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10 (i) Without using a calculator, solve the equation 6c 3 - 7c 2 + 1 = 0 . [5] It is given that y = tan x + 6 sin x . dy (ii) Find . [2] dx dy 3 2(iii) If = 7 show that 6 cos x - 7 cos x + 1 = 0 . [2] dx dy (iv) Hence solve the equation = 7 for 0 G x G r radians. [2] dx

Mark scheme: 10(i) 3 2 B1 Or correct division. Finding or c = 1 → 6 (1) − 7 (1) + 1 = 0 → ( c − 1) is a using one correct factor. factor. Attempt to factorise or use long division to M1 obtain 6 c 2 …± 1 or 6c 2 ± c … respectively 6 c − c − 1 = 0 ( c − 1)( 2 A1 ) ( c − 1)( 2 c − 1)( 3c + 1) =0 A1 1 1 A1 FT c = 1, , − From three different linear factors 2 3 10(ii) dy 2 B2 B1 for each term = sec x + 6cosx dx 10(iii) 1 B1 B1dep + 6cosx = 7 cos 2 x 2 1 Replaces sec x by cos 2 x → 6cos 3 x − 7cos 2 x + 1 = 0 B1 B1dep Answer given so all steps must be correct. 10(iv) 1 1 A2 A1 for 2 values awrt cosx = 1, , − . A1 for third value and no others in 2 3 range. No credit for answers in  π  → x = 0, 1.05  or  , 1.91 degrees  3 

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11 y y = mx + 8 B O A x y = 4 + 3x – x2 The diagram shows the curve y = 4 + 3 x - x 2 intersecting the positive x-axis at the point A. The line y = mx + 8 is a tangent to the curve at the point B. Find (i) the coordinates of A, [2] (ii) the value of m, [3] (iii) the coordinates of B, [2] (iv) the area of the shaded region, showing all your working. [5]

Mark scheme: 11(i) y = 0 → ( x − 4 )( x + 1) = 0 M1 Solve → A is ( 4,0 ) nfww A1 Indication somewhere that x = 4 when y = 0 11(ii) 4 + 3 x − x 2 = mx + 8 M1 Eliminate y . x 2 + ( m − 3 ) x + 4 = 0 2 2 M1 M1dep b − 4 ac ( = 0 ) → ( m − 3 ) = 16 Use of discriminant m = − 1 A1 Do not award if m = 7 is not discarded 11(iii) Obtain quadratic x 2 + ( m − 3 ) x + 4 = 0 using M1 Working must be seen for any marks to be awarded. their m and attempt to solve. Must not be awarded if m is not obtained correctly Point B (2, 6) A1 11(iv) 4 2 M1 Area under curve = ∫2 ( 4 + 3 x − x ) d x Integrate powers increased in at least 2 terms 4 A1  3 2 1 3  = 4 x + x − x    2 3  2  64   8  M1 M1dep = 16 + 24 − − 8 + 6 −     Insert limits of their 2 and 4 and  3   3  subtract in correct order. May be 1 = 7 2 3 implied by 18 −… 3 6 × 6 M1 Area of triangle using Intercept is (8,0) so area of triangle = = 18 2 ( their 8 − x B ) their B = × y B 2 or Attempt to find other suitable areas to result in a complete method. 1 2 A1 Accept 10.7. Must not be awarded Shaded area = 18 − 7 = 10 if point B is not obtained correctly. 3 3

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