Cambridge IGCSE Mathematics - Additional 0606 — 2021 Oct/Nov Paper 2 · Variant 2

1 y 20 15 10 5 x - 6 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 6 (a) On the axes, draw the graphs of y = 5 + 3 x - 2 and y = 11 - x . [4] (b) Using the graphs, or otherwise, solve the inequality 11 - x 1 5 + 3x - 2 . [2]

Mark scheme: Question Answer Marks Partial Marks 1(a) 4 M1 for ∨ shape of y = 5 + 3 x − 2 with vertex at  2   , 5   3  A1 for correct graph with y-intercept ( 0, 7 ) M1 for correct straight line for y = 11 − x A1 for correct straight line with y-intercept ( 0,1 1) 1(b) x > 2 or x < − 2 B2 Mark final answer for B2 B1 FT for exactly two correct critical values or correct FT critical values soi, FT dependent on at least M1 in (a)

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2 (a) Expand ( 2 - 3)x 4 , evaluating all of the coefficients. [4] 4 a (b) The sum of the first three terms in ascending powers of x in the expansion of ( 2 - 3)x 1 + b x l 32 is + b + cx , where a, b and c are integers. Find the values of each of a, b and c. [4] x

Mark scheme: 2(a) 16 − 96 x + 216 x 2 − 216 x 3 + 81x 4 B4 Mark final answer for B4 B3 for any 4 correct simplified terms in a sum or for all 5 simplified terms listed but not summed or for a correct simplified expansion that is not their final answer or B2 for any 3 correct simplified terms in a sum or for 4 correct simplified terms listed but not summed or B1 for any 2 correct simplified terms in a sum or for 3 correct simplified terms listed but not summed or M1 for correct unsimplified expansion 2 4 + 4 × 23 ( −3 x ) + 6 × 2 2 ( −3 x ) 2 3 4 +4 × 2 ( −3 x ) + ( −3 x ) 2(b) 2  a  B1 their 16 − 96 x + 216 x ….. ×  1 +  ( )  x  FT Expansion using their (a) a = 16 − 96 x + 16 − 96 a + 216 ax … soi x a = 2 B1 a FT their 16 x b = − 176 B1 c = 336 B1

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3 (a) Show that + = 2 cot x cosec x . [4] sec x - 1 sec x + 1 1 1 (b) Hence solve the equation + = 3 sec x for 0° 1 x 1 360° . [4] sec x - 1 sec x + 1

Mark scheme: 3(a) cos x cos x s ecx + 1 + sec x − 1 M1 + or 1 − cos x 1 + cos x sec 2 x − 1 cosx + cos 2 x + cos x − cos 2 x 2sec x A1 or 1 − cos 2 x tan 2 x 2cos x 2cos 2 x A1 or oe sin 2 x cos x sin 2 x Fully correct justification of given answer: 2cot xcosec x A1 3(b) 3tan 2 x = 2 oe or better, soi B1 or 5cos 2 x = 3 oe or better, soi or 5sin 2 x = 2 oe or better, soi 2 M1 FT an equation of the form tanx = [ ± ] oe or [±] 0.816[4…] 2 a tan x = b , a > 0, b > 0 3 or p sin 2 x = q or p cos 2 x = q 3 or cosx = [ ± ] oe or [±] 0.774[5…] where p > 0, q > 0 and p > q 5 2 or sinx = [ ± ] oe or [±] 0.632[4…] 5 39.2° or 39.2315… rot to 2 or more dp A2 no extras in range 140.8° or 140.7684… rot to 2 or more dp 219.2° or 219.2315… rot to 2 or more dp 320.8° or 320.7684… rot to 2 or more dp A1 for any two correct answers

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4 (a) Find the x-coordinates of the stationary points on the curve y = 3 ln x + x 2 - 7 x , where x 2 0 . [5] (b) Determine the nature of each of these stationary points. [3]

Mark scheme: 4(a) dy 3 B2 B1 for the first term correct and = + 2 x − 7 one other term correct dx x or for all terms correct with extra terms seen dy M1 Equates their to zero and rearranges to 3-term dx quadratic in x Solves their 3-term quadratic M1 Dep on previous M1 x = 0.5 , 3 nfww isw A1 no extra solutions 4(b) d 2 y 3 M1 dy = − + 2 FT their providing B1 dx 2 x 2 dx earned in (a) d 2 y d 2 y A1 x = 0.5 , < 0 → max or = −10 → max dx 2 dx 2 d 2 y d 2 y 5 A1 x = 3 , > 0 → min or = → min dx 2 dx 2 3 Alternative method Considers gradient at x ‒ h and x + h for x = 0.5 or (M1) dy FT their providing B1 x = 3 [where h is small] dx earned in (a) or Considers y-values at x ‒ h and x + h for x = 0.5 or x = 3 [where h is small] Correct conclusion for one turning point (A1) max at x = 0.5 or min at x = 3 Correct method and conclusion for second turning point (A1)

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5 (a) Solve the following simultaneous equations. e x + e y = 5 2 e x - 3e y = 8 [5] (b) Solve the equation e ( 2 t - 1 ) = 5e ( 5 t - 3 ) . [4]

Mark scheme: 5(a) Solves 3e x + 3e y = 15 and 2e x − 3e y = 8 oe M1 by elimination as far as 3 e x + 2e x = 23 or substitutes e y = 5 − e x into 2e x − 3e y = 8 oe OR Solves 2e x + 2e y = 10 and 2e x − 3e y = 8 oe by elimination as far as 2e y + 3e y = 2 or substitutes e x = 5 − e y into 2e x − 3e y = 8 oe x 23 y 2 A1 e = or e = oe 5 5 x = ln4.6 [ = 1.53] oe A1 If M0 scored SC1 for using their expression of the form cex = d or y = ln0.4 [ = −0.916 ] oe d d to give x = ln provided > 0 c c Finds the other value, e y or e x , by substituting their ex M1 FT their ex or ey or ey y = ln0.4 [ = −0.916 ] oe A1 or x = ln4.6 [ = 1.53] oe 5(b) 2 t −−1 ( 5 t − 3 ) 5 t −−3 ( 2 t 1) 1 M1 e = 5 or e −= oe 5 2 − 3 t 3 2 1 A1 e = 5 or e t−= 5 1 M1 FT their e a − bt = 5 or 2 − 3t = ln5 or 3t − 2 = ln 5 ct d 1 their e −= where a, b, c and 5 d are positive integers 2 − ln5 2 + ln0.2 A1 t = or t = or 0.13[0] oe 3 3 Alternative method ln e 2 t −1 = ln5 + lne 5 t − 3 oe (M1) (2t − 1)[lne] = ln5 + (5t − 3) [ lne ] oe (A1) 5t – 2t = 3 – 1 – ln5 oe (M1) Dep on one correct log law applied with at most one sign error 2 − ln5 2 + ln0.2 (A1) t = or t = or 0.13[0] oe 3 3

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6 DO NOT USE A CALCULATOR IN THIS QUESTION. All lengths in this question are in centimetres. You may use the following A trigonometrical ratios. 3 60° sin 60° = 2 6 – 2 1 cos 60° = 2 6 + 2 tan60° = 3 C B The diagram shows triangle ABC with AC = 6 - 2 , AB = 6 + 2 and angle CAB = 60° . (a) Find the exact length of BC. [3] 6 + 2 (b) Show that sinACB = . [2] 4 (c) Show that the perpendicular distance from A to the line BC is 1. [2]

Mark scheme: 6(a) 2 2 M1 6 − 2 + 6 + 2 − 2 6 − 2 6 + 2 cos60 ( ) ( ) ( )( ) 1 M1 Condone one error in expansion 6 + 2 − 2 12 + 6 + 2 + 2 12 −×2 ( 6 − 2 ) × of brackets 2 [ BC = ] 2 3 isw A1 6(b) their 2 3 6 + 2 their 2 3 6 + 2 M1 Condone other letters for ACB = or = sin60 sin ACB 3 sin ACB 2 3 1 6 + 2 A1 A0 if necessary brackets missing 6 + 2 sin ACB = × × = unless clearly recovered ( ) 2 2 3 4 6(c) 6 + 2 x M1 Complete method = 4 6 − 2 or 1 × their 2 3 × x = 2 1 × 6 − 2 × 6 + 2 × sin60 ( ) ( ) 2 [where x is the perpendicular from A to BC] A1 6 − 2 6 + 2 ( )( ) 6 − 2 x = = = 1 4 4 or ( 6 − 2 ) 3 4 x = × = = 1 2 3 2 4

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7 It is given that 2 2 x = e + 2 for x 2- 1. d x ( x + 1 ) d y d y (a) Find an expression for given that = 2 when x = 0 . [3] d x d x (b) Find an expression for y given that y = 4 when x = 0 . [3]

Mark scheme: 7(a)  dy  1 2 x −1 5 B3 1 2 x −1 = e − ( x + 1) + oe M2 for e − ( x + 1) + c oe    dx  2 2 2 or M1 for any two terms correct 1 2 x −1 from e , − ( x + 1) , + c 2 7(b) 1 2 x M1 [ y = ] e − ln ( x + 1) 4 5 M1 FT their c from (a), providing + their × x + d c ≠ 0 2 1 2 x 5 15 A1 [ y = ] e − ln ( x + 1) + x + oe 4 2 4

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8 Variables x and y are such that when y is plotted against log 2 ( x + 1), where x 2- 1, a straight line is obtained which passes through ( 2, 10.4) and ( 4 , 15.4) . (a) Find y in terms of log 2 ( x + 1) . [4] (b) Find the value of y when x = 15 . [1]

Mark scheme: 8(a) 15.4 − 10.4 M1 [Gradient =] oe soi 4 − 2 10.4 = their2.5 × 2 + c or 15.4 = their2.5 × 4 + c M1 FT their gradient or y − 10.4 y − 15.4 = their 2.5 or = their 2.5 x − 2 x − 4 [Gradient = ] 2.5 soi and [intercept =] 5.4 soi A1 y = 2.5log 2 ( x + 1) + 5.4 oe isw A1 Alternative method 10.4 = 2m + c and 15.4 = 4m + c (M1) and solving to find m or c Use their m or c to find their c or m (M1) m = 2.5 and c = 5.4 (A1) y = 2.5log 2 ( x + 1) + 5.4 oe isw (A1) 8(b) 5929 B1 or 237.16 25 8(c) 5 = their 2.5log 2 ( x + 1) + their 5.4 M1 FT their equation from (a) of correct form with m ≠ 1 or 0, and and rearrange to make log 2 ( x + 1) the subject c ≠ 0 Condone any base 4 A1 Condone any base − = log 2 ( x + 1) oe 25 x = − 0.105 or −0.1049[74…] rot to 4 or more sf A1

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(b) DO NOT USE A CALCULATOR IN THIS PART OF THE QUESTION. Find the exact x-coordinate of each of the two points where the normal cuts the curve again. [5]

Mark scheme: 9(a) dy 2 M2 M1 for any two terms correct = 3 x + 2 x − 4 dx dy A1 x = 1 → = 1 dx [ m⊥= ] −1 M1 −1 FT their1 y – 4 = −1(x – 1) oe isw A1 FT their m⊥ 9(b) x 3 + x 2 − 4 x + 6 = their ( −+x 5 ) M1 FT their linear equation of the form y = mx + c where m ≠ 0 and 3 2 c ≠ 0 → x + x − 3 x + 1 [ = 0 ] from (a) Correct quadratic factor: x2 + 2x – 1 B2 B1 for any two out of three terms correct Must be from the correct cubic Solves their (x2 + 2x – 1) = 0 using the formula or by M1 dep on M1 and valid attempt at completing the square finding quadratic factor M0 if their quadratic factor does not have real roots −±2 8 −±2 2 2 A1 isw or isw 2 2

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10 (a) The first three terms of an arithmetic progression are x, 5x - 4 and 8x + 2 . Find x and the common difference. [4] (b) The first three terms of a geometric progression are y, 5y - 4 and 8y + 2 . (i) Find the two possible values of y. [4] (ii) For each of these values of y, find the corresponding value of the common ratio. [2]

Mark scheme: 10(a) Eliminate one unknown using two correct equations M2 B1 for one correct equation seen, e.g. e.g. d = 4x – 4 oe d = 4x – 4 oe d = 3x + 6 oe or d = 3x + 6 oe and or 2d = 7x + 2 oe solve as far as x = … or d =… May come from the sum of terms, e.g. 11x – 3d = 2 x = 10 A1 d = 36 A1 10(b)(i) 5 y − 4 8 y + 2 M1 = oe y 5 y − 4 25 y 2 − 40 y + 16 = 8 y 2 + 2 y M1 → 17 y 2 − 42 y + 16 [ = 0 ] (17 y − 8 )( y − 2 )[ = 0 ] M1 Solves their 3-term quadratic 8 A1 Both values , 2 17 Alternative method Eliminates y from yr = 5y – 4 and yr2 = 8y + 2 (M1) and simplifies to 3-term quadratic in r → 2 r 2 + r − 21[ = 0 ] Solves their 3-term quadratic (M1) Substitutes their two r values to find two y values (M1) 8 (A1) , 2 17 10(b)(ii) 7 B2 B1 for one correct − , 3 2

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