Cambridge IGCSE Mathematics - Additional 0606 — 2023 May/June Paper 2 · Variant 2

1 (a) Solve the inequality 3x 2 - 12x + 16 2 3x + 4 . [3] (b) (i) Write 3x 2 - 12x + 16 in the form a ( x + b) 2 + c where a, b and c are integers. [3] (ii) Hence, write down the equation of the tangent to the curve y = 3x 2 - 12x + 16 at the minimum point of the curve. [1]

Mark scheme: Question Answer Marks Partial Marks 1(a) 3 x 2  15 x  12 [* 0] oe where * is any B1 inequality sign or = Factorises or solves their 3-term quadratic M1 FT their 3-term quadratic x < 1 or x > 4 mark final answer A1 1(b)(i) 2 3 2 3  x  2   4 B2 for 3  x  2  2 or B1 for  x  2  or a = 3, b = –2 and 2 B1 for a  x  b   4 with numerical values of a and b or c = 4 1(b)(ii) y = their 4 B1 STRICT FT their 4 from part (i)

More questions on Find the maximum or minimum value of the

! 0 .2 A curve has equation y = 32x 2 + 2 where x 8x (a) Find the coordinates of the stationary points of the curve. [5] (b) These stationary points have the same nature. Use the second derivative test to determine whether they are maximum points or minimum points. [3]

Mark scheme: 2(a) dy 2 x 3 B2  64 x  oe, isw dy 3 dx 8 B1 for  64 x  kx or dx dy 2 x 3  kx  where k is a non-zero d x 8 constant or dy 2 3 SC1 for  64 x  x  c dx 8 d y M1 FT their derivative providing it has two their = 0 and attempt to solve terms and at least one term is a correct dx power of x (0.25, 4), (0.25, 4) nfww, isw A2 A1 for either stationary point correct or for x =  0.25 nfww dy 3 or, if  64 x  16 x , then award dx  1 65  SC2 for  , oe or    2 4  SC1 for either of these stationary points 1 or x  oe 2 2(b) Correct second derivative: M1 d y 3 2 FT their = mx  nx where m  0 d y 3 4 dx 2  64  x oe, isw dx 4 and n  0 seen in part (a)  4  4 A2 dep on x = 0.25 nfww in part (a) 3  1  3  1  64  = 256 or 64  > 0     4  4  4  4  or A1 dep on x = 0.25 or x = –0.25 nfww in d 2 y d 2 y part (a) for correctly showing or stating when x = 0.25 2  256 or 2  0 oe 2 dx dx d y  3  is positive  64    and minimum [points] oe dx 2 4 x 4   OR 4  4 3  1  3  1  64    256 or 64    0     4  4  4  4  or d 2 y d 2 y when x = 0.25 2  256 or 2  0 oe dx dx and minimum [points] oe OR d 2 y 3 2  64  4 and this is positive for any dx 4 x value of x and minimum [points]

More questions on Use differentiation to find stationary points

3 DO NOT USE A CALCULATOR IN THIS QUESTION. (a) Show that x + 3 is a factor of - 12 + 23x + 3x 2 - 2x 3 . [1] (b) The curve y =- 5 + 33 x + 3x 2 - 2x 3 and the line y = 10x + 7 intersect at three points, A, B and C. These points are such that the x-coordinate of A has the least value and the x-coordinate of C has the greatest value. Show that B is the mid-point of AC. [7]

Mark scheme: 3(a) 12 – 69 + 27 + 54 = 0 B1 3(b) 10 x  7 2 x 3  3 x 2  33 x  5 oe, soi M1 Uses the correct factor x + 3 to find a M1 quadratic factor of the polynomial from part (a) oe with at least 2 terms correct 2 x 2  9 x  4 A1 or 2 x 2  9 x  4 Factorises or solves their 3-term quadratic DM1 dep on previous M1 factor = 0: (2 x  1)( x 4) or ( 2 x  1)( x  4) or (2 x  1)( x  4) oe x = 3, x = 0.5, x = 4 nfww A1 dep on at least M0 M1 A1 DM1 awarded A(3, 23), B(0.5, 12), C(4, 47) oe B2 dep on x = 3, x = 0.5, x = 4 nfww and correct method to show mid-point e.g.:  3 4 23  47   1  B1 dep on x = 3, x = 0.5, x = 4 nfww for ,  ,12 oe      2 2   2  A(3, 23), B(0.5, 12), C(4, 47) oe   0.5   3   3.5  AB   or            and  12   23   35  or   4   0.5   3.5  [x-coordinate of the mid-point ]  BC     47    12    35  oe 3 4 1  oe 3 4 1 2 2 OR [x-coordinate mid-point]  oe 2 2 and valid comment e.g. The points are collinear [so B is the mid-point of AC].

More questions on Solve simultaneous equations in two

4 Variables x and y are related by the equation y = 2 + tan ( 1 - x) where 0 G x G r . Given that x is 2 increasing at a constant rate of 0.04 radians per second, find the corresponding rate of change of y when y = 3 . [6]

Mark scheme: 4 dy 2 B2 must be seen  sec (1  x ) oe dx dy 2 B1 for  k sec (1  x ) oe, k  1 dx or dy 2 SC1 for  sec 1  x or dx dy 2  sec (1  x )  c dx Solves 3  2  tan(1  x ) as far as M1 1 – x = tan–11 π A1 1  x  isw or 0.7853[98…] 4 π x 1 isw or 0.2146[01…] 4 Correct use of chain rule and correctly writes M1 dep on at least B1 and an attempt to in terms of cosine or tangent: solve 3  2  tan(1  x ) their   1  0.04 oe, soi or 2  π  cos  their   4   2  π   their  1 1  tan  their    0.04 oe soi   4   0.08 oe, nfww A1 dep on all previous marks awarded

More questions on Apply differentiation to connected rates of

5 Variables P and T are known to be connected by the relationship P = AbT , where A and b are constants. Values of P are found for certain values of time, T. (a) Show that a graph of lgP against T will be a straight line. [2] (b) lg P 12 10 8 6 4 2 T 0 2 4 6 8 10 12 14 The diagram shows the graph of lgP against T. The graph passes through (0, 6) and (14, 12). Find the values of A and b. [4] (c) Using the graph or otherwise, find the length of time for which P is between 100 million and 1000 million. [3]

Mark scheme: 5(a) lg P  lg A  T lg b oe nfww B2 Must be seen and not from wrong working and correct comparison with y = mx + c soi B1 for lg P  lg A  T lg b isw, nfww 5(b) A = 106 oe isw and 4 B2 for A = 106 oe isw 3 or B1correct method which could be used b = 10 7 oe isw to find A e.g. 3 lgA = 6 or 12 =  14  lg A 7 3 B2 for b = 10 7 oe isw or B1correct method which could be used to find b e.g. 12  6 lgb = oe or 12 = 14lg b  6 14  0 5(c) lg P1 = 8 and lg P2 = 9 soi M1 If graph not used then allow M1 for substitution of their A and their b in the leading to exponential equation as far as 108 10 9  (theirb )T and  (theirb )T T1 = 4.6 to 4.8 theirA theirA or T2 = 6.8 to 7.2 OR substitution of their A and their b or their lgA and their lg b in the log equation lg108  their lg A  T (their lg b ) or better and lg109  their lg A  T (their lg b ) or better Difference of correct times: M1 T2  T1 where T2 = 6.8 to 7.2 T1 = 4.6 to 4.8 Answer in range 2.2 to 2.4 nfww A1 Alternative method 9  8 (M2) M1 for lg108 = 8 and lg109 = 9 and Change in T = 3 Change in lg P 3  7 Change in T 7 Answer in range 2.2 to 2.4 nfww (A1)

More questions on Use the equation of a straight line

6 (a) (i) Find the first three terms in the expansion of 1 + , in ascending powers of x. Simplify the 7 coefficient of each term. [2] 5 n x (ii) The expansion of 7 ( 1 + )x 1 + , where n is a positive integer, is written in ascending b 7 l powers of x. The first two terms in the expansion are 7 + 89x . Find the value of n. [2] (b) In the expansion of ( k - 2 x) 8 , where k is a constant, the coefficient of x4 divided by the coefficient of x2 is 5. The coefficient of x is positive. Form an equation and hence find the value of k. [5] 8

Mark scheme: 6(a)(i) 5 10 2 B2 B1 for any two correct terms or for the 1  x  x three terms listed but not summed 7 49 6(a)(ii) 12 B2  5  B1 for 7n + 5 = 89 or 7 n   89 oe    7  or nC1  12 6(b) 8C 4  k 4 ( 2) 4 [ x 4 ] oe or 1120k4 [x4] soi B1 8C 2  k 6 ( 2) 2 [ x 2 ] oe or 112k6 [x2] soi B1 1120 k 4 5 70  16  k 4 5 M1 FT providing at least B1 awarded and 6  or 6  oe, soi correct terms attempted 112 k 8 28 4 k 8 k 2 = 16 soi A1 [For coefficient of x to be positive k < 0, A1 therefore] k =  4

More questions on Use the binomial theorem for expansion of

7 (a) f ( x) = 3 + ( 4 x - 2) 5 where x 2 1. Find an expression for f l ( x) , giving your answer as a simplified algebraic fraction. [3] 5x (b) Variables x and y are related by the equation y = . Using differentiation, find the 3x + 2 approximate change in x when y increases from 10 by the small amount 0.01. [4] (c) (i) Differentiate y = x 3 ln x with respect to x. [2] x 2 (ii) Hence find ( 2 + 3 ln x) dx . [3] y e 6 o

Mark scheme: 7(a) 10(4 x  2) 4 10(4 x  2) 4 3 B2 for correct unsimplified form e.g. or isw 1 5 5 3  (4 x  2) 5 2 2  5  4 x – 2  4  4  3  (4 x  2)   1  3  (4 x  2)  1 2 or 1  5 2 10  3  (4 x  2)   4 x – 2  4 or B1 for 5(4x – 2)4 4 soi  1 5 1 or  3  (4 x  2)  2  g( x ) 2 7(b) dy 5(3 x  2)  3(5 x ) B1  2 oe isw or d x (3 x  2) dy 2 1  5 x   3 x  2   3  5  3 x  2  oe isw   dx [ y = 10] x = 0.8 B1 0.01  dy  M1 FT their x providing their x ≠ 10 or  their  oe 0.01and their genuine attempt at a δx  dx x 0.8  derivative using product or quotient rule 1 A1 dep on all previous marks awarded or 0.00016 or 1.6  104 isw 6250 Alternative method 2 y (B1) x  oe 3 y  5 dx 2(3 y  5)  3( 2 y ) (B1)  2 oe isw or dy (3 y  5) dx 2 1 2 y   3 y  5   3 ( 2)  3 y  5    dy   (M1) FT their genuine attempt at a derivative δ x d x  their oe using product or quotient rule  0.01  d y y 10   1 (A1) dep on all previous marks awarded or 0.00016 or 1.6  104 isw 6250 7(c)(i) 1 B2 1 3x2lnx + x3  or better, isw B1 for (their 3x2)lnx + x3  (their ) x x 7(c)(ii) x 3 ln x x 3 B3 must have arbitrary constant   c oe, isw 6 18 x 3 ln x x 3 B2 for  or 6 18 x 3 ln x kx 3  +c, k > 0 nfww 6 3 or B1 for 1  3 x 2 ln x  x 2  dx  1 x 2 dx soi 6  6  x 3 ln x x 2 or dx soi 6  6 or x 2 d x soi 3 ln x  3 x 2 ln x  d x  x    2 3 x 3 or soi 3 x ln x  dx  x ln x    3

More questions on Make use of the relationships in

8 A curve has equation y = cos where x is in radians. The normal to the curve at the point where 4 4 r x = cuts the x-axis at the point P. Find the exact coordinates of P. [7] 3

Mark scheme: 8 dy 1 x M2 dy x 1  sin M1 for  k sin , k < 0 or k = dx 4 4 dx 4 4 y = 0.5 B1 dy 1 3 3 M1 3   or  FT (their k)  providing at least M1 dx x  4π 4 2 8 2 3 awarded 8 M1 1 FT  mnormal   soi 3 their d y dx x  4π 3 8  4π  A1 dep on previous M1; y  0.5   x   oe must have exact values 3  3  FT their mnormal and their 0.5 providing both are non-zero, exact values  4π 3  A1   ,0  or exact equivalent;    3 16  mark final answer

More questions on Use differentiation to find gradients, tangents

9 A particle travels in a straight line so that, t seconds after passing a fixed point, its velocity, v ms -1 , is given by t v = e 4 for 0 G t G 4 , 16e v = 2 for 4 G t G k . t The total distance travelled by the particle between t = 0 and t = k is 13.4 metres. Find the value of k. [6]

Mark scheme: t9 t 16e 16e B3 B2 for either correct 4 4 t t e dt  4e (  c ) and 2 dt  (  c )   t t 4 4 or B1 for e dt  ae (  c )  where a is a constant, a > 0 16e b or 2 dt  (  c )  t t where b is a constant b > 0 Correct plan: M1 4 t k 16e 4 e dt  2 dt  13.4 soi 0 4 t Correct equation: A1 dep on B3; implies M1 1 0  16e 16e  4e  4e    13.4    k 4  OR [When t = 4 s = 4e – 4  16e When t = k s =  8e  4 and] k  16e 13.4 =  8e  4 k k = 10 or awrt 10.0 A1 dep on all previous marks awarded

More questions on Evaluate definite integrals and apply

10 B E D O C A The diagram shows a triangle OAB. The point C is the mid-point of OA. The point D lies on CB such that CD | DB = 2 | 3 . OC = c CB = b The point E lies on AB such that OE = m OD and AE = nAB where m and n are scalars. Find two expressions for OE, each in terms of b, c and a scalar, and hence find AE | EB . [8] Continuation of working space for Question 10.

Mark scheme: 10  c + 2 b  isw B2 B1 for   2  5  c + kb  where k  or 1, k > 0 5 2c   b  c  isw or B2 B1 for any of the following with n > 0 2c   b  n c  or nc   b  c  or b  (2  )c isw or c  b  (1  )( nc  b ) or c  b  (1  )( c  b ) isw n  c  b   (1  )(c  b ) Equates components at least once M1 FT providing of equivalent forms e.g.: 2  s c + t b  and xc   yb  z c  where s,  2   or   soi 5 t, x, y, z are scalars 2 A1  2   and   soi, nfww 5 4  10  A1     7  7  [AE : EB = ] 4 : 3 oe A1 must have earned all previous marks

More questions on Find the magnitude of a vector; add and