Cambridge IGCSE Mathematics - Additional 0606 — 2019 Oct/Nov Paper 2 · Variant 3

2 (i) Show that - = cosecx . [3] 1 - cos x cosecx - cot x (ii) Hence solve = 2 for 0° 1 x 1 180° . [2] 1 - cos x

Mark scheme: 2(i) sin 1 − cossin xx M1 express in terms of sinx and cosx 1 − cos x (1 − cos x ) A1 rewrite not as a fraction within a fraction sin (1 − cos x ) 1 A1 correct completion = cosec x answer given sin x 2(ii)  1  B1 sin x = x = 30°    2  x = 150º nfww B1 no extra answers

More questions on Prove trigonometric relationships involving the

3 The first four terms in the expansion of (1 + ax) 5 (2 + bx) are 2 + 32x + 210x 2 + cx 3 , where a, b and c are integers. Show that 3a 2 - 16a + 21 = 0 and hence find the values of a, b and c. [8]

Mark scheme: 3 (1 + ax)5 = 1+ 5ax + 10a2x2 + 10a3x3 soi B1 4 terms not nCr notation [2] + (10a + b)x + (5ab + 20a2)x2 M1 obtain expansion with 2 terms in x, 2 terms in x2 equate terms in x and x2 to give two M1 equations in a and b each consisting of three terms 10a + b = 32 A1 correct equations imply previous two 5ab + 20a2 = 210 M marks eliminate b M1 obtain 3a2 – 16a + 21 = 0 correctly A1 answer given a = 3 and b = 2 B1 c = 720 only B1 no additional answers

More questions on Use the binomial theorem for expansion of

4 (i) Given that y = 2x 2 - 4x - 7 , write y in the form a (x - b) 2 + c , where a, b and c are constants. [3] (ii) Hence write down the minimum value of y and the value of x at which it occurs. [2] (iii) Using your answer to part (i), solve the equation 2p - 4 p - 7 = 0 , giving your answer correct to 2 decimal places. [3]

Mark scheme: 4(i) y = 2(x – 1)2 – 9 B3 a = 2, b = 1, c = –9 in correct form. B1 for each 4(ii) minimum their –9 B1 FT from their correct form, with a > 0 when x = their 1 B1 FT from their correct form, with a > 0 4(iii) x = p or p = x 2 soi B1 9 M1 − c ( x − 1) = ( x − b ) = 2 a 9 − c or p − 1 = oe p − b = ( ) ( ) 2 a using their values of a, b, c from (i) p = 9.74 A1 completion not involving use of quadratic formula

More questions on Find the maximum or minimum value of the

5 (a) Solve 3 cot 2 y - r = 1 for 0 1 y 1 r radians. [4] 4 (b) Solve 7 cot z + tan z = 7 cosec z for 0° G z G 360° . [6]

Mark scheme: 5(a)  π  M1 ± 1.73… 3 tan  y −  = ( ± )  4  π π 2 π A1 1.04(7…) or 2.09(4…) y − = or 4 3 3 7 π A1 y = or 1.83 12 11π A1 y = or 2.88 12 5(b) correctly rewrite equation in terms of sinz M1 and cosz use sin2z = 1 – cos2z M1 appropriate use of Pythagorean identity for forming an equation in one trig ratio 6cos2z – 7cosz + 1 = 0 oe A1 (6cosz – 1) (cosz – 1) = 0 M1 solve three term quadratic in cosz 80.4º A1 279.6º A1

More questions on Solve, for a given domain, trigonometric

6 Do not use a calculator in this question. C 3– 3 A B 3 + 3 (i) Find tan ACB in the form r + s 3 , where r and s are integers. [3] (ii) Find AC in the form t u, where t and u are integers and t ! 1. [3]

Mark scheme: 6(i) 3 + 3 B1 [ tan ACB = ] 3 − 3 rationalise with 3 + 3 M1 simplify showing at least 3 terms in A1 numerator to 2 + 3 3 + 3 + 3 − 3 oe6(ii) ( AC ) 2 = ( 2 2 M1 Pythagoras ) ( ) at least 4 terms 12 + 6 3 + 12 − 6 3 A1

More questions on Know and use the six trigonometric functions

7 x = 2 y 6 y = x + 2 (3 x + 2) x O 6 The diagram shows part of the curve y = x + 2 and the line x = 2 . (3x + 2) (i) Find, correct to 2 decimal places, the coordinates of the stationary point. [6]

Mark scheme: 7(i) evidence of differentiation (3x + 2)–3 M1 –12(3x + 2)–3 × 3 A1 may use PR or QR on fraction part +1 B1 d y M1 1 –36(3x + 2)–3 = 0 set their = 0 d x x = 0.43 nfww A1 y = 0.98 only A1 7(ii) − 2 B1 oe 3 x + 2 1 2 B1 x 2  −2   −2  M1 insert correct limits into their two term + 2 −     integral and subtract two non-zero  6 + 2   2  terms in correct order 2.75 nfww A1 2.75 following B1 B1implies M1

More questions on Use differentiation to find stationary points

8 The roots of the equation x 3 + ax 2 + bx + 24 = 0 are 2, 3 and p, where p is an integer. (i) Find the value of p. [1] (ii) Show that a =-1 and find the value of b. [4] Given that a curve has equation y = x 3 - x 2 + bx + 24 find, using your value of b, dy (iii) , [1] dx (iv) the integer value of x for which the gradient of the curve is 2 and the corresponding value of y. [3] The coordinates of the point P on the curve are given by the values of x and y found in part (iv). (v) Find the equation of the tangent to the curve at P. [1]

Mark scheme: 8(i) p = –4 B1 8(ii) (x – 2) (x – 3) (x + 4) M1 FT (x – 2) (x –3) (x – p) (x2 – 5x + 6) (x + 4) A1 FT (x2 – 5x + 6) (x – p) multiply out two factors correctly obtain a = –1 A1 answer given x3 – x2 – 14x + 24 b = –14 stated B1 8(iii) d y 2 B1 FT their numerical b 3x2 – 2x + b = 3 x − 2 x − 14 d x 8(iv) d y M1 FT their numerical b set their equal to 2 d x x = 2 A1 y = 40 only A1 no additional answers 8(v) y – 40 = 2(x + 2) (y = 2x + 44) B1

More questions on Find factors of polynomials

9 B D X O A C The diagram shows points O, A, B, C, D and X. The position vectors of A, B, and C relative to O are OA = a, OB = 2b and OC = 3a . The vector CD = b . (i) Given that AX = m AD , find OX in terms of m, a and b. [2] (ii) Given that BX = n BC , find OX in terms of n, a and b. [2] (iii) Hence find the value of m and of n. [4] AX(iv) Find the ratio . [1] XD

Mark scheme: JJJG 9(i) AD = 2a + b B1 JJJG OX = a + λ( 2a + b ) B1 JJJG 9(ii) BC = 3a − 2b B1 JJJG OX = 2b + µ( 3a − 2b ) B1 JJJG JJJG 9(iii) OX = OX and equate for a or b M1 1 + 2λ= 3µ and λ= 2 − 2µ A1 solve correct equations for λor µ M1 4 5 A1 λ= and µ= 7 7 9(iv) 4 B1 FT λ/(1 – λ) 0 < λ < 1 or 4 : 3 3

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

10 The functions f and g are defined by 2 f (x) = ln (3x + 2) for x 2 - , 3 g (x) = e 2x - 4 for x ! R . (i) Solve gf ()x = 5 . [5] (ii) Find f -1 ()x . [2] (iii) Solve f -1 ( x) = g ( x) . [4]

Mark scheme: 10(i) 2 ( ln ( 3 x + 2 ) ) B1 gf ( x ) = e − 4 their gf = 5 M1 use lnap = plna or elna = a or lnea = a B1 correct use of log/exponential relationship seen anywhere 3x + 2 = 3 or (3x + 2)2= 9 A1 3 may take the form of e0.5ln9 9 may take the form of eln9 1 A1 x = only 3 10(ii) e y − 2 M1 find x in terms of y x = 3 e x − 2 ( −1 A1 interchange x and y = f ( x ) or = y ) correct completion 3 10(iii) e x − 2 2 x M1 their f -1 ( x ) = g ( x ) = e − 4 3 3e 2 x − e x − 10 ( = 0 ) A1 obtain quadratic in ex must be arranged as a three term quadratic in order shown 3e x + 5 e x − 2 = 0 ) M1 solve for ex ( )( ) ( x = ln2 or 0.693 only A1

More questions on Find the inverse of a one–one function