Know the conditions for f(x) = 0 to have — Paper 2 · IGCSE Mathematics - Additional 0606

4 Find the set of values of k for which the line y = 3x + k and the curve y = 2x 2 - 3 x + 4 do not intersect. [4]

4 marks

This question in 0606/22 Feb/March 2017

9 (i) Express 5x 2 - 14x - 3 in the form p x + q 2 + r , where p, q and r are constants. [3] (ii) Sketch the graph of y = 5x 2 - 14x - 3 on the axes below. Show clearly any points where your graph meets the coordinate axes. [4] y O x (iii) State the set of values of k for which 5 x 2 - 14 x - 3 = k has exactly four solutions. [2]

9 marks

Mark scheme: 9(i) 2 B3 B1 for each of p, q, r correct in correct format;  7  64 5  x −  − allow correct equivalent values.  5  5 If B0, then  7  64 SC2 for 5  x −  −  5  5 or SC1 for correct values but incorrect format 9(ii) B4 B2 for fully correct shape in correct position or B1 for fully correct shape translated parallel to the x-axis B1 for y-intercept at (0, 3) marked on graph 3 B1 for roots marked on graph at −0.2 and 3 −0.2 O 3 9(iii)  64  B2 FT their (i) 0 < k < their −   64  5  B1 for any inequality using their or max y 5 value is their 12.8soi

This question in 0606/21 May/June 2018

9 (i) Express 5x 2 - 14x - 3 in the form p x + q 2 + r , where p, q and r are constants. [3] (ii) Sketch the graph of y = 5x 2 - 14x - 3 on the axes below. Show clearly any points where your graph meets the coordinate axes. [4] y O x (iii) State the set of values of k for which 5 x 2 - 14 x - 3 = k has exactly four solutions. [2]

9 marks

Mark scheme: 9(i) 2 B3 B1 for each of p, q, r correct in correct format;  7  64 5  x −  − allow correct equivalent values.  5  5 If B0, then  7  64 SC2 for 5  x −  −  5  5 or SC1 for correct values but incorrect format 9(ii) B4 B2 for fully correct shape in correct position or B1 for fully correct shape translated parallel to the x-axis B1 for y-intercept at (0, 3) marked on graph 3 B1 for roots marked on graph at −0.2 and 3 −0.2 O 3 9(iii)  64  B2 FT their (i) 0 < k < their −   64  5  B1 for any inequality using their or max y 5 value is their 12.8soi

This question in 0606/23 May/June 2018

4 (a) Find the values of x for which 2x + 1 G 3x + 4 . [3] x 2 2 (b) Show that, whatever the value of k, the equation + kx + k + 1 = 0 has no real roots. [3] 4

6 marks

Mark scheme: 4(a) Expands, rearranges to form a M1 3-term quadratic on one side 4 x 2 + x − 3 [*0 ] 3 A1 Critical values and −1 4 3 A1 FT their critical values − 1 - x - final answer 4 4(b) 2  1  2 M1 k − 4 k + 1   ( )  4  −1 A1 discriminant independent of k and negative oe A1 FT their −1

This question in 0606/22 Feb/March 2019

6 Find the values of k for which the line y = kx - 7 and the curve y = 3x 2 + 8x + 5 do not intersect. [6]

6 marks

Mark scheme: 6 3x2 + 8x + 5 = kx – 7 M1 3x2 + (8 – k) x + 12 [= 0] soi A1 (8 – k)2 – 4 (3) (12) M1 k2 – 16k – 80*0 M1 Critical values: A1 –4 and 20 soi –4 < k < 20 A1 Alternative method: M1 for k = 6x + 8 oe M1 for y = (6x + 8)x – 7 M1 for 3x2 + 8x + 5 = (6x + 8)x – 7 A1 for x = ±2 A1 for k = –4, k = 20 A1 for –4 < k < 20

This question in 0606/21 May/June 2020

3 Find the values of k for which the line y = x - 3 intersects the curve y = k 2 x 2 + 5kx + 1 at two distinct points. [6]

6 marks

Mark scheme: 3 x −=3 k 2 x 2 + 5kx + 1 M1 k 2 x 2 + ( 5k − 1) x + 4 = 0 soi A1 (5 k − 1) 2 − 4( k 2 )(4) M1 9k 2 − 10k + 1 ∗ 0 M1 Critical values: A1 1 and 1 soi 9 1 A1 k < or k > 1 9

This question in 0606/22 May/June 2020

3 Find the values of k for which the equation x 2 + ( k + 9) x + 9 = 0 has two distinct real roots. [4]

4 marks

Mark scheme: 3 use b 2 − 4 ac ( k + 9 ) 2 −×4 9 ( > 0) M1 k 2 + 18 k + 45 ( > 0 ) A1 k =−15 k =−3 A1 k < −15 or k > −3 no isw A1 not ‘and’ mark final answer A0 if combined as one statement

This question in 0606/23 Oct/Nov 2020

2 Find the values of the constant k for which the equation kx 2 - 3 ( k + 1) x + 25 = 0 has equal roots. [4]

4 marks

Mark scheme: 2 Uses b 2 − 4 ac with at most one M1 error in substitution: ( −3( k + 1)) 2 − 4( k )(25)*0 9 k 2 − 82 k + 9*0 A1 Factorises or solves their 3-term M1 quadratic 1 A1 k = or 9; mark final answer 9

This question in 0606/22 Feb/March 2021

6 Find the values of m for which the line y = m x - 2 does not touch or cut the curve y = ( m + 1) x 2 + 8x + 1. [6]

6 marks

Mark scheme: 6 2 B1 (m +1) x + (8 − m) x + 3 = 0 oe, soi 2 M1 (8 − m) − 4(m +1)(3) m 2 − 28 m + 52 [*0] oe M1 dep on previous M1; condone one sign error where * is = or any inequality sign Factorises or solves their 3-term quadratic M1 dep on use of b2 – 4ac expression or equation for CVs Finds correct CVs: 2, 26 A1 2 < m < 26 A1 Mark final answer

This question in 0606/21 Oct/Nov 2021

3 Find the possible values of k for which the equation kx 2 + ( k + 5) x - 4 = 0 has real roots. [5]

5 marks

Mark scheme: 3 Uses b 2  4 ac oe: M1 ( k  5) 2  4 k ( 4) [* 0, where * could be = or any inequality sign] Forms a correct 3-term expression: A1 k 2  26 k  25 Factorises k 2  26 k  25 or solves M1 dep on first M1, FT their 3-term 2 quadratic in k k  26 k  25  0 oe Correct critical values 1,  25 soi A1 k   25 , k   1 A1 mark final answer

This question in 0606/22 May/June 2022

4 (a) Find the range of values of x satisfying the inequality ( 5x - 1)( 6 - x) 1 0 . [2] 2 1 (b) Show that the equation ( 2k + 1) x - 4kx + 2k - 1 = 0 , where k !- , has distinct, real roots. 2 [3]

5 marks

Mark scheme: 4(a) 1 M1 CVs , 6 5 1 A1 mark final answer x < , x > 6 5 4(b) (4k)2 – 4(2k + 1)(2k – 1) M1 16k2 – 4(4k2 – 1) A1 or 16 k 2  16 k 2  8k  8k  4 or better 4 > 0 A1

This question in 0606/23 May/June 2022

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]

4 marks

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

This question in 0606/23 May/June 2024