Find the maximum or minimum value of the — Paper 2 · IGCSE Mathematics - Additional 0606

9 A function f is defined, for x G , by f ( x) = 2 x 2 - 6x + 5 . 2 (i) Express f ( x) in the form a ( x - b) 2 + c , where a, b and c are constants. [3] (ii) On the same axes, sketch the graphs of y = f ( x) and y = f -1 ( x) , showing the geometrical relationship between them. [3] y O x (iii) Using your answer from part (i), find an expression for f -1 ( x) , stating its domain. [3]

9 marks

Mark scheme: 9(i) 2 B3 or B3 for a = 2 and b = 1.5 and c = 0.5 2 ( x − 1.5 ) + 0.5 isw provided not from wrong format isw or B2 for 2 ( x − 1.5 ) 2 + c where c ≠ 0.5 or a = 2 and b = 1.5 or SC2 for 2 ( x − 1.5 ) + 0.5 or  2 1  seen 2  ( x − 1.5 ) +   4  or B1 for ( x − 1.5 ) 2 seen or for b = 1.5 or for c = 0.5 or SC1 for 3 correct values seen in incorrect format e.g. 2 ( x − 1.5 x ) + 0.5 or 2 ( x 2 − 1.5 ) + 0.5 9(ii) y B3 B1 for correct graph for f over correct domain or correct graph for f – 1 over 5 correct domain B1 for vertex marked for f or f – 1 and intercept marked for f or f – 1 B1 for idea of symmetry – either symmetrical by eye, ignoring any scale or 1.5 line y = x drawn and labelled 0.5 x Maximum of 2 marks if not fully correct 0 0.5 1.5 5 9(iii) x – 0.5 2 M1 FT their a,b,c, provided their a ≠1 and = ( y − 1.5 ) a,b,c are all non-zero constants 2 y – 0.5 2 or = ( x − 1.5 ) and reverses 2 variables at some point −1 x – 0. 5 A1 must have selected negative square root f ( x ) = 1.5 − oe only; condone y = ... etc.; must be in terms 2 of x −1 6 − 8 x – 4 If M0 then SC2 for f ( )x = 4 oe or SC1 for − 1 −−( 6) ± 36 – 4( 2)(5 − x ) f ( x ) = oe 2(2) 1 B1 x ≥ oe 2

This question in 0606/22 May/June 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

3 (i) Write 8 + 7x - x 2 in the form a - (x - b) 2 , where a and b are constants. [3] (ii) Hence state the maximum value of 8 + 7 x - x 2 and the value of x at which it occurs. [2] (iii) Using your answer to part (i), or otherwise, solve the equation 8 + 7z 2 - z 4 = 0 . [3]

8 marks

Mark scheme: 3(i) 2 3 7 81  7  B1 b = −  x −  2 4  2   7  2 M1 ± 8 ±   seen  2  or expand given form and equate for 8 or 7 A1 fully correct 3(ii) 81 2 maximum their B1 4 7 when x = their B1 2 from their correct form 3(iii)  2 7  2 81 M1 replace x by z 2 in their (i) and  z −  = oe  2  4 equate to zero. 7 9 M1 z2 = ± 2 2 z = ± 8 A1

This question in 0606/23 Oct/Nov 2018

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]

8 marks

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

This question in 0606/23 Oct/Nov 2019

2 (a) Write 9x 2 - 12x + 5 in the form p ( x - q) 2 + r, where p, q and r are constants. [3] (b) Hence write down the coordinates of the minimum point of the curve y = 9x 2 - 12x + 5. [1]

4 marks

Mark scheme: 2(a) 2 B3 B1 for each of p, q, r correct in  2  9  x −  + 1 oe correctly formatted expression; allow  3  correct equivalent values  2  If B0 then SC2 for 9  x −  + 1 or  3  SC1 for correct values but other incorrect format 2(b)  2  B1 FT their (a) their  ,1  oe  3 

This question in 0606/21 May/June 2020

2 DO NOT USE A CALCULATOR IN THIS QUESTION. Variables x and y are related by the equation y = kx2 . When x = 1 + 2 , y = 1 - 2 . Find the value of k, giving your answer in the form a + b c , where a, b and c are integers. [4]

4 marks

Mark scheme: 2  2  1 2  3  2 2 M1  1  2   leading to   3  2 2  1  2 3  2 2 M1 FT their 3  2 2 if of equivalent  difficulty 3  2 2 3  2 2 3  2 2  3 2  4 DM1 FT their 3  2 2 if of equivalent Correctly expands 9  8 difficulty 7  5 2 A1 Alternative method (M1) 1  2 1  2 1  2     1  2 1  2 1  2 1  2       (M1) 3  2 2 1  2    1  2  2 Correctly expands (DM1) FT their 3  2 2 if of equivalent 3  3 2  2 2  4 difficulty [(  1) 2 ] 7  5 2 (A1)

This question in 0606/23 May/June 2022

6 (a) Write 3x 2 + 15 x - 20 in the form a ( x + b) 2 + c where a, b and c are rational numbers. [4] (b) State the minimum value of 3x 2 + 15x - 20 and the value of x at which it occurs. [2] 2 1 (c) Use your answer to part (a) to solve the equation 3y 3 + 15y 3 - 20 = 0 , giving your answers correct to three significant figures. [3]

9 marks

Mark scheme: 6(a) 2 B4 2  5  155  5  3  x +  − B2 for 3  x +  or 3 ( x + 2.5 )2  2  4  2   5  2 or B1 for  x +  or ( x + 2.5 )2  2  155 B2 for c = − or −38.75 4 25 or B1 for − −3 20 oe 4 6(b) 155 5 B2 FT their c from part a and −their b from Min value − when x is − 4 2 (a) B1 for either without contradiction 6(c) 1 2 M1 FT an expression of correct form from (a)   155 3 5 3  y +  = soi  2  4 1 A1 3 5 155 Rearranges as far as: y = −  2 12 soi y = 1.31 or −226 A1

This question in 0606/21 Oct/Nov 2022

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]

7 marks

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)

This question in 0606/22 May/June 2023

1 (a) Write 19 - 12x - 3x 2 in the form a ( x + b) 2 + c where a, b and c are integers. [4] (b) Hence find the maximum value of 19 - 12x - 3 x 2 and the value of x at which this maximum occurs. [2] (c) Use your answer to part (a) to solve the equation 19 - 12 u - 3 u = 0 . [3]

9 marks

Mark scheme: Question Answer Marks Partial marks 1(a) −3(x + 2)2 + 31 B4 B2 for −3(x + 2)2 or B1 for (x + 2)2 or a = –3 and b = 2 B2 for c = 31 or B1 for –4  –3 + 19 soi 1(b) Maximum value 31 when x = –2 B2 Strict FT their c from part (a) and −their b from part (a) B1 for either without contradiction 1(c) 2 M1 FT an expression of correct form from part (a) −3 u + 2 = −31 oe ( ) 31 A1 Rearranges as far as u = −2 3 1.48 cao or 1.475[13…] rot to 3 or more dp A1 43 − 4 93 or isw 3

This question in 0606/21 Oct/Nov 2023

2 (a) Write 3 + 4x - 2 x 2 in the form a + b ( x + c) 2 , where a, b and c are integers. [3] (b) Hence write down the range of the function f ( )x = 3 + 4x - 2x 2 , where x ! R . [1]

4 marks

Mark scheme: 2(a) 5 – 2(x  1)2 B3 Mark final expression B2 for – 2(x  1)2 or B1 for (x  1)2 or b = –2, c = –1 and B1 for 5 + b(x + c)2 oe with numerical values of b and c or a = 5 2(b) f  their 5 B1 STRICT FT of their 5 from (a)

This question in 0606/21 May/June 2024