Link to past paper on OCR website: www.ocr.org.uk The above link takes you to OCR s website. From there you click QUALIFICATIONS, QUALIFICATIONS BY TYPE, AS/A LEVEL GCE, MATHEMATICS (MEI), VIEW ALL DOCUMENTS, PAST PAPERS JUNE SERIES 2010, QUESTION PAPER UNIT 4751/01 CORE MATHEMATICS 1 These solutions are for your personal use only. DO NOT photocopy or pass on to third parties. If you are a school or an organisation and would like to purchase these solutions please contact Chatterton Tuition for further details. Section A Question 1 The gradient of the line y = 3x + 1 is 3. So the gradient of the new line will also be 3. We now have the gradient and a point (4, 5) that the line goes through so we can work out the equation of the line using the formula y y 1 = m(x x 1 ) y - 5 = 3(x 4) expand the brackets y 5 = 3x 12 add 5 to both sides y = 3x - 7 www.chattertontuition.co.uk 0775 950 1629 Page 1
Question 2 i) 5 3 x a 2x3 x b 3 x 2b 4 = 125 x a 6 x b 3 x 2 x b 4 125 x 2 x a 6 x b 3+4 = 250a 6 b 7 ii) a negative power has the effect of turning the fraction upside down and changing the negative power to a positive power ( )1 = 16 iii) is the same as x 3 16 3/2 = (16 1/2 ) 3 = ( 16) 3 = 4 3 = 64 www.chattertontuition.co.uk 0775 950 1629 Page 2
Question 3 multiply both sides by c ac = - 5 add 5 to both sides ac + 5 = = ac + 5 square both sides y = (ac + 5) 2 www.chattertontuition.co.uk 0775 950 1629 Page 3
Question 4 i) first expand the brackets 2 2x 6x + 5 add 2x to both sides 2 8x + 5 subtract 5 from both sides -3 8x divide both sides by 8 x rewrite the other way around x ii) no need to expand the brackets as we want to find when the inequality equals 0 it has already been factorised If the quadratic were equal to 0 then the two x values would be -4 and Now we need to decide if we want x to be between these two values or either side of them Try x = 0 (which is in between), when x = 0: we get -4 which is less than 0 so we do want x to be between the two values Alternatively if we know the way that a quadratic graph lies (like a big ) then we can also see that this curve dips below 0 when x is between the two values -4 x www.chattertontuition.co.uk 0775 950 1629 Page 4
Question 5 i) 48 = 16 x 3 = 16 x 3 = 4 3 and 27 = 9 x 3 = 9 x 3 = 3 3 so we have 4 3 + 3 3 = 7 3 ii) we need to rationalise the denominator, so multiply the numerator and the denominator by (3-2) x = when we expand the denominator we get 3 2 3 2 = 9-3 2 + 3 2-2 = 7 expanding the numerator we get 5 2 (3 + 2) = (5 x 3 x 2) + (5 x 2 x 2) = 15 2 + (5 x 2) = 15 2 + 10 so we have www.chattertontuition.co.uk 0775 950 1629 Page 5
Question 6 (5 + 2x 2 )(x 3 + kx + m) I have shown the two terms that give x 3 with a curved link (shown above) We have 5x 3 + 2kx 3 and we know this should equal 29x 3 5 + 2k = 29 subtract 5 from both sides 2k = 24 divide both sides by 2 k = 12 As we are dividing by (x 3) we need to input x = 3 into x 3 + kx + m and the remainder should be 59 Using the remainder theorem f(3) = 59 (3 3 ) + 3k + m = 59 27 + 3k + m = 59 subtract 27 from both sides 3k + m = 32 we already know that k = 12 (3 x 12) + m = 32 36 + m = 32 subtract 36 from both sides m = -4 k = 12 and m = -4 www.chattertontuition.co.uk 0775 950 1629 Page 6
Question 7 First set out pascal s triangle. The power is 4 so we only need 5 rows (one more than the power) 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 The final row of the triangle corresponds to the coefficients that we need to use in our expansion I find it easier to work down the page. The first term (1) increases in powers from 0 to 4 whilst the second term ( ) decreases in powers from 4 to 0. On any row the sum of the powers should be 4. 1 x (1) 0 x ( )4 + 4 x (1) 1 x ( )3 + 6 x (1) 2 x ( )2 + 4 x (1) 3 x ( )1 + 1 x (1) 4 x ( )0 now we need to tidy up each term (1 x 1 x ( )) + (4 x 1 x ( )) + (6 x 1 x ( )) +(4 x 1 x ( )) + (1 x 1 x 1) + + + 2x + 1 www.chattertontuition.co.uk 0775 950 1629 Page 7
Question 8 first we need to factorise out the 5 (no need to factorise the whole thing, just the first two terms will do) 5(x 2 + 4x) + 6 now complete the square on the bracket 5((x + 2) 2 4) + 6 now multiply by the 5 again 5(x + 2) 2 20 + 6 group the units 5(x + 2) 2-14 www.chattertontuition.co.uk 0775 950 1629 Page 8
Question 9 means that one side implies the other means the left implies the right x 5 = 0 x = 5 x 2 = 25 but this doesn t work the other way if x 5 = 0, then x must be 5 and so x 2 = 5 2 = 25 x 2 = 25 x = 5 (not just +5) this does not imply that x 5 = 0 www.chattertontuition.co.uk 0775 950 1629 Page 9
Section B Question 10 i) we need to first multiply 2 by -3 to get -6 then we need to find two numbers that multiply to give -6 but add to give -1 (the coefficient of x) these two numbers are -3 and +2 rewrite the equation splitting the middle x term into -3x and +2x 2x 2 3x + 2x 3 + 0 factorise in pairs x(2x 3) + 1(2x 3) = 0 factorise again (2x 3)(x + 1) = 0 So (2x 3) = 0 or (x + 1) = 0 2x = 3 or x = -1 x = 1.5 or x = -1 ii) we know the values of x where the curve meets the x axis from part i) when x = 0, y = 2(0) 2 0 3 = -3 we have the three coordinates (-1, 0), (0, -3) and (1.5, 0) the curve is a quadratic (with a leading positive x 2 )so will have the usual shape www.chattertontuition.co.uk 0775 950 1629 Page 10
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iii) the discriminant tells us the number of real roots there are the discriminant is part of the quadratic formula and is equal to b 2 4ac in this case a = 1, b = -5 and c = 10 discriminant = b 2 4ac = (-5) 2 (4 x 1 x 10) = 25 40 = -15 discriminant 0 so there are no real roots as this is negative there are no real roots (as in the quadratic formula we would need to square root this and we cannot square root a negative) iv) this is simultaneous equations both equations are equal to y, so put them equal to each other 2x 2 x 3 = x 2 5x + 10 subtract x 2 from both sides x 2 x 3 = -5x + 10 add 5x to both sides x 2 + 4x 3 = 10 subtract 10 to both sides x 2 + 4x - 13 = 0 we can t factorise this so we could use the quadratic formula or complete the square I shall complete the square as the numbers are fairly easy for this (x + 2) 2 2 2-13 = 0 (x + 2) 2-17 = 0 add 17 to both sides (x + 2) 2 = 17 square root both sides x + 2 = 17 subtract 2 from both sides x = -2 17 The question only asked for the x values not the y values so we are done www.chattertontuition.co.uk 0775 950 1629 Page 12
Question 11 i) first we can find the gradient of the line through A and B gradient = gradient AB = = = - We now have the gradient and a point (A or B) that the line goes through so we can work out the equation of the line AB using the formula y y 1 = m(x x 1 ) (I will use point A) y - 3 = - (x-- 1) multiply both sides by 3 3y 9 = -(x + 1) 3y 9 = -x 1 add x to both sides x + 3y 9 = -1 add 1 to both sides x + 3y - 8 = 0 www.chattertontuition.co.uk 0775 950 1629 Page 13
ii) area of a triangle is ½ base x height we need to find where the line crosses the x and the y axis so we can determine the size of the base and the height of the triangle line AB meets the x axis when y = 0 x + (3 x 0) 8 = 0 x 8 = 0 add 8 to both sides x = 8 base = 8 line AB meets the y axis when x = 0 0 + 3y 8 = 0 3y 8 = 0 add 8 to both sides 3y = 8 divide both sides by 3 y = height = area of triangle = x 8 x = square units www.chattertontuition.co.uk 0775 950 1629 Page 14
iii) the gradient of the perpendicular bisector will be the negative reciprocal of the gradient of line AB gradient of bisector will be 3 midpoint of AB is (, ) = (2, 2) We now have the gradient and a point that the line goes through so we can work out the equation of the line using the formula y y 1 = m(x x 1 ) y 2 = 3(x 2) expand the brackets y -2 = 3x 6 add 2 to both sides y = 3x - 4 iv) we need to find where the line x = 3 meets the perpendicular bisector of AB substitute x =3 into y = 3x - 4 y = (3 x 3) 4 = 9 4 = 5 centre is (3, 5) length from centre to A will be the radius of the circle length 2 = (-1 3) 2 + (3 5) 2 = (-4) 2 + (-2) 2 = 16 + 4 = 20 r = 20 equation of circle is (x 3) 2 + (y -5) 2 = ( 20) 2 = 20 (x 3) 2 + (y - 5) 2 = 20 www.chattertontuition.co.uk 0775 950 1629 Page 15
Question 12 i) the factor theorem states that if (x a) is a factor then f(a) = 0 we need to try some simple small numbers (such as 1, 2, 3, -1, -2, -3) to see which gives f(x) = 0 try x = 1 f(1) = 1 3 + (6 x 1 2 ) 1 30 = -24 try x = 2 f(2) = 2 3 + (6 x 2 2 ) 2 30 = 8 + 24 2 30 = 0 f(2) = 0 so that means that (x 2) is a factor now we know that (x 2) is a factor we can work out how many times this goes into the cubic equation we can do this by inspection, by algebra or by polynomial division algebraically the quadratic equation that we will get will be of the form Ax 2 + Bx + C we can easily see that A must be 1 as the leading term is x 3 (x 2)(x 2 +Bx + C) = x 3 + 6x 2 x - 30 Just looking at the x 2 terms (x 2)(x 2 +Bx + C) = x 3 + 6x 2 x 30 Bx 2 2x 2 = 6x 2 B 2 = 6 add 2 to both sides B = 8 www.chattertontuition.co.uk 0775 950 1629 Page 16
just looking at the x terms (x 2)(x 2 +Bx + C) = x 3 + 6x 2 x - 30-2Bx +Cx = -x -2B + C = -1 we know that b = 8-16 + C = -1 add 16 to both sides C = 15 check with the units (x 2)(x 2 +Bx + C) = x 3 + 6x 2 x - 30-2C = - 30 divide both sides by -2 C = 15 A = 1, B = 8, C = 15 (x 2)(x 2 + 8x + 15) now we can factorise the quadratic Two numbers multiply to give 15 and add to give 8. The two numbers are 3 and 5. (x 2)(x + 3)(x + 5) www.chattertontuition.co.uk 0775 950 1629 Page 17
By Polynomial division x 2 + 8x + 15 x - 2 x 3 + 6x 2 x - 30 x 3 2x 2 8x 2 - x 8x 2 16x 15x - 30 15x - 30 0 (x 2)(x 2 + 8x + 15) now we can factorise the quadratic Two numbers multiply to give 15 and add to give 8. The two numbers are 3 and 5. (x 2)(x + 3)(x + 5) www.chattertontuition.co.uk 0775 950 1629 Page 18
ii) now we have factorised we know the three points where the graph crosses the x axis they are -5, -3 and 2 if we put x as 0 then we can also see that the graph crosses the y axis at -30 www.chattertontuition.co.uk 0775 950 1629 Page 19
iii) this means that the graph is picked up and moved 1 unit in the positive x direction we can replace all the x values with x 1 (when we change the x values we do them in the opposite way to what we would expect hence we subtract 1 rather than adding 1) f(x - 1) = (x - 1) 3 + 6(x - 1) 2 (x - 1) 30 y = (x 1)(x 1)(x 1) + 6(x -1)(x 1) (x 1) 30 expand the brackets to get y = (x 1)(x 2 2x + 1) + 6(x 2 2x + 1) (x 1) 30 expand again y = x 3-3x 2 + 3x - 1 + 6x 2-12x + 6 x + 1-30 group terms y = x 3 + 3x 2-10x - 24 If you found these solutions helpful and would like to see some more then visit our website http://www.chattertontuition.co.uk/a-level-maths-papers It should be noted that Chatterton Tuition is responsible for these solutions. The solutions have not been produced nor approved by OCR. In addition these solutions may not necessarily constitute the only possible solutions. www.chattertontuition.co.uk 0775 950 1629 Page 20