Mathematics Edexcel Advanced Subsidiary GCE Core 1 (6663) January 2010

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Delphia Butler
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1 Link to past paper on Edexcel website: 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. Question 1 to differentiate you multiply by the power and then reduce the power by 1 = 4x3 + x-2/3 Question 2 a) (7 + 5)(3-5) = (7 x 3) ( 5 x 5) + (3 x 5) (7 x 5) = b) to get rid of the surds from the denominator we multiply the numerator and denominator by (3-5) ( ) ( ) ( ) ( ) x = ( ) ( )( ) numerator: 7+ 5 (3 5 ) = = (also this is what we did in part a) denominator: (3+ 5 )(3 5 ) = = 4 so we have ( ) = = Page 1

2 Question 3 a) we first need to rearrange the equation so that it is in the form y = mx + c then the gradient is the number in front of the x subtract 3x from both sides 5y 2 = -3x add 2 to both sides 5y = -3x + 2 divide both sides by 5 y = + so the gradient is b) the gradient of l 2 will be the negative reciprocal of the gradient of l 1 gradient = We now have the gradient and a point (3, 1) 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 1 = (x 3) expand the brackets y 1 = - 5 add 1 to both sides y = Page 2

3 Question 4 Before we can integrate we need to rewrite with single powers of x = 5x-1/2 + x 3/2 We want y = (5 / + / ) dx to integrate we add a power and divide by the new power y = / + / + c = / / 10x1/2 + / + c now we know that y = 35 when x = 4 so we can use this information to work out what c is 35 = 10(4 1/2 )+ ( ) 35 = (10 x 2) + 35 = c + c + c 35 = c subtract 20 from both sides 15 = + c subtract from both sides = c now put back into the equation for y y = 10x 1/2 + / Page 3

4 Question 5 we can easily rearrange the first equation to make y the subject, then we can substitute this into the second equation add 3x to both sides y + 2 = 3x subtract 2 from both sides y = 3x 2 substitute this into the second equation (3x 2) 2 - x 6x 2 = 0 (3x 2)(3x 2) x 6x 2 = 0 expand the brackets 9x x 6x x 6x 2 = 0 group terms 3x 2 13x + 4 = 0 to factorise we need to multiply 3 by 4 to get 12 then we need to find two numbers that multiply to give 12 but add to give -13 these two numbers are -1 and -12 rewrite the quadratic with -13x split into -1x and -12x 3x 2 1x 12x + 4 = 0 factorise in pairs x(3x 1) 4(3x 1) = 0 we have the same thing in both brackets as we should factorise again (3x 1)(x 4) = 0 so either 3x 1 = 0 add 1 to both sides 3x = 1 divide by 3 x = or x 4 = 0 add 4 to both sides x = Page 4

5 put both of these x values into one of the original equations to get the corresponding y values x = y (3 x ) + 2 = 0 y = 0 y + 1 = 0 subtract 1 from both sides y = -1 x = 4 y (3 x 4) + 2 = 0 y = 0 y 10 = 0 add 10 to both sides y = 10 so we have x =, y = -1 or x = 4, y = 10 we should check by putting back into the other equation (-1) (6 x ( )2 ) = = 0 and (6 x 4 2 ) = = Page 5

6 Question 6 a) we need to expand the numerator and then divide through by x (x + 3)(x 8) = x x 8x = x 2 5x 24 = x 5 24x -1 now we can differentiate to differentiate we multiply each term by the power of x and then reduce the power by 1 = x-2 = 1 + b) if we put x = 2 into we will have the gradient of the tangent = 1 + = = 7 we need to find the corresponding y coordinate when x = 2 substitute x = 2 into the equation for c y = ()( ) = = -15 We now have the gradient (7) and a point (2, -15) that the tangent goes through so we can work out the equation of the tangent using the formula y y 1 = m(x x 1 ) y = 7(x 2) expand the bracket y + 15 = 7x 14 subtract 15 from both sides y = 7x Page 6

7 Question 7 a) the arithmetic sequence is 150, 160, 170, 180 so the first term (a) is 150 the difference (d) is 10 a = 150, d = 10 we want to know the 10 th term (u 10 ) let n = 10 using the formula u n = a + (n 1)d u 10 = ((10 1)x 10) u 10 = = 240 b) we want to know s 20 let n = 20 using the formula s 20 = ½n(2a + (n 1)d) s 20 = ½ x 20((2 x 150) + ((20 1) x 10)) s 20 = 10( ) = 4900 c) Kevin s sequence was A, A + 30, A + 60, A + 90 a = A, d = 30, n = 20 s 20 = ½ x 20(2A + ((20 1) x 30)) s 20 = 10 x (2A + 570) now we know that Kevin s s 20 is twice Jill s s 20 so 10 x (2A + 570) = 2 x 4900 expanding the bracket 20A = 9800 subtract 5700 from both sides 20A = 4100 divide both sides by 20 A = Page 7

Question 8 a) this is a translation of 2 units in the positive y direction all the y coordinates are increased by 2 the asymptote will be at y = 3 the maximum point will be at (-2, 7) b) this is a

8 Question 8 a) this is a translation of 2 units in the positive y direction all the y coordinates are increased by 2 the asymptote will be at y = 3 the maximum point will be at (-2, 7) b) this is a stretch of scale factor 4 parallel to the y axis all the y coordinates will be 4 times what they were before the asymptote will be at y = 4 the maximum point will be at (-2, 20) Page 8

9 c) this is a translation of 1 unit in the negative x direction (when the x is affected then it is always in the opposite way to what you would think) all the x coordinates are decreased by 1 the asymptote will still be at y = 1 the maximum point will be at (-3, 5) Page 9

10 Question 9 a) a clue in the question is that we must factorise completely so it will be more than just the x that factorises first factorise out the x x(x 2 4) we have here the difference of the squares so we can factorise again x(x 2)(x + 2) b) c) we must first work out the full coordinates of A and B set x = -1 y = (-1) 3 (4 x -1) = = = 3 A is (-1, 3) set x = 3 y = 3 3 (4 x 3) = = 15 B is (3, 15) Page 10

11 we need to work out the gradient of ab gradient = gradient = = = 3 we now have the gradient (3) and a point (3, 15) 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 15 = 3(x 3) expand the bracket y 15 = 3x 9 add 15 to both sides y = 3x + 6 d) length 2 = (y 2 y 1 ) 2 + (x 2 x 1 ) 2 AB 2 = (15 3) 2 + (3 - -1) 2 = = = 160 square root both sides AB = 160 = 16 x 10 = 16 x 10 = 4 10 k = Page 11

12 Question 10 a) this is completing the square f(x) = (x + 2k) 2 (2k) 2 + (3 + 11k) f(x) = (x + 2k) 2 4k k p = 2k q = -4k k b) if f(x) = 0 then q must be positive we cannot square root a negative number so this would lead to no real roots (x + p) 2 + q = 0 subtract q from both sides (x + p) 2 = -q square root both sides (x + p) = which is impossible if q is positive (as q will be negative) So we need -4k k > 0 (so that we have no real roots) I prefer to work with a positive k 2 so add 4k 2 to both sides and subtract the other terms 4k 2 11k 3 < 0 we can work out where the lhs = 0 by factorising first find two numbers that multiply to make -12 (4 x -3) but add to make -11 these two numbers are -12 and +1 rewrite the quadratic splitting the -11k into -12k and +1k 4k 2 12k + 1k 3 factorise in pairs 4k(k 3) + 1(k 3) factorise again (k 3)(4k + 1) this equals 0 when k 3 = 0 or 4k + 1 = 0 so either k = 3 or 4k + 1 = 0 k = 3 or 4k = -1 k = 3 or k = we now need to work out whether we want to be between these two values or either side of them the positive quadratic curve is shaped like a so it is below 0 between the two values < k < Page 12

13 c) using the completed square version and setting k = 1 the equation of the curve will be f(x) = (x + 2) f(x) = (x + 2) this will cross the y axis when x = 0 y = = 14 (0, 14) We know there are no real roots as < k < 3 and k = 1 So this means that the curve does not cross the x axis If you found these solutions helpful and would like to see some more then visit our website It should be noted that Chatterton Tuition is responsible for these solutions. The solutions have not been produced nor approved by Edexcel. In addition these solutions may not necessarily constitute the only possible solutions Page 13

Mathematics (MEI) Advanced Subsidiary GCE Core 1 (4751) May 2010

Mathematics (MEI) Advanced Subsidiary GCE Core 1 (4751) May 2010 Link to past paper on OCR website: http://www.mei.org.uk/files/papers/c110ju_ergh.pdf These solutions are for your personal use only. DO NOT photocopy or pass on to third parties. If you are a school or