Mathematics AQA Advanced Subsidiary GCE Core 1 (MPC1) January 2010

Size: px

Start display at page:

Download "Mathematics AQA Advanced Subsidiary GCE Core 1 (MPC1) January 2010"

Tiffany Cain
5 years ago
Views:

1 Link to past paper on AQA 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 a) We set (x + 3) = 0 to get x = -3 and then use the remainder theorem p(-3) = (-3) 3 - (13 x -3) - 12 p(-3) = = = 0 as p(-3) = 0 this means that (x + 3) is a factor of p(x) b) now we know that (x + 3) 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 + 3)(x 2 +Bx + C) = x 3-13x - 12 Just looking at the x 2 terms (x + 3)(x 2 +Bx + C) = x 3-13x - 12 Bx 2 + 3x 2 = 0x 2 B + 3 = 0 subtract 3 from both sides B = -3 just looking at the x terms (x + 3)(x 2 +Bx + C) = x 3-13x Bx +Cx = -13x 3B + C = -13 we know that B = C = -13 add 9 to both sides C = Page 1

2 check with the units (x + 3)(x 2 +Bx + C) = x 3-13x C = - 12 divide both sides by 3 C = -4 A = 1, B = -3, C = -4 (x + 3)(x 2-3x - 4) now we can factorise the quadratic Two numbers multiply to give -4 and add to give -3. The two numbers are -4 and +1. (x + 3)(x - 4)(x + 1) By Polynomial division We must make sure that each power of x in the polynomial is included (so we need 0x 2 ) x 2-3x - 4 x + 3 x 3 + 0x 2 13x - 12 x 3 + 3x 2-3x 2-13x -3x 2 9x -4x x (x + 3)(x 2-3x - 4) now we can factorise the quadratic Two numbers multiply to give -4 and add to give -3. The two numbers are -4 and +1. (x + 3)(x - 4)(x + 1) Page 2

3 Question 2 a) the gradient of a line is calculated by dividing the difference in the y values by the difference in the x values gradient = i) gradient AB = = = 2 ii) we will know that angle ABC is a right angle if lines AB and BC are perpendicular to each other gradient BC = = = if the product of the two gradients is -1 then the lines are perpendicular this is the same as saying the gradients are the negative reciprocal of each other 2 x = -1 so angle ABC is a right angle b) i) to find the midpoint of two points you add the two x values and divide by 2 and add the two y values and divide by 2 M = (, ) = (0, 6) ii) length AB 2 = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 AB 2 = (3 1) 2 + (7 3) 2 = = = 20, length AB = 20 BC 2 = (3 - -1) 2 + (7 9) 2 = (-2) 2 = = 20, length BC = 20 length AB = length BC iii) the line of symmetry through triangle ABC will be the line that goes through B and M we have the two coordinates for B and M (3,7) and (0, 6) so we can calculate the gradient gradient BM = = we have the gradient and a point that the line goes through so we can use y y 1 = m(x x 1 ) to get the equation y 6 = (x 0) add 6 to both sides y = Page 3

4 Question 3 to differentiate we multiply by the power and then reduce the power by 1 a) i) = t3 4t + 4 = t3 4t + 4 ii) = t2 4 b) stationary values are minimum or maximum, they happen when = 0 substituting t = 2 into we should get 0 = 2)3 (4 x 2) + 4 = = 0 To determine whether the point is a maximum or a minimum we use If this comes out positive then the point is a minimum If this comes out negative then the point is a maximum When t = 2 = 2)2 4 = 6 4 = 2 which is positive The stationary point is a minimum c) i) we want when t = 1 = (1)3 (4 x 1) + 4 = = ii) is positive which means that y is increasing is a measure of the gradient of the slope and if this is positive then y (the depth) is increasing) ( Page 4

5 Question 4 i) 50 = 25 x 2 = 25 x 2 = 5 2 and 18 = 9 x 2 = 9 x 2 = 3 2 and 8 = 4 x 2 = 4 x 2 = 2 2 so we have = = = 4 ii) we need to rationalise the denominator, so multiply the numerator and the denominator by (2 7-5) x = when we expand the numerator we get = (2 x 2 x 7 x 7) = = expanding the denominator we get = (2 x 2 x 7 x 7) = = 3 so we have = = Page 5

6 Question 5 a) we first need to expand the brackets (x 5)(x 3) + 2 = x 2 5x 3x = x 2-8x + 17 now we can complete the square (x 4) = (x 4) p = 4, q = 1 b) i) the minimum point will be when x = 4 and when y = 1 (4, 1) this is because the smallest (x 4) 2 can ever be is 0 (since it is a square of something) and (x 4) 2 is 0 when x = 4 the graph crosses the y axis when x = 0 this happens when y = 0 2 (8 x 0) + 17 = 17 crosses y axis at (0, 17) ii) the tangent to the curve at the minimum point is just the straight line y = 1 c) we know from part a) that y = (x 4) the changes from y = x 2 to (x 4) are two translations (since we are adding or subtracting values) one of the changes is inside the brackets so affects the x values. When the x values are affected the change is in the opposite direction to what you expect. This is a translation of 4 units in the positive x direction. The other change is outside of the brackets so affects the y values. This is a translation of 1 unit in the positive y direction Overall we have vector translation of Page 6

7 Question 6 a) i) to find the gradient we need to differentiate to differentiate we multiply by the power and then reduce the power by 1 = 24x 19 6x2 To find the gradient at the point (2, -6) substitute x = 2 into Gradient = (24 x 2) 19 (6 x 2 2 ) = = 5 ii) the normal is perpendicular to the tangent the gradient of the tangent at A is 5 the gradient of the normal at A will be the negative reciprocal of 5 which is - We now have the gradient and a point (2, -6) 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 --6 = - (x 2) multiply by 5 5(y + 6) = -(x 2) expand the brackets 5y + 30 = -x + 2 add x to both sides x + 5y + 30 = 2 subtract 2 from both sides x + 5y + 28 = 0 b) i) to integrate you increase the power by 1 and then divide by the new power = (4(2 3 ) ) (0) = = -14 ii) what we have done in part i) is work out the whole area enclosed by the curve and the x axis between the lines x = 0 and x = 2 the area came out as negative because it is below the x axis we need to work out the area of the triangle and subtract this from 14 area of triangle = ½ x base x height = ½ x 2 x 6 = 6 shaded area = 14 6 = 8 square units Page 7

8 Question 7 a) we need to complete the square then when the equation is in the form (x x 1 ) 2 + (y y 1 ) 2 = r 2 we will be able to work out the centre (x 1, y 1 ) and radius r x 2 4x + y y + 15 = 0 complete the square on both x and y (x 2) (y + 6) = 0 group the units (x 2) 2 + (y + 6) 2-25 = 0 add 25 to both sides (x 2) 2 + (y + 6) 2 = 25 (= 5 2 ) i) centre will be (2, -6) ii) radius will be 5 b) the centre is below the x axis (as the y coordinate of centre is -6) the radius is 5 the highest the circle will reach will be = -1 and as this is negative it is still below the x axis c)i) length PC 2 = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 PC 2 = (5 2) 2 + (k -6) 2 = (k + 6) 2 = 9 + (k + 6)(k + 6) expand the brackets PC 2 = 9 + k 2 + 6k + 6k + 36 group terms PC 2 = k k + 45 ii) we know that P is outside of the circle so PC must be more than the radius which is 5 PC 5 so also PC 2 25 k k subtract 25 from both sides k k Page 8

iii) we can factorise if we can find two numbers that multiply to make 20 and add to make 12 these two numbers are +2 and +10 (k + 2)(k + 10) 0 If we set this equal to 0 then the critical values will

9 iii) we can factorise if we can find two numbers that multiply to make 20 and add to make 12 these two numbers are +2 and +10 (k + 2)(k + 10) 0 If we set this equal to 0 then the critical values will be k = -2 and k = -10 Now we need to decide if we want k to be between these two values or either side of them Try k = -3 (which is in between), when k = -3: we get = -7 which is less than 0 (not what we want) so we don t want k to be between the two values Alternatively if we know the way that a quadratic graph (with a leading positive k 2 term lies (like a big ) then we can also see that this curve is above 0 when k is either side of the two critical values k -10, k -2 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 AQA. In addition these solutions may not necessarily constitute the only possible solutions Page 9

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