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

Transcription

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 75= 25 3= 25 x 3=5 3 27= 9 x 3= 9 x 3= = = Page 1

2 Question 2 to integrate you increase the power by 1 and then divide by the new power remember to include the constant c if there are no limits = ( ) = + + = To divide by a fraction you flip the fraction over and multiply by it 6 =6 x = = Page 2

3 Question 3 a) expand the brackets 3 6<8 2 Add 2 to both sides of the inequality 5 6<8 Add 6 to both sides 5 <14 Divide both sides by 5 < b) we don t need to expand as the inequality is already factorised and the RHS is 0 (2 7)( +1)<0 The critical values (where the inequality equals 0) are when = = 1 We need to work out whether we want the numbers between these two values or either side of them.set out the two numbers on a number line The curve is a positive quadratic so has the usual shape. We can see that if the curve is <0 then we want the numbers between the two critical values 1< < c) this is easiest seen on a number line put the two inequalities on a number line and then see where they overlap 1< < Page 3

4 Question 4 a) this is completing the square ( +3) 3 +11=( +3) 9+11 ( +3) +2 =3, =2 b) this is the usual positive quadratic shape we can use the completed square form to tell us the minimum point at ( 3,2) c) discriminant is 4 =1 =6 =11 4 =(6) (4 x 1 x 11)=36 44= 8 This is negative as expected (we can see this from the fact that the curve does not cross or touch the axis Page 4

5 Question 5 a) =2 substitute this value into the equation = +3= 2 +3= 7 = +3= 7+3= 10 b) = +3= 10+3= 13 = +3= 13+3= 16=4 as required Page 5

6 Question 6 a) this is a translation of 3 units in the negative direction (all the coordinates are now 3 less than what they were) Vector translation ( ) (2 ) ( 2,3) ( 5,3) (0,0) ( 3,0) (3, 5) (0, 5) Page 6

7 b) this is a stretch of scale factor 2 parallel to the y axis (all the coordinates are now double what they were) ( ) (2 ) ( 2,3) ( 2,6) (0,0) (0,0) (3, 5) (3, 10) c) = ( )+ is a translation of a units in the positive y direction if the minimum is now at (3,0) then it must have moved up 5 units = Page 7

8 Question 7 a) to differentiate you multiply by the power and then reduce the power by 1 but first we must rewrite y so that all the terms are on the top and are expressed as powers not as square roots etc = =(3)8 4 +(1)3 +( 1)2 = Page 8

9 Question 8 a)the gradient of a line is given by Gradient of AB = = Equation of a line is given by = ( ) where (, ) is a point that the line goes through Could use point A or point B but have used point B Equation of line AB is 0= ( 2) Multiply both sides by 5 5 =4 8 Subtract 5y from both sides 4 5 8=0 b) length of a line is given by ( ) +( ) Length AB is (7 2) +(4 0) = 5 +4 = Page 9

10 c) length AC equals length AB so length AC = 41 (7 2) +(4 ) = 41 Square both sides 5 +(4 ) =41 25+(4 ) =41 Subtract 25 from both sides (4 ) =16 Square root both sides 4 = ±4 Subtract 4 from both sides = +4 4=0 = 4 4= 8 Multiply both sides by -1 =0 =8 We are told that >0 so t must be 8 t = 8 d) this is helped by visualising the triangle Area of triangle is given by ½ base x height Let height be 8, base be 7 2=5 Area = x 5 x 8=20 units squared Page 10

11 Question 9 a) this is an arithmetic sequence first term is a difference is d we are given that =40.75 from formulae book = +( 1) 40.75= +29 b) we are given that =1005 =40.75 from formulae book = ( + ) 1005= (30)( )=15( ) 15( )=1005 as required Page 11

12 c) 15( )=1005 divide both sides by 5 (easier than dividing by 15) 3( )=201 Divide both sides by =67 Subtract from both sides =. Now substitute this value into equation from part a) 40.75= = Subtract from both sides 14.50=29 Divide both sides by 29 = Page 12

13 Question 10 a) The red curve is the cubic curve. It is the typical shape for a negative cubic curve (where the leading term is a negative. It crosses the axis when =0 and =7 there is a turning point when =0 because this is the root that is repeated The blue curve is the quadratic curve. It is the typical shape for a negative quadratic (where the leading term is a negative ). It crosses the axis when =0 and = Page 13

14 b) both equations equal y so we can put the two equations equal to each other (4 )= (7 ) Expand both sides 4 =7 Add to both sides +4 =7 Subtract 7 from both sides +4 8 = =0 Factorise ( 8 +4)=0 as required Page 14

15 c) we need to solve this equation in order to find A one solution is =0 but as this isn t positive it isn t what we want 8 +4=0 This won t factorise. We can complete the square or use the quadratic formula. I am going to complete the square. ( 4) 4 +4=0 ( 4) 16+4=0 ( 4) 12=0 Add 12 to both sides ( 4) =12 Square root both sides 4= ± 12 Add 4 to both sides =4± 12 We want this to be positive, but both values for will be positive From our sketch in part a) we can see that the curves meet 3 times (once at the origin) and then in the first quadrant and then again in the fourth quadrant. From this we can see that the value we want is the smaller of these two. Alternatively if you didn t spot this you could substitute both values for into one of the curves to see which value gives you the positive value. =4 12=4 2 3 Substitute this into = (4 ) =(4 2 3)( ) =(4 2 3)(2 3) = = A is (4 2 3, ) Page 15

16 Question 11 ( ) just means the same as we have so we can integrate to get back to = ( ) the point that they have given us means we can work out what the constant c is first we need to rewrite with all the powers on the top =3 5 2 = (3 5 2 ) To integrate you add one to the power and then divide by the new power = (3 5 2 ) = = ( )= ( ) + + = To divide by a fraction we flip the fraction over and multiply by it 5 =5 x = =10 Now we know that =5 h =4 Substitute these values in to get c 5= (4) 10(4) 2(4)+ 5= = 4+ Add 4 to both sides =9 = ( )= Page 16

17 b) Equation of a line is given by = ( ) where (, ) is a point that the line goes through and m is the gradient We must first calculate the gradient at the point P We already have so just substitute =4 into this =3(4) 2=12 2= Equation of line AB is 5= ( 4) Multiply both sides by =15 60 Subtract 2y from both sides = Page 17

18 If you found these solutions helpful and would like to see some more then visit our website Chatterton Tuition provide tuition in all subjects in the North Yorkshire area. However we also provide online tuition via Skype. Please call, or visit our website for more information. 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 18