Core1 Worked solutions May 2006 paper

Link to worked solutions for May 2006 C1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYTMzZmRkZjgtZjVmZC00ODQ0LWFhN2QtZmMwMTlkMDJhY2U3&hl=en

Notes supporting worked solutions:

1. Integrate each term - increase the power and divide by that - don't forget the +C

2. factorise, identify roots, sketch graph, identify part of graph needed (in this case "outside" the roots then state inequality solution (in this case two separate inequalities

3. (a) basic quadratic translated 3 left - substitute x=0 to get crossing point on y-axis
(b) from (a) to (b) is translated "k" units up - so crosses y-axis "k" units higher than in (a)

4. (a) substitute a1 into the equation to find a2 and subtitute a2 into the equation to find a3
(b) question means find the sum of the first five terms - so first find a4 and a5 in the same way as a2 and a3 were found - then add the terms

5. (a) first change root x to power of 1/2 then differentiate - bring power down and reduce power
(b) expand and simplify the numerator then put into separate fraction terms and simplify each term - then differentiate

6. (a) just expand and simplify
(b) to rationalise the denominator multiply top and bottom by the expression in the denominator but with the opposite sign

7. Using the information that the 11th term is 9, an equation in terms of a and d can be set up. Using the information that the sum of the first 11 terms is 77, another equation in terms of a and d can be set up. These are then solved simultaneously

8. (a) equal roots means discriminant = 0, identify a,b,c and substitute then solve for p (the question states p>0 so p=4 is only solution)
(b) substitute p=4 from (a) into the original equation and solve - expecting equal roots solution as stated in the original problem

9. (a) expand and simplify then factorise common factor of x
(b) factorise the remaining quadratic
(c) basic cubic shape with roots at 0,3,5

10. (a) integrate - increase power and divide by that - don't forget +C - then use point to substitute values of x and y to find C - then restate equation with value for C
(b) just substitute x=2 into equation found in (a)
(c) substitute x=-2 into original equation of f '(x) to find gradient of tangent at that point then use either straight line equation method to find equation - finally stating in correct form with integer coefficients

11. (a) Find gradient from two points. The use either straight line equation method with either point to find the equation.
(b) perpendicular so gradient goes from 1/2 to -2 and point known so use either straight line equation method to find equation - then equate l1 and l2 for x coordinate of intersection (substitute into either equation for y coordinate
(c) pythagoras using two points to find distance
(d) pythagoras again to find length PQ and then sketch identifies triangle (l1 and l2 known to be perpendicular)