Core1 Worked solutions January 2007 paper

Link to worked solutions for Jan 2007 C1 paper:



https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZTkwMDg5NWEtNzc2MS00ODYxLWJlNDItYThkNWUyZDQwZDc1&hl=en



Notes supporting worked solutions:



1. Just differentiate - bring power down and reduce power



2. (a) the form of the answer required gives away that 3 is one of the factors to split 108 into

(b) expand and simplify



3. f(x) + 3 is translate 3 up - to find point crossing x axis make equation equal to zero



4. substitute equation 1 into 2 and solve the quadratic to find solutions for x - then put these solutions into equation 1 to find solutions for y



5. no real roots means discriminant is less than zero - identify a,b,c from original equation, substitute and solve the inequality

6. (a) expand and simplify ensuring the required form is reached
(b) use the answer to part (a) as the starting point then change the root x to power of 1/2 and finally integrate - increase the power and divide by that power - don't forget the +C

7. (a) first change the last term of f '(x) to power of -2 then integrate to find f(x) don't forget the +C then use the given point to substitute in values for x and y to find C then finally restate f(x) with the value for C
(b) substitute x=2 into original equation for f '(x) to find gradient of tangent at that point then use either straight line equation method to find the equation - ensuring it is given in the required from (y=mx+c)

8. (a) just differentiate bring power down and reduce the power
(b) just substitute x=4 into the original equation for y
(c) substitute x=4 into dy/dx found in (a) to find gradient of tangent at P
then if gradient of tangent is -3 then gradient of normal is 1/3 (perpendicular) - then using m=1/3 and point P in either straight line equation method to find equation of normal
(c) find x coordinate of Q by substituting y=0 into the equation then use pythagoras to find length PQ

9. (a) method 1 is using formula given in formula sheet, method 2 is basically old GCSE style method
(b) n,a&d are known so just substitute into formula for Sn
(c) a&d are known and use n=k in formula for Sn and then make less than 1750 and simplify to reach required expression
(d) negative value for k is not possible to must be the positive value and k must be an integer as it represents a number of rows so 33 is the most rows possible

10. (a) obtain roots as shown - for (i) a basic cubic shape with double root at zero and single root at 2 - for (ii) a quadratic but negative coefficient of x squared term so upside down and roots at 0 and 6
(b) equate the two equations and solve - ensure the solutions make sense from the graph sketches