Core1 Worked solutions January 2010 paper

Link to worked solutions for January 2010 C1 paper:


https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZDFmOGVhODgtMDk2NC00ZGQ0LTg5MDgtYTg3ZDRlYjcxOGQ5&hl=en


Further supporting notes for worked solutions:


1. differentiate - bring the power down and reduce the power

2. (a) just expand and simplify
(b) to rationalise the denominator multiply the top and bottom by the expression in the denominator with the opposite sign - use the answer to part (a) for the numerator

3. (a) find the gradient of l1 by rearranging the equation to the form y=mx+c and indentifying th m
(b) gradient of l1 is -3/5 so gradient of perpendicular l2 is 5/3 - then use this gradient and the given point with either straight line equation method to find the equation of l2

4. - first change root x to power of 1/2 and simplify the second term
- then to find y need to integrate dy/dx - increase the power and divide by that - don't forget the +C
- then use the given point to substitue values of x and y to find the C
- then finally retstate the equation for y with the value for C included

5. first rearrange equation 1 to make y the subject and substitute into equation 2 and solve to find solutions for x - substitute these solutions into the rearranged equation to find the solutions for y

6. (a) - expand and simplify the top
- break into separate terms
- simplify
- then differentiate - bring the power down and reduce the power
(b) - substitute x=2 into dy/dx to find the gradient of the tangent at that point
- substitue x=2 into the original equation for y to find the y coordinate of the point
- then use this gradient and point with either straight line equation method to find the equation of the tangent

7. from the given information identify u1,u2,u3 and then a & d
(a) Year 10 is 9 steps of the common difference on from Year 1
(b) just substitute values for a,n,d into the sum formula
(c) subsitute values for n& d and the unknown A and make equal to double the previous total - solve for A

8. (a) f(x)+2 outside the bracket so translation up 2 - don't forget the equation of the asymptote
(b) 4f(x) outside the bracket so stretch 4 vertically
(c) f(x+1) inside the bracket so horizontal and opposite direction to sign - so translate 1 left

9. (a) common factor of x first then the remaining quadratic is a difference of two squares
(b) identify roots from answer to (a) and basic cubic shape
(c) separately substitue x=-1 and x=3 into the equation to find the y-coordinates of A and B then find the gradient of AB by difference in y over difference in x then use this gradient and one of the points A or B in either straight line equation method to find the equation of the line
(d) length AB is found using pythagoras on the two pairs of coordinates for A and B

10. (a) completing the square method
(b) no real roots means discriminant is less than zero - identify a,b,c from original equation and substitute into discriminant (or alternatively identify discriminant from answer to (a)) - then solve the inequality
(c) substitue k=1 into the answer to (a) and identify the two translations from basic quadratic (left 2 and up 10) - y intercept found by substituting x=0