Core1 Worked solutions May 2005 paper

The link is to worked solutions of the May 2005 C1 paper and there are some additional explanations below.

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMDI5MzIzNjAtZDdkZS00YzE1LTgwNmMtZWU3NmVlMTA4MzJi&hl=en

1. (a) power of 1/3 is cube root
(b) negative power gives one over - the numerator of 2 in the power is squared and the denominator of 3 in the power is cube root

2. (a) first change over x squared to power of -2 then differentiate (bring power down and reduce power by one)
(b) integrate - increase power and divide by that and don't forget the +C

3. (a) completing the square
(b) method 1 is to follow on from part (a) using completing the square - method 2 is quadratic formula

4. (a) y=3f(x) is stretch s.f. 3 vertical
(b) y=f(x+2) is translation 2 left

5. make x the subject of first equation and substitute into second (could have made y the subject but this is more efficient)
then solve the quadratic to obtain two solutions for y
substitute the values of y into the equation with x as the subject

6.(a) solve linear inequality (the method used avoids any divison by a negative which would require a change of sign)
(b) solve quadratic inequality
- factorise
- identify roots
- sketch
- identify outside of roots
(c) combine solutions to (a) and (b) - number line clearly identifies the overlap (between 1/4 and 1/2 and greater than 3)

7. (a) expand the top, simplify, split into separate fractions, simplify
(b) integrate (increase power and divide by that) don't forget +C then use given value to find C and restate equation

8. (a) given gradient and point - use either straight line equation method
(b) the given info for l2 gives the equation as y = -2x
lines intersect - equate and solve for x and substitue into either equation to find y
(c) sketch of triangle identifies that one side is on y-axis so half x base x height is simple from coordinates

9. (a) show list of terms then backwards to show that 2Sn can be simplfies to n lots of (2a + (n-1)d)
(b) formally or informally use a=149 and d=-2
(c) a=149, d = -2 and Sn=5000 so only unknown is n
(d) just factorise and solve
(e) a=149 and d=-2 so after 75 terms the value of the term is negative which does not make sense for the given situation - loan repayments

10. (a) just subst. x=3 into the equation for y
(b) differentiate and substitute x=3 to find gradient of tangent to curve at P then use either straight line equation method given gradient of -7 and point (3,0)
(c) parallel lines have same gradient - make dy/dx equal to -7 and solve - the solution of x=3 gives the point P already known so the other solution if for point Q - substitute into original equation for y