Core1 Worked solutions January 2005 paper

Link to worked solutions for January 2005 C1 paper and then further explanatory notes below.



https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpN2NhODI0MjktYTlkZS00ZjNhLTkwN2MtYzFiOGVjNzdmYjdk&hl=en

1. (a) power of 1/2 is square root

(b) the minus in the power gives one over; the over two in power is square root; and the three is cubed

2. (i) basic rules of differentiation of differentiation - bring the power down and reduce the power - and for part (b) repeat to obtain second differential

(ii) first change the root to power 1/2 and the one over x squared to power of -2 then basic rules of integration - increase the power and divide by it - and don't forget the plus C - remember dividing by 3/2 is same as multiplying by 2/3

3. remember equal roots means discriminant is equal to zero then substitue values and solve for k

4. rearrange first equation (could make x the subject rather than y as shown) then substitute into second equation - solving quadratic obtains two solutions for x then substitute back into the equation which has y as its subject to find the solutions for y

5. (a) just substitute r=1,2,3 in turn into the expression

(b) common difference identified from part (a) results

(c) just substituting a=-3 and d=2 as identified in (a) and (b) into the formula for the sum to n terms (which is given on the formula sheet)

6. (a) reflection in x axis - because the minus is outside the transformation is in the vertical direction

(b) squash s.f.2 in x direction - the 2 is inside the bracket so in x direction and squash rather than stretch

7. (a) first rearrange to split the fraction and simplify then differentiate then substitute to find the value of the gradient of the tangent at the given point

(b) substitute given value of x into original equation for y to find y coordinate of P. Then use either straight line equation method with gradient of 3 and point (1,8).

(c) just substituting y=0 into equation found in (b)

8. (a) identify coordinates so that D is midpoint of AC

(b) calculate gradient of AC using difference in y over difference in x (being careful with negative signs) - then gradient of line l is opposite sign and fraction upside down as they are perpendicular - then use eitherstraight line equation method and leave in correct form

(c) substitute y=7 (because AB is the line y=7 and E is on AB) into equation found in (b) to find x coordinate

9. (a) substitute x = 1 into given dy/dx to find gradient of tangent at P - gradient of normal to P is minus one over that - then use either straight line method to find equation of normal at P

(b) integrate given dy/dx to find y - don't forget + C and then use point (1,3) to find C and then restate equation for y

(c) just showing that there are no solutions for the given equation for dy/dx to be negative (would need to be -2 to be parallel to y=1-2x)

10. (a) complete the square

(b) from y=x^2 just a translation right 3 and up 9 and substitute x=0 to get point on y-axis

(c) solve by completing the square (could also use quadratic formula)