Core1 Worked solutions January 2009 paper

Link to the worked solutions for the Jan 2009 C1 paper:



https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOGZkNWU0YzItZDk4My00ODMxLWEyYjEtMzNlZTU5OTBiZTQ1&hl=en


Supporting notes for the worked solutions:



1. (a) power of 1/3 is cube root
(b) negative power is one over, and 2/3 is cube root and squared

2. integrate - increase power and divide by that - don't forget +C

3. just expand and simplify

4. integate f '(x) to find f(x) - increase the power and divide by that - don't forget the +C - then substitute the values from the given point to find C - and finally restate the equation for f(x) with the value for C

5. (a) f(x+3) - inside the bracket means x direction and opposite way - so translate 3 left
(b) f(-x) - inside the bracket means x direction so horizontal change so reflection in y-axis

6. (a) change root x to power of 1/2 then split into separate terms and simplify
(b) differentiate the answer to (a) - bring the power down and decrease the power

7. 2 different real solutions means discriminant is greater than zero
(a) identify a,b,c then substitute into the discriminant and simplify
(b) solve the quadratic inequality - factorise, identify roots, sketch graph and identify required part of graph which in this case is outside of the roots so the solution is two separate inequalities

8. (a) to find the y coordinate of point P just substitute x=1 into the equation
(b) (i) is a cubic with negative coefficient of the cubed term (because of the -x in the second bracket) with a double root at x=-1 and another root at x=2
(ii) is a basic reciprocal shape
(c) the two equations equated finds points of intersection - so the two points of intersection on the diagram means there are two solutions

9. (a) one equation in terms and a & d can be written from each given term
(b) solve the simultaneous equations from (a)
(c) n is the only unknown - can substitute Sn, a & d into the formula - and then rearrange to the required equation
(d) solve the equation by factorising - n must be >0

10. (a) gradient and point given for l1 so just use either straight line equation method to find the equation for l1
(b) subsitute x=-2 into the answer from (a) and show y=7
(c) length AB found by using pythagoras in the two given pairs of coordinates
(d) subsitute x=p into the equation to find the y-coordinate then find an expression for the length by using pythagoras - make it equal to 5 and rearrange

11. (a)
-find y coordinate of point P by substituting x=2 into the equation
-differentiate and substitute x=2 to find gradient of tangent at P
-use gradient and point in either straight line equation method to find equation of tangent
(b) if gradient of tangent is -2 then gradient of normal is 1/2 as they are perpendicular - then use point and gradient in either straight line equation method to find equation of normal
(c) substitute y=0 into the two equations found in (a) and (b) to find the x coordinates of points A and B - diagram shows the triangle and how the area is calculated