Core1 Worked solutions May 2010 paper

Link to worked solutions for May 2010 C1 paper:


https://docs.google.com/leaf?id=0B2J1T-insOmpZWE2ZDU5YzItZjU2ZS00NWFlLTg1ODMtMGU4N2JhYTM3MDNi&hl=en


Further supporting notes for worked solutions:


1. both root 75 and root 27 can be simplified to multiples of root 3

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

3. (a) solve the linear inequality - the steps used here avoid the need to divide by a negative
(b) solve the quadratic inequality by identifying the roots, sketching the graph, identifying which part of the graph is required which is between the roots in this case so state the final solution as a double inequality
(c) show the solutions to (a) and (b) on a number line and state an inequality for the overlap

4. (a) completing the square method
(b) using the answer to (a) the curve is a basic quadratic translated 3 left and 2 up - the point of intersection on the y-axis is found by substituting x=0 into the equation
(c) from the original equation identify a,b,c substitute into the discriminant (expecting the answer to this to be negative as there are no real roots

5. (a) substitute a1 into the equation to find a2 and then substitute a2 into the equation to find a3
(b) substitute a3 into the equation to find a4 and then substitute a4 into the equation to find a5

6. (a) f(x+3) is inside the bracket so horizontal and opposite direction to sign so translate 3 left
(b) 2f(x) is outside the bracket so stretch vertical scale factor 2
(c) for the minimum to be a (3,0) then the graph would have been translated 5 up so outside the bracket giving f(x)+5 so a=5

7. first change to root x to power of 1/2, split the fraction into separate terms and simplify then differentiate - bring power down and reduce the power

8. (a) find the gradient of AB by using difference in y / difference in x then use this gradient and either of the points in either straight line equation method - leaving in correct form with all coefficient as integers
(b) length of AB is found by using pythagoras on the two pairs of coordinates
(c) use pythagoras with the length AC and the unknown and make equal to the answer to (b) and solve to find the unknown t
(d) sketch of the coordinates and triangle demonstrates how to find the area of the triangle

9. (a) just substituting values into the formula
(b) in a similar way substitue values into the sum formula then rearrange the answer to (a) so that it can be substituted into the sum formula and rearranged to the required equation
(c) solve the answer to (b) to find a and then substitute this value into the answer to (a) to find d

10. (a) (i) quadratic shape but upside down because of the -x in the bracket - roots at 0 and 4
(ii) cubic shape but upside down because of the -x in the bracket - a double roots at 0 and another root at 7
(b) x-coordinates of the points of intersection found by equating the two equations and rearranging
(c) one solution to the equation given in (b) is x=0 so interested in solutions to the quadratic inside the bracket - use either quadratic formula or completing the square to solve - the two solutions for the x-coordinates are both positive but either by substituting to find y or with reference to the graph, the lower value of x gives a positive value of y

11.(a) first change the second term to power of -1/2 then integrate to find f(x) - don't forget the +C - then substitute values from the given point to find the C and restate the equation for f(x) with the value for C included
(b) substitute x=4 into the original equation for dy/dx to find the gradient of the tangent then use this gradient and the given point in either straight line equation method to find the equation of the tangent - leaving in the correct form with coefficients as integers