Core1 Worked solutions June 2008 paper

Link to worked solutions for June 2008 C1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOWQ1OTUwM2MtMTVlOS00MGI4LWJhNDMtYjZhZmVhMTczNGIz&hl=en

Further supporting notes for the worked solutions:

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

2. first factorise the common factor of x then the remaining quadratic is a difference of two squares

3. (a) f(x) + 3 is translate 3 up (the +3 is outside the bracket)
(b) f(2x) is squash s.f.2 horizontally (the multiply by 2 is inside the bracket)

4. (a) differentiate - bring the power down and reduce the power
(b) make the answer to part (a) equal to 15 then solve

5. (a) substitute x1 into the equation to find x2
(b) substitute x2 into the equation to find x3
(c) make the answer to (b) equal to 7 and solve

6. (a) separately substitute values of x=0 and y=0 into each equation to find points crossing axes
(b) equate the two equations and solve to find x coordinates of the intersection points then substitute each x coordinate into either equation to find the y coordinates

7. first identify a & d from the given information
(a) three steps from the first term to the fourth term
(b) method 1 is using the formula from the formula book and method 2 is basically an old gcse type method
(c) substitue the values for a & d into the formula for Sn and simplify
(d) most efficient method is to use the answer to (b), make it equal to 43 and solve
(e) use the answer to part (c) rather than starting from scratch

8. no real roots means discriminant is less than zero
(a) identify a,b,c and substitute into discriminant
(b) solve the quadratic inequality by factorising, identifying roots, sketching and identifying part of the graph which in this case is between the roots so solutions is a double inequality

9. (a) differentiate - bring power down and reduce power
(b) find an expression for the gradient of the tangent at A by substituting x=-1/2 into the equation for dy/dx - the gradient of the given straight line equation is found by rearranging into the form y=mx+c - find k by equating these
(c) substitute x=-1/2 and the value of k into the original equation

10. (a) find the length QR by using pythagoras on the two pairs of coordinates
(b) find the gradient of l1 from the two given points - found to be -1/2 so gradient of l2 is 2 as it is perpendicular to l1 - use this gradient and coordinates of Q in either straight line equation method to find an equation for l2
(c) P is the y intercept of l2 so coordinates identified from y=mx+c answer to part (b)
(d) diagram shows the triangle - PQ and QR are perpendicular as l1 and l2 are perpendicular so lengths PQ and QR needed to find area of triangle - QR was found in (a) & PQ now found in same way then half x base x height for area of triangle

11.(a) expand the numerator and then split into separate terms and simplify
(b) integrate from the answer to part (a) - don't forget the +C - then use the coordinates from the given point to find C - then restate the equation for f(x) with the value found for C