Core1 Worked solutions June 2009 paper

Link to worked solutions for June 2009 C1 paper:


https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOTgyYWZkOGMtODcyNi00MWE3LTkxNDQtMTgwNTBiZTIwNmMy&hl=en


Further supporting notes for worked solutions:


1. (a) just the term inside the bracket squared
(b) expand and simplify


2. express the 32 and the root 2 as powers of 2 and then multiply together (so add the indices together)


3. first change the second term to power of -2
(a) differentiate - bring the power down and reduce the power
(b) integrate - increase the power and divide by that - don't forget the +C


4. (a) solve the linear inequality - the steps used avoid any dividing by a negative number
(b) solve the quadratic inequality by factorising, identifying roots, sketching the graph, identifying which part of the graph is required - in this case between the roots so finally state the solution as a double inequality
(c) show the solutions to (a) and (b) on a number line and give the overlap as an inequality


5. (a) 1960 is the 10th year and 1990 the 40th year so the difference between them is 30 steps of the common difference - the answer is negative because the values are decreasing
(b) to find the first term need to go back 9 steps from 10th term
(c) last term is known (from year 1990) so either sum formula can be used


6. equal roots means discriminant is equal to zero - identify a,b,c then substitute into the discriminant an solve for p (the questions states that p is non-zero)


7. (a) substitute a1 into the equation to find a2
(b) substitute a2 into the equation to find a3
(c) the question means the sum of the first four terms = 43 so substitute a3 into the equation to find a4 then add the terms, equate to 43 and then solve for k


8. (a) find the gradient of AB using the given points as this is -5/2 the gradient of line l is 2/5 as they are perpendicular - then use this gradient and the coordinates of A with either straight line equation method to find the equation of line l - giving in correct form with all coefficients as integers
(b) method shown is to rearrange the equation into the form y=mx+c to find the y-intercept - could also have substituted x=0 into the equation to find y
(c) OC is the base of the triangle and the x coordinate of B is the height of the triangle


9. (a) expand and simplify the top, split into separate terms, make root x power of 1/2 and simplify each term
(b) differentiate the answer to (a) - bring the power down and decrease the power
(c) substitute 9 into the answer to (b)


10. (a) common factor of x then factorise the remaining quadratic
(b) basic cubic shape and from (a) double root at 3 and another root at 0
(c) comparing the original equation to this one all the x's have become x-2 so it is a translation 2 right


11. (a) substitute x=2 into the equation so get y=7 to show P is on the curve
(b) differentiate and subsitute x=2 to find the gradient of the tangent at P and then use either straight line equation method to find the equation of the tangent
(c) gradient of tangent at P is 3 to gradient of tangent at Q is -1/3 as they are perpendicular - make the equation for dy/dx equal to -1/3 and solve to find the x coordinate of P - from the quadratic use either quadratic formula or completing the square method