OCR Core 1 Module RevisionSheet - Mathshelper

1 OCR core 1 Module Revision SheetThe C1 exam is 1 hour 30 minutes long. You you go into the exam make sure you are fully aware of thecontents of the formula bookletyou receive. Also be sure not to panic; it is not uncommon to get stuck on a question (I vebeen there!). Just continue with what you can do and return atthe end to the question(s)you have found hard. If you have time check all your work, especially the first question youattempted.. always an area prone to Changing the subject of an equation. For example iny= x+ 6,yis the subject of theequation. To change the subject toxwe merely need to re arrange tox=y2 6.

2 Aharder example is make to makegthe subject ofT= 2 lg=T lg=(T2 )2 g=l(2 T)2=4 2lT2. To solve simultaneous equations, isolatexoryfrom one of the equations and substituteinto the other. For example, solvex+ 2y= 1,x2 2y2= the first we findx= 1 2yand putting into the second we find (1 2y)2 2y2= 31,which simplifies toy2 2y 15 = 0. This givesy= 3 ory= 5. To calculate thexvalues we put theysolutions into either original equation. The solutions are( 9,5) and(7, 3). [Give your solutions as coordinates to show whichxandyvalues go together.] If you ever need to find where two lines or curves cross, then merely view it as a pair ofsimultaneous equations to be solved.

3 You must also know how to handle algebraic fractions and how to write two algebraicfractional expressions as one fraction. The general rules areab cd=ad bcbdandab cd= to writex xx+1as a single fraction we do the followingx xx+ 1=x1 xx+ 1=x(x+ 1) xx+ 1=x2x+ 1. Anyline or curve crosses thex-axis wheny= 0. Similarly, any line or curve crossesthey-axis whenx= 0. So to find wherey=x2+x 12 crosses thex-axis we solve0 =x2+x 12 and find (3,0) and ( 4,0). To find where it crosses they-axis we put inx= 0 to discover (0, 12). Always, always, always draw a sketch in any problem that is even vaguely geometric; thesooner you do, the sooner you ll get full my own school a few candidates were disqualified recently because they had it with them in the examdespite the fact they were not using it.

4 Don t bring it in at all! Also, don t bring in any paper with formulae onthat they give with certain pencil , Points and Lines Mid point of (x1, y1), (x2, y2) is(x1+x22,y1+y22). Average thex-coordinates andaverage they-coordinates. Distance from (x1, y1) to (x2, y2) is (by Pythagoras) (x2 x1)2+ (y2 y1)2. Be carefulabout negatives! Remember (2 ( 3))2= (2 + 3)2. Gradient is defined to bedifference inydifference inx=y2 y1x2 you need the gradient between two points you should visualise them first to see if youshould be getting a positive or negative answer. This shouldalso give you an idea ofwhether to expect a big (steep) of small (shallow) gradient.

5 Two lines with gradientsm1andm2are at right angles (perpendicular) ifm1 m2= if a line has gradient 3 then the line perpendicular to it has gradient13. Lines can be written in many forms, the most common beingy=mx+candax+by= form can be converted to any other. For example write 3x 2y= 4 in the formy=mx+ 2y= 42y= 3x 4y=32x 2. Given one point (x1, y1) and a gradientmthe line is given byy y1=m(x x1).Surds Know and understand the laws a b= a band ab= a particular know how to deal with 442; it isnot 22! It is 442= 44 4= 444= 11. This comes up in solving quadratics by the formula; checkthat when you solvex2+ 4x 2 = 0 by the formula you obtainx= 2 6.

6 You also need to be able to rationalise the denominator of certain types of surd example to rationalise9 3is easy; just multiply by 3 3to obtain9 33= 3 3. In harderexamples you must multiply the top and bottom of the fractionby the denominator withthe sign flipped . For example2 + 2 35 2 3=2 + 2 35 2 3 5 + 2 35 + 2 3=10 + 4 3 + 10 3 + 1225 + 10 3 10 3 12=22 + 14 Important Graphs Know the shape of the graphy=xnforn={1,2,3,4..}. If the power is even, then the graph will be -shaped. They all pass through the points( 1,1), (0,0) and (1,1). The bigger the power, the faster it goes to infinity.

7 Slightly moresubtle is the point that in the range 1< x <1 then the higher the power, thesmallery-value (because < ). They are all even functions with they-axisas a line of symmetry. If the power is odd then they will (with the exception ofy=x1=x, which is a straightline) be shaped likey=x3. They all pass through ( 1, 1), (0,0) and (1,1). Similararguments as for even powers exist here. They are all odd functions with the origin beinga point of rotational symmetry. The family of curvesy=ax2+bx+care parabolas. Ifais positive then you get a happy -type curve. Ifais negative then you get a sad -type curve.

8 They have a line ofsymmetry and a vertex (turning point) that you can discover by completing the square(see later). If you have a curve that is factorised then you can sketch it easily. For exampley= (x 1)(x+ 4)(x 3)is a cubic curve that crosses thex-axis at (1,0), ( 4,0) and (3,0). It crosses they-axiswhenx= 0, which gives (0,12). Ifxis huge,yis huge and positive and ifxis massivelynegative, then so isy. So-4-224-40-2020 If a factor is repeated, then it merely touches thex-axis at that point. Soy= (x 3)2(x+ 1)(x 7)is a quartic curve that crosses thex-axis at ( 1,0) and (7,0), but only touches at (3,0).

9 Factorising quadratics. To check whether a given quadraticfactorises calculate the dis-criminantb2 4ac; if it is a perfect square (4,49,81 etc.) then it factorises. When thex2coefficient (the number in front of thex2) is one this is easy. Just spot twonumbers which multiply to the constant and add to thexcoefficient. For example withx2+ 8x+ 15 we need to find two numbers which multiply to 15 and sum to 8;clearly 3and 5. Sox2+ 8x+ 15 = (x+ 3)(x+ 5). If thex2coefficient is not one then more work is required. You need to multiply thex2coefficient by the constant term and then find 2 numbers which multiply to this and sumto thexcoefficient.

10 For example with 6x2+x 12 we calculate 6 12 = 72 so thetwo numbers are clearly 9 and 8. So6x2+x 12 = 6x2+ 9x 8x 12= 6x2 8x+ 9x 12= 3x(2x+ 3) 4(2x+ 3) = 2x(3x 4) + 3(3x 4)= (3x 4)(2x+ 3)= (2x+ 3)(3x 4).Notice that it does not matter which way round we write the 9xand 8x. For quadratics that cannot be factorised we need to use the formula. Forax2+bx+c= 0the solution isx= b b2 4ac2a. Theb2 4acpart is called thediscriminant. If it is positive then there are twodistinctroots. If it is zero then there exists only one root and it isrepeated. If it is negative thenthere are no roots.