Core Maths C1. Revision Notes


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1 Core Maths C Revision Notes November 0
2 Core Maths C Algebra... Indices... Rules of indices... Surds... 4 Simplifying surds... 4 Rationalising the denominator... 4 Quadratic functions... 4 Completing the square Method... Factorising quadratics... 6 Solving quadratic equations by factorising by completing the square... 6 by using the formula... 6 The discriminant b 4ac... 7 Miscellaneous quadratic equations... 7 Quadratic graphs... 8 Simultaneous equations... 9 Two linear equations... 9 One linear and one quadratic Inequalities... 0 Linear inequalities... 0 Quadratic inequalities... 0 Coordinate geometry... Distance between two points... Gradient... Equation of a line... Parallel and perpendicular lines... 4 Sequences and series... Definition by a formula n = f(n)... Definitions of the form n+ = f( n )... Series and Σ notation... Arithmetic series... Proof of the formula for the sum of an arithmetic series... Curve sketching... 4 Standard graphs... 4 Transformations of graphs... Translations... Stretches... Reflections in the ais... 8 Reflections in the y ais... 8 C 4/04/0 SDB
3 6 Differentiation... 0 General result... 0 Tangents and Normals... 0 Tangents... 0 Normals... 7 Integration... Indefinite integrals... Finding the arbitrary constant... Inde... C 4/04/0 SDB
4 Algebra Indices Rules of indices a m a n = a m+n (a ) n = a mn a m a n = a m n a 0 = a n = n a a n m n = n a n ( ) n m a = a = a m Eamples: (i) 4 = + 4 = = = = 7 4 = 7 6 = (ii) = = + = = = 9 0 = = = + = = (iii) (6 ) 4 = 6 4 = 6 = 6. (iv) = = ( 4) = 6 (v) = since minus means turn upside down =, since on bottom of fraction is cube root, = = Eample: (6 a ) (8 b ) = ( 4 ) a ( ) b = 4a b = 4a b. Eample: Find if 9 = 7 +. Solution: First notice that 9 = and 7 = and so 9 = 7 + ( ) = ( ) + 4 = + 4 = + =. C 4/04/0 SDB
5 Surds A surd is a nasty root i.e. a root which is not rational Thus 64 8 =, 7 =, 4 = are rational and not surds and, 4, 7 are irrational and are surds. Simplifying surds Eample: To simplify 0 we notice that 0 = = 0 = = =. Eample: To simplify 40 we notice that 40 = 8 = 40 = 8 = 8 =. Rationalising the denominator Rationalising means getting rid of surds. We remember that multiplying (a + b) by (a b) gives a b squaring both a and b at the same time!! which has the effect of Eample: Rationalise the denominator of +. Solution: + = = = + 4. Quadratic functions A quadratic function is a function a + b + c, where a, b and c are constants and the highest power of is. Completing the square. Method Rule: ] The coefficient of must be. ] Halve the coefficient of, square it then add it and subtract it. Eample: Complete the square in Solution: ] The coefficient of is already, 4 C 4/04/0 SDB
6 ] the coefficient of is 6, halve it to give then square to give 9 and add and subtract = = ( ). Notice that the minimum value of the epression is when =, since the minimum value of ( ) is 0. Eample: Complete the square in + 4. Solution: ] The coefficient of is not, so we must fiddle it to make it, and then go on to step ]. + 4 = ( + 8) = ( ) = ( + 4) 48 = ( + 4) Notice that the minimum value of the epression is when = 4, since the minimum value of ( + 4) is 0. Thus the verte of the curve is at ( 4, ). Method Eample: Rewrite + 7 in completed square form Solution: We know that we need 6 (half the coefficient of ) and that ( 6) = = ( 6) 7 Notice that the minimum value of the epression is 7 when = 6, since the minimum value of ( 6) is 0. Thus the verte of the curve is at (6, 7). Eample: Epress + in completed square form. Solution: The coefficient of is not so we must take out + = ( + 4) and we know that ( + ) = , so + 4 = ( + ) 4 + = ( + 4) = [( + ) 4] = ( + ) 4 Notice that the minimum value of the epression is 4 when =, since the minimum value of ( ) is 0. Thus the verte of the curve is at (, 4). C 4/04/0 SDB
7 Factorising quadratics All the coefficients must be whole numbers (integers) then the factors will also have whole number coefficients. Eample: Factorise Solution: Looking at the 0 and the 6 we see that possible factors are (0 ± ), (0 ± ), (0 ± ), (0 ± 6), ( ± ), ( ± ), ( ± ), ( ± 6), ( ± ), ( ± ), ( ± ), ( ± 6), ( ± ), ( ± ), ( ± ), ( ± 6), Also the 6 tells us that the factors must have opposite signs, and by trial and error or common sense = ( + )( ). Solving quadratic equations. by factorising. Eample: + 6 = 0 ( )( ) = 0 = 0 or = 0 = or =. Eample: + 8 = 0 ( + 8) = 0 = 0 or + 8 = 0 = 0 or = 8 N.B. Do not divide through by first: you will lose the root of = 0. by completing the square Eample: 6 = 0 6 = and ( ) = = + 9 ( ) = ( ) = ± = ± = or by using the formula always try to factorise first. Eample: a + b + c = 0 = b ± b 4ac a = 0 will not factorise, so we use the formula with a =, b =, c = = ± ( ) 4 ( ) =. or C 4/04/0 SDB
8 The discriminant b 4ac In the formula for the quadratic equation a + b + c = 0 = b ± b 4ac a i) there will be two distinct real roots if b 4ac > 0 ii) there will be only one real root if b 4ac = 0 iii) there will be no real roots if b 4ac < 0 Eample: For what values of k does the equation k + = 0 have i) two distinct real roots, ii) eactly one real root iii) no real roots. Solution: The discriminant b 4ac = ( k) 4 = k 60 i) for two distinct real roots k 60 > 0 k > 60 k < 60, or k > + 60 and ii) for only one real root k 60 = 0 k = ± 60 and iii) for no real roots k 60 < 0 60 < k < Miscellaneous quadratic equations a) sin sin = 0. Put y = sin to give y y = 0 (y )(y + ) = 0 y = or y = ½ sin = or ½ = 90, 0 or 0 from 0 to 60. b) = 0 Notice that = ( ) and put y = to give y 0y + 9 = 0 (y 9)(y ) = 0 y = 9 or y = = 9 or = = or = 0. c) y y + = 0. Put y = to give + = 0 and solve to give = or y = = 4 or. C 4/04/0 SDB 7
9 Quadratic graphs a) If a > 0 the parabola will be the right way up i) if b 4ac > 0 the curve will meet the ais in two points. ii) if b 4ac = 0 the curve will meet the ais in only one point (it will touch the ais) iii) if b 4ac < 0 the curve will not meet the ais. b) a) If a < 0 the parabola will be the wrong way up i) if b 4ac > 0 the curve will meet the ais in two points. ii) if b 4ac = 0 the curve will meet the ais in only one point (it will touch the ais) iii) if b 4ac < 0 the curve will not meet the ais. Note: When sketching the curve of a quadratic function you should always show the value on the yais and if you have factorised you should show the values where it meets the ais, if you have completed the square you should give the coordinates of the verte. 8 C 4/04/0 SDB
10 Simultaneous equations Two linear equations Eample: Solve y = 4 and 4 + 7y =. Solution: Make the coefficients of (or y) equal then add or subtract the equations to eliminate (or y). Here 4 times the first equation gives 8y = 6 and times the second equation gives + y = 4 Subtracting gives 9y = 9 y = Put y = in equation one = 4 + y = 4 + = Check in equation two: L.H.S. = = = R.H.S. One linear and one quadratic. Find (or y) from the Linear equation and substitute in the Quadratic equation. Eample: Solve y =, y y = Solution: From first equation = y + Substitute in second (y + ) y y = 4y + y + 9 y y = y + 9y + 4 = 0 (y + )(y + 4) = 0 y = ½ or y = 4 = or = from the linear equation. Check in quadratic for =, y = ½ L.H.S. = ( ½) ( ½) = = R.H.S. and for =, y = 4 L.H.S. = ( ) ( 4) ( 4) = + = = R.H.S. C 4/04/0 SDB 9
11 Inequalities Linear inequalities Solving algebraic inequalities is just like solving equations, add, subtract, multiply or divide the same number to, from, etc. BOTH SIDES EXCEPT  if you multiply or divide both sides by a NEGATIVE number then you must TURN THE INEQUALITY SIGN ROUND. Eample: Solve + < Solution: sub from B.S. < + 4 sub 4 from B.S.  < divide B.S. by  and turn the inequality sign round > .. Eample: Solve > 6 Solution: We must be careful here since the square of a negative number is positive giving the full range of solutions as < 4 or > +4. Quadratic inequalities Always sketch a graph and find where the curve meets the ais Eample: Find the values of which satisfy 0. Solution: = 0 y ( + )( ) = 0 = / or We want the part of the curve which is above or on the ais / or  0 C 4/04/0 SDB
12 Coordinate geometry Distance between two points Distance between P (a, b ) and Q (a, b ) is ( a a ) + ( b b ) Gradient Gradient of PQ is m b = b a a Equation of a line Equation of the line PQ, above, is y = m + c and use a point to find c or the equation of the line with gradient m through the point (, y ) is y y = m( ) or the equation of the line through the points (, y ) and (, y ) is y y = y y (you do not need to know this one!!) Parallel and perpendicular lines Two lines are parallel if they have the same gradient and they are perpendicular if the product of their gradients is. Eample: Find the equation of the line through (4½, ) and perpendicular to the line joining the points A (, 7) and B (6, ). Solution: 7 Gradient of AB is = 4 6 gradient of line perpendicular to AB is ¼, (product of perpendicular gradients is ) so we want the line through (4½, ) with gradient ¼. Using y y = m( ) y = ¼ ( 4½) 4y =  ½ or + 8y + = 0. C 4/04/0 SDB
13 4 Sequences and series A sequence is any list of numbers. Definition by a formula n = f(n) Eample: The definition n = n gives = =, = = 7, = =,. Definitions of the form n+ = f( n ) These have two parts: (i) a starting value (or values) (ii) a method of obtaining each term from the one(s) before. Eamples: (i) The definition = and n = n + defines the sequence,,, 07,... (ii) The definition =, =, n = n + n defines the sequence,,,,, 8,,, 4, 6,.... This is the Fibonacci sequence. Series and Σ notation A series is the sum of the first so many terms of a sequence. For a sequence whose nth term is n = n + the sum of the first n terms is a series S n = n = (n + ) n i i= This is written in Σ notation as S n = n = ( i + ) and is a finite series of n terms. i= An infinite series has an infinite number of terms S = i= Arithmetic series An arithmetic series is a series in which each term is a constant amount bigger (or smaller) than the previous term: this constant amount is called the common difference. Eamples:, 7,,, 9,,.... with common difference 4 8,,, 9, 6,,... with common difference. i. Generally an arithmetic series can be written as S n = a + (a + d) + (a + d) + (a + d) + (a + 4d) +... upto n terms, where the first term is a and the common difference is d. The nth term n = a + (n )d The sum of the first n terms of the above arithmetic series is C 4/04/0 SDB
14 n S n = ( a + ( n ) d), or S n = ( L ) n a + where L is the last term. Eample: Find the nth term and the sum of the first 00 terms of the arithmetic series with rd term and 7 th term 7. Solution: 7 = + 4 d 7 = + 4 d d = = d = 6 = nth term n = a + (n )d = + (n ) and S 00 = 00 / ( + (00 ) ) = 470. Proof of the formula for the sum of an arithmetic series You must know this proof. First write down the general series and then write it down in reverse order S n = a + (a + d) + (a + d) (a + (n )d) + (a + (n )d S n = (a + (n )d + (a + (n )d) + (a + (n )d) + + (a + d) + a ADD S n = (a + (n )d) + (a + (n )d) + (a + (n )d) + (a + (n )d) + (a + (n )d) S n = n(a + (n )d) S n = n / (a + (n )d), Which can be written as S n = n / (a + a + (n )d) = n / (a + L), where L is the last term. C 4/04/0 SDB
15 Curve sketching Standard graphs y y y y = y =  y = y y y y = y = / y = / y y y = The eponential curve y = y = is like y = but steeper: similarly for y = and y = 7 /, etc. a b c > a b > y = ( a)( b)( c) y = ( a) ( b) a b c > a b > y = (a )( b)( c) y = (a )( b) 4 C 4/04/0 SDB
16 Transformations of graphs Translations (i) If the graph of y = + is translated through + in the ydirection the equation of the new graph is y = + + ; and in general if we know that y is given by some formula involving, which we write as y = f (), then the new curve after a translation through + units in the y direction is y = f () +. (ii) If the graph of y = + is translated through + in the direction the equation of the new graph is y = (  ) + (  ); and if y = f (), then the new curve after a translation through + units in the direction is y = f (  ): i.e. we replace by (  ) everywhere in the formula for y: note the minus sign, , which seems wrong but is correct! Eample: 4 Y The graph of y = (  ) + is the graph of y = + after a translation of + y = y = (  ) X In general y = or y = f () becomes y = (  a) + b a or y = f (  a) + b after a translation through b Stretches (i) If the graph of y = + is stretched by a factor of + in the ydirection.then the equation of the new graph is y = ( + ); and in general if y is given by some formula involving, which we write as y = f (), then the new curve after a stretch by a factor of + in the ydirection is y = f (). C 4/04/0 SDB
17 Eample: y =  4 Y The graph of y =  becomes y = (  ) after a stretch of factor in the ydirection X y = (  ) 6 In general y = or y = f () becomes y = a or y = af () after a stretch in the ydirection of factor a. (ii) If the graph of y = + is stretched by a factor of + in the direction then the equation of the new graph is y = + and in general if y is given by some formula involving, which we write as y = f (), then the new curve after a a stretch by a factor of + in the direction is y = f i.e. we replace by everywhere in the formula for y. 6 C 4/04/0 SDB
18 Eample: In this eample the graph of y = 4 has been stretched by a factor of in the direction to form a new graph with equation y = 4 y = y = 4 8 Y X (iii) Note that to stretch by a factor of ¼ in the direction we replace by that y = f () becomes y = f (4) = so 4 4 Eample: In this eample the graph of y =  9 has been stretched by a factor of in the direction to form a new graph with equation ( ) 9( ) y = = 7 7, The new equation is formed by replacing by = in the original equation. Y y = X ( ) 9( ) y = C 4/04/0 SDB 7
19 Reflections in the ais When reflecting in the ais all the positive ycoordinates become negative and vice versa and so the image of y = f () after reflection in the ais is y = f (). Eample: The image of y = + after reflection in the ais is y = f () = ( + ) = + y = +. y y = +. Reflections in the y ais When reflecting in the yais the ycoordinate for = + becomes the ycoordinate for = and the ycoordinate for = becomes the ycoordinate for = +. Thus the equation of the new graph is found by replacing by and the image of y = f () after reflection in the y ais is y = f ( ). Eample: The image of y = + after reflection in the yais is y = ( ) + ( ) = y y = + y = C 4/04/0 SDB
20 Old equation Transformation New equation y = f () y = f () Translation through Stretch with factor a in the ydirection. a b y = f (  a) + b y = a f () y = f () y = f () Stretch with factor a in the direction. Stretch with factor in the direction. a y = f a y = f( a ) y = f () Reflection in the ais y =  f () y = f () Reflection in the yais y = f () C 4/04/0 SDB 9
21 6 Differentiation General result Differentiating is finding the gradient of the curve. y = n dy d = n n, or f () = n f () = n n Eamples: a) y = dy = 6 7 d b) 7 f () = 7 = 7 f ( ) 7 = = c) 8 dy 4 4 y = = 8 = 8 = 4 d d) f () = ( + )( ) multiply out first = f ( ) = 4 e) 7 4 y = split up first 7 4 dy 4 = = 4 = 6 = d Tangents and Normals Tangents Eample: Find the equation of the tangent to y = 7 + at the point where =. Solution: We first find the gradient when =. y = 7 + dy d = 6 7 and when =, dy d = 6 7 =. so the gradient when = is. The equation is of the form y = m + c where m is the gradient so we have y = + c. To find c we must find a point on the line, namely the point on the curve when =. When = the y value on the curve is 7 + =, i.e. when =, y =. Substituting these values in y = + c we get = + c c = 7 the equation of the tangent is y = 7. 0 C 4/04/0 SDB
22 Normals (The normal to a curve is the line at 90 o to the tangent at that point). We first remember that if two lines with gradients m and m are perpendicular then m m =. Eample: Find the equation of the normal to the curve y = + at the point where =. Solution: We first find the gradient of the tangent when =. y = + dy d = = when = gradient of the tangent is m = 4 = ½. If gradient of the normal is m then m m = ½ m = m = Thus the equation of the normal must be of the form y = + c. From the equation y = + the value of y when = is y = + = Substituting = and y = in the equation of the normal y = + c we have = + c c = 7 equation of the normal is y = + 7. C 4/04/0 SDB
23 7 Integration Indefinite integrals n+ n d = + C provided that n n + N.B. NEVER FORGET THE ARBITRARY CONSTANT + C. Eamples: = d 4 4 = + C = + C ( )( + ) d = + d a) d 6 b) = + ½ + C. + 4 d = + d = + + C = + C 9 c) multiply out first split up first Finding the arbitrary constant If you know the derivative (gradient) function and a point on the curve you can find C. dy Eample: Solve =, given that y = 4 when =. d Solution: y = d = + C but y = 4 when = 4 = + C C = 6 y = + 6. C 4/04/0 SDB
24 Inde Differentiation, 0 Distance between two points, Equation of a line, Gradient, Graphs Reflections, 8 Standard graphs, 4 Stretches, Translations, Indices, Inequalities linear inequalities, 0 quadratic inequalities, 0 Integrals finding arbitrary constant, indefinite, Normals, Parallel and perpendicular lines, Quadratic equations b 4ac, 7 completing the square, 6 factorising, 6 formula, 6 miscellaneous, 7 Quadratic functions, 4 completing the square, 4 factorising, 6 verte of curve, Quadratic graphs, 8 Series Arithmetic, Arithmetic, proof of sum, Sigma notation, Simultaneous equations one linear equations and one quadratic, 9 two linear equations, 9 Surds, 4 rationalising the denominator, 4 Tangents, 0 C 4/04/0 SDB
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