Pure Core Revision Notes Ma 06
Pure Core Algebra... Indices... Rules of indices... Surds... 4 Simplifing surds... 4 Rationalising the denominator... 4 Quadratic functions... 5 Completing the square.... 5 Factorising quadratics... 6 Solving quadratic equations.... 6 B factorising.... 6 B completing the square... 6 B using the formula... 7 The discriminant, b 4ac... 7 Miscellaneous quadratic equations... 8 Quadratic graphs... 8 Simultaneous equations... 9 Two linear equations... 9 One linear equation and one quadratic.... 0 Inequalities... Linear inequalities... Quadratic inequalities, > k or < k... Quadratic inequalities, a + b + c > k or < k... Coordinate geometr... Distance between two points... Gradient... Equation of a line... Parallel and perpendicular lines... 4 Sequences and series... Definition b a formula n = f(n)... Definitions of the form n+ = f( n )... Series and Σ notation... Arithmetic series... 4 Proof of the formula for the sum of an arithmetic series... 4 5 Curve sketching... 5 Standard graphs... 5 Cubic graphs... 6 Transformations of graphs... 7 Translations... 7 Stretches... 8 Reflections in the -ais... 9 Reflections in the ais... 9 Summar of transformations... 0
6 Differentiation... General result... Tangents and Normals... Tangents... Normals... 7 Integration... Indefinite integrals... Finding the arbitrar constant... Appendi... 4 Quadratic equation formula proof... 4 Differentiation from first principles... 4 General formulae... 5 Inde... 6 C MAY 06 SDB
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) 5 5 4 = 5 + 4 = 5 = 5. 7 4 7 = 7 4 + = 7 4 = 7 6 = (ii) 5 = 5 = 5 + = 7. 9 4 9 6 = 9 4 6 = 9 0 = 9 0 7 6. 5 = 5 = + 5 = = (iii) (6 ) 4 = 6 4 = 6 = 6. (iv) 64 64 = = ( 4) = 6 (v) 5 = = = since minus means turn upside down 5, since on bottom of fraction is cube root, 5 = 5 5 5 Eample: Epress (6 a ) (8 b ) as power of. Solution: (6 a ) (8 b ) = ( 4 ) a ( ) b = 4a b = 4a b. Eample: Find if 9 = 7 +. C MAY 06 SDB
Solution: First notice that 9 = and 7 = and so 9 = 7 + ( ) = ( ) + 4 = + 4 = + =. Surds A surd is a nast root i.e. a root which is not rational. Thus 64 = 8, 7 =, 5 4 = are rational and not surds 5 and 5, 45, 7 are irrational and are surds. Simplifing surds Eample: To simplif 50 we notice that 50 = 5 = 5 50 = 5 = 5 = 5. Eample: To simplif 40 we notice that 40 = 8 5 = 5 40 = 8 5 = 8 5 = 5. Rationalising the denominator Rationalising means getting rid of surds. We remember that multipling (a + b) b (a b) gives a b effect of squaring both a and b at the same time!! which has the Eample: Rationalise the denominator of Solution: + 5 5 = = + 5 5 6+ 5 5+ 9 5+ 5 5 + 5 + 5 + 5 5. = + 5 4. 4 C MAY 06 SDB
Quadratic functions A quadratic function is a function a + b + c, where a, b and c are constants and the highest power of is. The numbers a and b are called the coefficients of and, and c is the constant term. Completing the square. a) The coefficient of must be +. b) Halve the coefficient of, square it then add it and subtract it. Eample: Complete the square in 6 + 7. Solution: a) The coefficient of is alread +, b) the coefficient of is 6, halve it to give then square to give 9 which is then added and subtracted 6 + 7 = 6 + ( ) 9 + 7 = ( ). Notice the minimum value of 6 + 7 is when =, since the minimum value of ( ) is 0 the verte of the graph of = 6 + 7 (parabola) is at (, ). Eample: Complete the square in 4 + 5. Solution: a) The coefficient of is not +, so we must take out a factor of first, and then go on to step b). 4 + 5 = ( + 8) + 5 b) = ( + 8 + 4 6) + 5 = ( + 4) + 48 + 5 = ( + 4) + 5 Notice the maimum value of 4 + 5 is +5 when = 4, since the maimum value of ( + 4) is 0 the verte of the graph of = 4 + 5 (parabola) is at ( 4, 5). C MAY 06 SDB 5
Factorising quadratics If all the coefficients are integers (whole numbers) then if we can find factors with rational (fraction) coefficients, then we could have found factors with integer coefficients. There is no point in looking for factors with fractions as coefficients. Eample: We could factorise 5 + 6 using fractions as ( 4), BUT ( 4) = ( ) = ( ) = ( )( ), factors with integer coefficients, so no point in using fractions. More complicated epressions Eample: Factorise 0 + 6. Solution: Looking at the 0 and the 6 we see that possible factors are (0 ± ), (0 ± ), (0 ± ), (0 ± 6), (5 ± ), (5 ± ), (5 ± ), (5 ± 6), ( ± ), ( ± ), ( ± ), ( ± 6), ( ± ), ( ± ), ( ± ), ( ± 6), Also the 6 tells us that the factors must have opposite signs, and b trial and error and the factor theorem, or using common sense 0 + 6 = ( + )(5 ). Solving quadratic equations. B factorising. Eample: + 8 = 0 ( 4)( + ) = 0 4 = 0 or + = 0 = 4 or =. Eample: + 8 = 0 ( + 8) = 0 = 0 or + 8 = 0 = 0 or = 8 N.B. Do not divide through b first: ou will lose the root of = 0. B completing the square Eample: 6 4 = 0 6 = 4 coefficient of is +, halve 6, square and add 9 to both sides 6 + 9 = 4 + 9 ( ) = ( ) = ± = ± = 0 6 or 6 6 to D.P. 6 C MAY 06 SDB
B using the formula For a proof of the formula, see the appendi Alwas tr to factorise first. a + b + c = 0 = b ± b 4ac a. Eample: Solve the equation 5 = 0 Solution: 5 will not factorise, so we use the formula with a =, b =, c = 5 = ( ) ± ( ) 4 ( 5) = 5 or + 468 to D.P. The discriminant, b 4ac The quadratic equation a + b + c = 0 = b ± b 4ac a will have i) two distinct real roots if b 4ac > 0 ii) onl one real root if b 4ac = 0 (or two coincident real roots) iii) no real roots if b 4ac < 0 Eample: For what values of k does the equation k + 5 = 0 have i) two distinct real roots, ii) eactl one real root iii) no real roots. Solution: The discriminant b 4ac = ( k) 4 5 = k 60 i) for two distinct real roots k 60 > 0 k > 60 k < 60, or k > + 60 and ii) for onl one real root k 60 = 0 k = ± 60 and iii) for no real roots k 60 < 0 60 < k < + 60. C MAY 06 SDB 7
Miscellaneous quadratic equations Eample: Solve 0 + 9 = 0 Solution: Notice that = ( ) so put = to give 0 + 9 = 0 ( 9)( ) = 0 = 9 or = = 9 or = = or = 0. Eample: Solve + = 0. Solution: Put = to give + = 0 ( )( ) = or = = 4 or. Quadratic graphs a) If a > 0 the parabola will be the right wa up b 4ac > 0 distinct real roots b 4ac = 0 onl real root b 4ac < 0 no real roots 8 C MAY 06 SDB
b) If a < 0 the parabola will be the wrong wa up b 4ac > 0 distinct real roots b 4ac = 0 onl real root b 4ac < 0 no real roots Note: When sketching the curve of a quadratic function ou should alwas show the value on the -ais, if ou have factorised ou should show the values where it meets the -ais, if ou have completed the square ou should give the co-ordinates of the verte. Simultaneous equations Two linear equations Eample: Solve = 4 4 + 7 = 5 I II Solution: Make the coefficients of (or ) equal then add or subtract the equations to eliminate (or ). 4 I 8 = 6 II + = 45 Subtracting 9 = 9 = Put = in I = 4 + = 4 + = Check in equation two: L.H.S. = 4 + 7 = 5 = R.H.S. Answer : =, = C MAY 06 SDB 9
One linear equation and one quadratic. Find (or ) from the linear equation and substitute in the quadratic equation. Eample: Solve = = 5 I II Solution: From I = + Substitute in II ( + ) = 5 4 + + 9 = 5 + 9 + 4 = 0 ( + )( + 4) = 0 = ½ or = 4 = or = 5 using I (do not use II unless ou like doing etra work!) Check in the quadratic equation When = and = ½ L.H.S. = ( ½) ( ½) = 5 = R.H.S. and when = 5 and = 4 L.H.S. = ( 5) ( 4) ( 4) = 5 + = 5 = R.H.S. Answer: =, = ½ or = 5, = 4 0 C MAY 06 SDB
Inequalities Linear inequalities Solving algebraic inequalities is just like solving equations, add, subtract, multipl or divide the same number to, from, etc. BOTH SIDES EXCEPT - if ou multipl or divide both sides b a NEGATIVE number then ou must TURN THE INEQUALITY SIGN ROUND. Eample: Solve + < 8 + 4 Solution: sub from B.S. < 5 + 4 sub 4 from B.S. < 5 divide B.S. b and turn the inequalit sign round > 5. Quadratic inequalities, > k or < k 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, a + b + c > k or < k Alwas sketch a graph and find where the curve meets the -ais Eample: Find the values of which satisf 5 0. Solution: solving 5 = 0 ( + )( ) = 0 = / or We want the part of the curve = 5 which is above or on the -ais / or =² 5 / 4 C MAY 06 SDB
Coordinate geometr 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 Q (a, b ) Equation of the line PQ, above, is = m + c and use a point to find c or the equation of the line with gradient m P (a, b ) through the point (, ) is = m( ) or the equation of the line through the points (, ) and (, ) is = (ou do not need to know this one!!) (a a ) (b b ) Parallel and perpendicular lines Two lines are parallel if the have the same gradient and the are perpendicular if the product of their gradients is. Eample: Find the equation of the line through (4, ) and perpendicular to the line joining the points A (, 7) and B (6, 5). Solution: Gradient of AB is 7 5 = 4 6 gradient of line perpendicular to AB is 4 so we want the line through (4, ) with gradient 4. Using = m( ) = ¼ ( 4 ), product of perpendicular gradients is 4 = or 8 = 0. C MAY 06 SDB
4 Sequences and series A sequence is an list of numbers. Definition b a formula n = f(n) Eample: The definition n = n 5 gives = 5 =, = 5 = 7, = 5 =,. Definitions of the form n+ = f( n ) These have two parts: (i) (ii) a starting value (or values) a method of obtaining each term from the one(s) before. Eamples: (i) The definition = and n = n + defines the sequence,, 5, 07,... (ii) The definition =, =, n = n + n defines the sequence,,,, 5, 8,,, 4, 65,.... This is the Fibonacci sequence. Series and Σ notation A series is the sum of the first so man 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 = + + + 4... + n = 5 + 7 + 9 + +... + (n + ) n i i= This is written in Σ notation as S n = terms. n = ( i + ) and is a finite series of n i= An infinite series has an infinite number of terms S = i= i. C MAY 06 SDB
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,, 5, 9,,.... with common difference 4 8, 5,, 9, 6,,... with common difference. Generall 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 n S n = ( a + ( n ) d), or S n = ( a + l) n where l is the last term. 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) S n = (a + (n )d) + (a + (n )d) + (a + (n )d) + + S n = (a + (n )d) + (a + (n )d) + (a + (n )d) + (a + (n )d) S n = n(a + (n )d) S n = n (a + (n )d). I a ADD This can be written as S n = n {a +( a + (n )d)} S n = n (a + l), where l is the last term. II You should know both I and II. Eample: Find the nth term and the sum of the first 00 terms of the arithmetic series with rd term 5 and 7 th term 7. Solution: 7 = + 4d add d four times to get from the third term to the seventh 7 = 5 + 4d d = = d = 5 6 = nth term n = a + (n )d = + (n ) and S 00 = 00 ( ( ) + (00 ) ) = 4750. 4 C MAY 06 SDB
5 Curve sketching Standard graphs 4 4 = = 5 4 4 4 4 4 =² 4 4 4 =³ 4 =/ 5 4 =/² 4 4 4 4 4 4 4 =( )( )( 4) 4 = 5 4 = 4 5 6 4 5 6 4 4 4 4 = is like = but steeper: similarl for = 5 and = 7 /, etc. C MAY 06 SDB 5
Cubic graphs Some tpes of cubic graph are shown below: Coefficient of is positive Coefficient of is negative =k(a )( b)( c) a b c a b c =k( a)( b)( c) =k( a)²( b) a b a b =k( a)²(b ) =k( a)³ =k(a )³ a a 6 C MAY 06 SDB
Transformations of graphs Translations (i) If the graph of = + is translated through 0, +5 in the -direction, the equation 5 of the new graph is = + + 5; and, in general, the graph of = f () after a translation through 0 has equation b = f () + b. (ii) If the graph of = + is translated through 5, +5 in the -direction, the 0 equation of the new graph is = ( 5) + ( 5). and, in general, the graph of = f () after a translation of a has equation = f ( a). 0 5 =²+ =( 5)²+( 5) 4 5 4 4 5 6 We replace b ( a) everwhere in the formula for : note the minus sign, a, which seems wrong but is correct! Eample: 6 The graph of = ( ) + is the graph of = after a translation of + + =² 5 4 =( )²+ 4 5 6 In general The equation of the graph of =, or = f (), becomes = ( a) + b, or = f ( a) + b, after a translation through a b. C MAY 06 SDB 7
Stretches (i) If the graph of = f () is stretched b a factor of + in the -direction then the new equation is = f (). Eample: 5 The graph of = f () = becomes = f () = ( ) after a stretch of factor in the -direction 4 =² In general = or = f () becomes = a or = af () after a stretch in the -direction of factor a. =(² ) (ii) If the graph of = f () is stretched b a factor of + in the -direction then the new equation is = f. We replace b everwhere in the formula for. Eample: In this eample the graph of = 4 has been stretched b a factor of in the -direction to form a new graph with equation = f () = f = 4 4 =³ 4 5 4 4 5 =(/)³ 4(/) 4 (iii) Note that to stretch b a factor of = f () becomes = f () in the -direction we replace b = so that 8 C MAY 06 SDB
Eample: In this eample the graph of = 0 5 5 =0.5()² () has been stretched b a factor of in the -direction to form a new graph with equation = f () = f = f () = 0 5 () () = 4 =0.5² 4 5 The new equation is formed b replacing b = in the original equation. Reflections in the -ais When reflecting in the -ais all the positive -coordinates become negative and vice versa the image of = f () after reflection in the -ais is = f (). Eample: The image of = f () = + after reflection in the -ais is = f () = f () = ( + ) = + 4 =²+ 4 = (²+ ) Reflections in the ais When reflecting in the -ais the -coordinate for = + becomes the -coordinate for = and the -coordinate for = becomes the -coordinate for = +. Thus the equation of the new graph is found b replacing b the image of = f () after reflection in the ais is = f ( ). 4 Eample: The image of = f () = + after reflection in the -ais is = f () = f ( ) = ( ) + ( ) = =²+ 4 =( )²+( ) C MAY 06 SDB 9
Summar of transformations Old equation Transformation New equation = f () Translation through a = f ( a) + b b = f () Stretch with factor a in the -direction. = a f () = f () Stretch with factor a in the -direction. = f a = f () Enlargement with factor a centre (0, 0) = a f a = f () Stretch with factor a in the -direction. = f ( a ) = f () Reflection in the -ais = f () = f () Reflection in the -ais = f ( ) 0 C MAY 06 SDB
6 Differentiation General result Differentiating is finding the gradient of the curve. = n Eamples: d d = n n a) = 7 + 4,, or f () = n f () = n n d = 6 7 d b) f () = 7 = 7 f () = 7 = 8 c) = = 8 7 d d = 8 ( 4 ) = 4 4 d) f () = ( + )( ) multipl out first = 5 f () = 4 5 e) = 7 4 5 split up first = 7 5 4 5 = 4 d d = 4 ( 4 ) = 6 + 4 Tangents and Normals Tangents Eample: Find the equation of the tangent to = 7 + 5 at the point where =. Solution: We first find the gradient when =. = d 7 + 5 = 6 7 and when =, d so the gradient when = is 5. d d = 6 7 = 5. To find the equation of the line we need the -coordinate of the point where =. When =, = 7 + 5 =. the equation of the tangent is = 5( ) using = m( ) = 5 7 C MAY 06 SDB
Normals (The normal to a curve is the line which is perpendicular 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 = + at the point where =. Solution: We first find the gradient of the tangent when =. = + d d = = when = gradient of the tangent is m T = 4 =. If gradient of the normal is m N then m T m N = m N = m N = When =, = + = the equation of the normal is = ( ) using = m( ) = + 7. C MAY 06 SDB
7 Integration Indefinite integrals n d = n+ n+ + C provided that n N.B. NEVER FORGET THE ARBITRARY CONSTANT + C. Eamples: a) 4 4 6 d = 6 d = 4 5 5 + C = 4 5 5 + C. b) ( )( + ) d = + d multipl out irst = + + C. c) 9 + 5 5 d = 9 5 + 5 5 d = 4 + 5 d split up irst = 5 5 + 5 + C = 5 5 5 + C. Finding the arbitrar constant If ou know d and the coordinates of a point on the curve ou can find the arbitrar constant, C. d Eample: Solve d d = 5, given that the curve passes through the point (, 4). Solution: = 5 d = 5 + C = 4 when = 4 = 5 + C C = 6 = 5 + 6. C MAY 06 SDB
Appendi Quadratic equation formula proof a + b + c = 0 a + b = c + b = c a a b a to make coefficient of = + + b b + a a = + b a = b c 4a a add (half coefficient of ) to both sides b 4ac 4a complete square + b a = ± b 4ac 4a = ± b 4ac a square root, do not forget ± = b ± b 4ac a. Differentiation from first principles If P and Q are close together, the gradient of PQ will be nearl equal to the gradient of the tangent of the curve at P. = f () Let the -coordinates of M be, and of N be ( + h). PM = f () and QN = f ( + h) P (,) Q L QL = QN PM = f ( + h) f () and PL = h. M N + h The gradient of PQ = increase in = δ increase in δ = QL PL = f ( + h) f () h As h 0, Q gets closer and closer to P, and the gradient of PQ gets closer and closer to the gradient of the curve at P. We write f () to mean the gradient of the curve (or tangent) at P, f () = limit h 0 f ( + h) f () h and we also write the gradient as d d = limit δ 0 δ δ. 4 C MAY 06 SDB
Eample: Find, from first principles, the gradient of f () =. Solution: f () = f () = limit h 0 f () = limit h 0 f ( + h) f () h (+h) h h+h +h h = limit h 0 + h+h +h h = limit = limit h 0 h 0 + h + h cancelling h f () = + 0 + 0 putting h = 0 (which we can do after cancelling h) f () =, or d d =. Eample: Find, from first principles, the gradient of f () = 5. Solution: f () = 5 f () = f () = limit h 0 f () = limit δ 0 δ δ = 5 +h 5 h = limit 5h h 0 h(+h) 5 (+0) limit f ( + h) f () h 0 h = limit h 0 = limit h 0 f () = 5, or d d = 5. 5 5(+h) h(+h) 5 cancelling h (+h) putting h = 0 (which we can do after cancelling h) General formulae For an function f (), the derivative is f () = or f () f ( + h) f () h f ( + h) f ( h) h limit δ 0 δ δ = limit h 0 = limit h 0 The second formula comes from taking two points, Q and Q, equall placed on either side of P. Q has -coord h, and Q has -coord + h. C MAY 06 SDB 5
Inde differentiation, first principles general formulae, 5 from first principles, 4 distance between two points, equation of a line, gradient, graphs cubic graphs, 6 reflections, 9 standard graphs, 5 stretches, 8 translations, 7 indices, inequalities linear inequalities, quadratic inequalities, integration, finding arbitrar constant, normals, parallel and perpendicular lines, quadratic equations completing the square, 6 discriminant, b 4ac, 7 factorising, 6 formula, 7 miscellaneous, 8 proof of formula, 4 quadratic functions, 5 completing the square, 5 factorising, 6 quadratic graphs, 8 series arithmetic, 4 arithmetic, proof of sum, 4 sigma notation, simultaneous equations one linear equations and one quadratic, 0 two linear equations, 9 surds, 4 rationalising the denominator, 4 tangents, 6 C MAY 06 SDB