Core Maths C2. Revision Notes


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1 Core Maths C Revision Notes November 0
2 Core Maths C Algebra... Polnomials: +,,,.... Factorising... Long division... Remainder theorem... Factor theorem... 4 Choosing a suitable factor... 5 Cubic equations... 6 Trigonometr... 6 Radians... 6 Connection between radians and degrees... 6 Arc length, area of a sector and area of a segment... 6 Trig functions... 6 Basic results... 6 Eact values for 0 o, 45 o and 60 o... 7 Graphs of trig functions... 7 Graphs of = sin n, = sin( ), = sin( + n) etc Sine & Cosine rules and area of triangle... 8 Identities... 8 Trigonometric equations... 9 Coordinate Geometr... 0 Mid point... 0 Circle... 0 Centre at the origin... 0 General equation... 0 Equation of tangent... 4 Sequences and series... Geometric series... Finite geometric series... Infinite geometric series... Proof of the formula for the sum of a geometric series... Binomial series for positive integral inde... 4 Pascal s triangle... 4 Factorials... 4 n Binomial coefficients or n C r or... 4 r 5 Eponentials and logarithms... 5 Graphs of eponentials and logarithms... 5 Rules of logarithms... 5 Changing the base of a logarithm... 6 Equations of the form a = b... 7 C 4/04/0 SDB
3 6 Differentiation... 7 Increasing and decreasing functions... 7 Stationar points and local maima and minima (turning points) Using second derivative... 8 Using gradients before and after... 9 Maimum and minimum problems Integration... Definite integrals... Area under curve... Numerical integration: the trapezium rule... Inde... C 4/04/0 SDB
4 Algebra Polnomials: +,,,. A polnomial is an epression of the form a n n + a n n +... a + a + a 0 where all the powers of the variable,, are positive integers or 0. +,, of polnomials are eas, must be done b long division. Factorising General eamples of factorising: ab + 6ac = a(b + c ) = ( )( ) 6 = ( 6) 6 0 = ( + )( 5) Standard results: = ( )( + ), difference of two squares ( + ) = + +, ( ) = + Long division See eamples in book Remainder theorem If 67 is divided b 6 the quotient is 04 and the remainder is. This can be written as 67 = In the same wa, if a polnomial P() = a 0 + a + a +... a n n is divided b (c + d) to give a quotient, Q() with a remainder r, then r will be a constant (since the divisor is of degree one) and we can write P() = (c + d) Q() + r If we now choose the value of which makes (c + d) = 0 = d / c then we have d P( c ) = 0 Q() + r d P( c ) = r. C 4/04/0 SDB
5 Theorem: If we substitute = d / c in the polnomial we obtain the remainder that we would have after dividing the polnomial b (c + d). Eample: The remainder when P() = a  4 is divided b (  ) is 7. Find a. Solution: Choose the value of which makes (  ) = 0, i.e. = /. Then the remainder is P( ) = 4 ( )  ( ) + a ( ) a ( )  4 = a ( ) = 7 a = = 7 You ma be given a polnomial with two unknown letters, a and b. You will also be given two pieces of information to let ou form two simultaneous equations in a and b. Factor theorem Theorem: If, in the remainder theorem, r = 0 then (c + d) is a factor of P() d P( c ) = 0 (c + d) is a factor of P(). Eample: A quadratic equation has solutions (roots) = / and =. Find the quadratic equation in the form a + b + c = 0 Solution: The equation has roots = ½ and = it must have factors ( + ) and ( ) b the factor theorem the equation is ( + )( ) = 0 5 = 0. Eample: Show that ( ) is a factor of P() = and hence factorise the epression completel. Solution: Choose the value of which makes ( ) = 0, i.e. = remainder = P() = = = 0 ( ) is a factor b the factor theorem. We have a cubic and so we can see that the other factor must be a quadratic of the form (6 + a ) and we can write = ( ) (6 + a ) Multipling out we see that the 6 The term is and +6 terms are correct. 4 C 4/04/0 SDB
6 Multipling out the term comes from must come to + a which a = 7. We must now check the term which is 9. Multipling out with a = 7 the term is ( 7) + ( ) 6 = 7 = 9, which works! = ( )(6 7 ) which is now factorised completel. = ( )( )( + ) Choosing a suitable factor To choose a suitable factor we look at the coefficient of the highest power of and the constant (the term without an ). Eample: Factorise Solution: is the coefficient of and has factors of and. 6 is the constant and 6 has factors of,, and 6 so the possible linear factors of are ( ± ), ( ± ), ( ± ), ( ± 6) ( ± ), ( ± ), ( ± ), ( ± 6) But ( ± ) = ( ± ) and ( ± 6) = ( ± ), so the are not new factors. We now test the possible factors using the factor theorem until we find one that works. Test ( ), put = giving Test ( + ), put = giving ( ) + ( ) ( ) Test ( ), put = giving = = 0 and since the result is zero ( ) is a factor. We now divide in, as in the previous eample, to give = ( )( + 5 ) = ( )( )( + ). C 4/04/0 SDB 5
7 Cubic equations Factorise using the factor theorem then solve. N.B. The quadratic factor might not factorise in which case ou will need to use the formula for this part. Eample: Solve the equation + = 0. Solution: Possible factors are ( ± ) and ( ± ). Put = we have + = 0 ( ) is not a factor Putting = we have + = = 0 ( ) is a factor + = ( )( + ) = 0 = or + = 0 this will not factorise so we use the formula = or = ± ( ) 4 = 0.68 or. Trigonometr Radians A radian is the angle subtended at the centre of a circle b an arc of length equal to the radius. Connection between radians and degrees 80 o = π c Degrees Radians π / 6 π / 4 π / π / π / π / 4 5π / 6 π π / π Arc length, area of a sector and area of a segment Arc length s = rθ and area of sector A = ½r θ. Area of segment = area sector area of triangle = ½r θ ½r sinθ. Trig functions Basic results sin A tan A= ; cos A sin( A) = sin A; cos( A) = cos A; tan( A) = tan A; 6 C 4/04/0 SDB
8 Eact values for 0 o, 45 o and 60 o From the equilateral triangle of side we can read off sin 60 o = / sin 0 0 = ½ cos 60 o = ½ cos 0 o = / 0 o tan 60 o = tan 0 o = / 60 o and from the isosceles right angled triangle with sides,, we can read off sin 45 0 = / cos 45 o = / tan 45 o = 45 o Graphs of trig functions = sin = cos = tan Graphs of = sin n, = sin( ), = sin( + n) etc. You should know the shapes of these graphs = sin is like = sin but repeats itself times between 0 o and 60 o = sin( ) = sin and = tan( ) = tan are the graphs of = sin and = tan reflected in the ais. = cos( ) = cos is just the graph of = cos. = sin( + 0) is the graph of = sin translated through 0. 0 C 4/04/0 SDB 7
9 Sine & Cosine rules and area of triangle a b c = = : be careful the sine rule alwas gives ou two answers for each sin A sin B sinc angle so if possible do not use the sine rule to find the largest angle as it might be obtuse; or ou might want both answers if it is possible to draw two different triangles from the information given. a = b + c bc cosa etc. Unique answers here! Area of a triangle = ab sinc = bcsin A = acsin B. Identities sin A tan A = cos A Eample: Solve sin = 4 cos. Solution: First divide both sides b cos sin = 4 tan = 4 cos = 5. o, or. o. tan = 4 / sin A + cos A = Eample: Given that cos A = 5 / and that 70 o < A < 60 o, find sin A and tan A. Solution: We know that sin A + cos A = sin A = cos A sin A = ( 5 / ) sin A = ± /. But 70 o < A < 60 o sin A is negative sin A = /. sin A Also tan A = cos A tan A = 5 = / 5 =.4. 8 C 4/04/0 SDB
10 Eample: Solve sin + sin cos = Solution: Rewriting cos in terms of sin will make life easier so using sin + cos = cos = sin sin + sin cos = sin + sin ( sin ) = sin + sin = 0 ( sin )(sin + ) = 0 sin = / or = 4.8 o, 8.9 o, or 70 o. N.B. If asked to give answers in radians, ou are allowed to work in degrees as above and then convert to radians b multipling b π / 80 So answers in radians would be = 4.80 π / 80 = 0.70, or π / 80 =.4, or 70 π / 80 = π /. Trigonometric equations Eample: Solve sin ( π / 4 ) = 0.5 for 0 c π c, giving our answers in radians in terms of π. Solution: First we know that sin 60 o = 0.5, and 60 o = π / radians π / 4 = π / or π π / = π / = 7π / or π  /. Eample: Solve sin = 0.47 for 0 o 60 o, giving our answers to the nearest degree. Solution: First put X = and find all solutions of sin X = 0.47 for 0 o X 70 o X = 8., or = 5.9 or = 88., or = 5.9 i.e. X = 8., 5.9, 88., 5.9 = 4 o, 76 o, 94 o, 56 o to the nearest degree. There are several eamples in the book. C 4/04/0 SDB 9
11 Coordinate Geometr Mid point The mid point of the line joining P (a, b ) and Q (a, b ) is ( ( a a ), ( b b )) Circle + +. Centre at the origin Take an point, P, on a circle centre the origin and radius 5. Suppose that P has coordinates (, ) P (, ) Using Pthagoras Theorem we have + = 5 + = 5 r which is the equation of the circle. and in general the equation of a circle centre (0, 0) and radius r is + = r. General equation In the circle shown the centre is C, (a, b), and the radius is r. CQ = a and PQ = b and, using Pthagoras r P (, ) b CQ + PQ = r ( a) + ( b) = r, C (a, b) a Q which is the general equation of a circle. Eample: Find the centre and radius of the circle whose equation is = 0. Solution: First complete the square in both and to give = = 5 ( ) + ( + ) = 5 which is the equation of a circle with centre (, ) and radius 5. 0 C 4/04/0 SDB
12 Eample: Find the equation of the circle on the line joining A, (, 5), and B, (8, 7), as diameter. Solution: The centre is the mid point of AB is ( 8), (5 7) ) ( and the radius is ½ AB = ½ (8 ) + ( 7 5) = 6. 5 equation is ( 5.5) + ( + ) = = (5½, ) Equation of tangent Eample: Find the equation of the tangent to the circle = 64 which passes through the point of the circle ( 6, 4). Solution: First complete the square in and in to give ( + ) + ( ) = 69. Net find the gradient of the radius from the centre (, ) to the point ( 6, 4) which is / 5 gradient of the tangent at that point is 5 /, since the tangent is perpendicular to the radius and product of gradients of perpendicular lines is equation of the tangent is 4 = 5 ( + 6) 5 = 98. Eample: Find the intersection of the line = + 4 with the circle + = 5. Solution: Put = + 4 in + = 5 to give + ( + 4) = = = 0 (5 + )( + ) = 0 =. or = 0.4 or line intersects circle at (., 0.4) and (, ) If the two points of intersection are the same point then the line is a tangent. Note. You should know that the angle in a semicircle is a right angle and that the perpendicular from the centre to a chord bisects the chord (cuts it eactl in half). C 4/04/0 SDB
13 4 Sequences and series Geometric series Finite geometric series A geometric series is a series in which each term is a constant amount times the previous term: this constant amount is called the common ratio. The common ratio can be or, and positive or negative. Eamples:, 6, 8, 54, 6, 486,..... with common ratio, 40, 0, 0, 5, ½, ¼,..... with common ratio ½, ½,, 8,, 8, 5,.... with common ratio 4. Generall a geometric series can be written as S n = a + ar + ar + ar + ar up to n terms where a is the first term and r is the common ratio. The nth term is u n = ar n. The sum of the first n terms of the above geometric series is n r S n = a r a r n = r. Eample: Find the n th term and the sum of the first terms of the geometric series whose rd term is and whose 6 th term is 6. Solution: 6 = r 6 = r r = 8 r = Now = r = r = ( ) = ½ n th term, n = ar n = ½ ( ) n and the sum of the first terms is S = ( ) = S = 4 ½ Infinite geometric series When the common ratio is between and + the series converges to a limit. S n = a + ar + ar + ar + ar up to n terms n r and S n = a. r Since r <, r n 0 as n and so a S n S = r C 4/04/0 SDB
14 Eample: Show that the following geometric series converges to a limit and find its sum to infinit. S = ¾ + Solution: Firstl the common ratio is / 6 = ¾ which lies between  and + therefore the sum converges to a limit. a 6 The sum to infinit S = = r S = 64 4 Proof of the formula for the sum of a geometric series You must know this proof. S n = a + ar + ar + ar + ar n + ar n, multipl through b r r S n = ar + ar + ar + ar n + ar n + ar n subtract S n r S n = a ar n ( r) S n = a ar n = a( r n ) n n r r S n = a = a. r r Notice that if < r < + then r n 0 and S n S = a r. C 4/04/0 SDB
15 Binomial series for positive integral inde Pascal s triangle When using Pascal s triangle we think of the top row as row 0. row 0 row row row row 4 row 5 row To epand (a + b) 6 we first write out all the terms of degree 6 in order of decreasing powers of a to give a 6 + a 5 b + a 4 b + a b + a b 4 + ab 5 + b 6 and then fill in the coefficients using row 6 of the triangle to give a 6 + 6a 5 b + 5a 4 b + 0a b + 5a b 4 + 6ab 5 + b 6 = a 6 + 6a 5 b + 5a 4 b + 0a b + 5a b 4 + 6ab 5 + b 6 Factorials Factorial n, written as n! = n (n ) (n ).... So 5! = 5 4 = 0 Binomial coefficients or n n C r or r If we think of row 6 starting with the 0 th term we use the following notation 0 th term st term nd term rd term 4 th term 5 th term 6 th term C 0 6 C 6 C 6 C 6 C 4 6 C 5 6 C where the binomial coefficients n C r or n are defined b r 4 C 4/04/0 SDB
16 n n n! C r = = r ( n r)! r! This is particularl useful for calculating the numbers further down in Pascal s triangle e.g. The fourth term in row 5 is 5 5 5! 5! 5 4 C 4 = = = = 5 7 = 65 4 (5 4)!4!!4! 4 You ma find an n C r button on our calculator. Eample: Find the coefficient of in the epansion of ( ) 5. Solution: The term in is 5 C ( ) = 0 ( 8) = 80 so the coefficient of is Eponentials and logarithms Graphs of eponentials and logarithms = is an eponential function and its inverse is the logarithm function = log. Remember that the graph of an inverse function is the reflection of the original graph in = = = log Rules of logarithms log a = = a log a = log a + log a log a ( ) = log a log a log a n = n log a log a = 0 log a a = Note: log 0 is often written as lg C 4/04/0 SDB 5
17 Eample: Find log 8. Solution: Write log 8 = 8 = = 4 log 8 = 4. To solve log equations we can either use the rules of logarithms to end with log a = log a = or log a = = a Eample: Solve log a 40 log a = log a 5 Solution: log a 40 log a = log a 5 log a 40 log a 5 = log a log a (40 5) = log a log a 8 = log a = 8 =. Eample: Solve log 5 = + ½ log 5. Solution: log 5 = + ½ log 5 log 5 ½ log 5 =.5 log 5 = log 5 = = 5 = 5. Changing the base of a logarithm log a b = log log c c b a Eample: Find log 4 9. log Solution: log 9 = = =. 4 log A particular case logb b log a b = log a 4. 0 b = This gives a source of eam questions. log a b Eample: Solve log 4 6 log 4 = Solution: log 4 6 = log 4 (log 4 ) log 4 6 = 0 (log 4 )( log 4 + ) = 0 log 4 = or = 4 or 4 = 64 or / 6. 6 C 4/04/0 SDB
18 Equations of the form a = b Eample: Solve 5 = Solution: Take logs of both sides log 0 5 = log 0 log 0 5 = log 0 log0.9 = = =.59. log Differentiation Increasing and decreasing functions is an increasing function if its gradient is positive, is an increasing function if its gradient is negative, Eample: d > 0; d d < 0 d For what values of is = + 7 an increasing function. Solution: = + 7 d = d For an increasing function we want values of for which > 0 Find solutions of = 0 ( + )( ) = 0 =  / or so graph of meets ais at  / and and is above ais for <  / or > So = + 7 is an increasing function for < / or >. C 4/04/0 SDB 7
19 Stationar points and local maima and minima (turning points). An point where the gradient is zero is called a stationar point. minimum maimum Turning points Stationar points of inflection Local maima and minima are called turning points. The gradient at a local maimum or minimum is 0. Therefore to find ma and min firstl differentiate and find the values of which give gradient, d, equal to zero: d secondl find second derivative d and substitute value of found above d second derivative positive minimum, and second derivative negative maimum: N.B. If d = 0, it does not help! In this case ou will need to find the gradient d just before and just after the value of. Be careful: ou might have a stationar point of inflection thirdl substitute to find the value of and give both coordinates in our answer. Using second derivative Eample: Find the local maima and minima of the curve with equation = Solution: = First find d d = At maima and minima the gradient = d d = = = 0 ( + 4) = 0 8 C 4/04/0 SDB
20 ( + 4)( ) = 0 = 4, 0 or. d Second find d = +4 6 When = 4, When = 0, When =, d = = 80, positive min at = 4 d d = 6, negative, ma at = 0 d d = = 0, positive, min at =. d Third find values: when = 4, 0 or = 5, 7 or 0 Maimum at (0, 7) and Minimum at ( 4, 5) and (, 0). d N.B. If = 0, it does not help! You can have an of ma, min or stationar d point of inflection. Using gradients before and after Eample: Find the stationar points of = Solution: = d = 4 + = 0 for stationar points d ( + ) = 0 ( ) = 0 = 0 or. d d = which is (positive) when = 0 minimum at (0, 7) and which is 0 when =, so we must look at gradients before and after. = 0.9. d d = stationar point of inflection at (, ) N.B. We could have ma, min or stationar point of inflection when the second derivative is zero, so we must look at gradients before and after. C 4/04/0 SDB 9
21 Maimum and minimum problems Eample: A manufacturer of cans for baked beans wishes to use as little metal as possible in the manufacture of these cans. The cans must have a volume of 500 cm : how should he design the cans? Solution: h cm We need to find the radius and height needed to make cans of volume 500 cm using the minimum possible amount of metal. Suppose that the radius is cm and that the height is h cm. The area of top and bottom together is π cm and the area of the curved surface is πh cm the total surface area A = π + πh cm. (I) cm We have a problem here: A is a function not onl of, but also of h. But we know that the volume is 500 cm and that the volume can also be written as π h cm π h = 500 h = 500 π and so (I) can be written A = π + π 500 π A = π da d = π = 4π 000 = 4π. For stationar values of A, the area, da d = 0 4π = π = 000 = = = π = 4.0 to S.F. h = 500 = 860. π We do not know whether this value gives a maimum or a minimum value of A or a stationar point of inflection d A 000 so we must find = 4 π = 4 π +. d Clearl this is positive when = 4.0 and thus this gives a minimum of A minimum area of metal is 49 cm when the radius is 4.0 cm and the height is 8.60 cm. 0 C 4/04/0 SDB
22 7 Integration Definite integrals When limits of integration are given. Eample: Find d Solution: d = [ 4 + ] no need for +C as it cancels out = [ 4 + ] [ 4 + ] put top limit in first = [] [ ] =. Area under curve The integral is the area between the curve and the ais, but areas above the ais are positive and areas below the ais are negative. Eample: Find the area between the ais, = 0, = and = 4. Solution: 0 4 d 8 = = [ 8] ] [0 0] 0 6 = which is negative since the area is below the ais required area is Eample: Find the area between the ais, =, = 4 and =. Solution: First sketch the curve to see which bits are above (positive) and which bits are below (negative). = = ( ) A meets ais at 0 and 4 A so graph is as shown. Area A, between and, is above ais: area A, between and 4, is below ais 5 so we must find these areas separatel. C 4/04/0 SDB
23 A = d = = [4.5] [ ] = / 6. and 4 5 d = = [ ] [4.5] = 6 and so area A (areas are positive) = + 5 / 6 so total area = A + A = / + 5 / 6 = 5 / 6. Note that 4 4 = [ ] [ ] = 6 d which is A A = / 5 / 6 = /. 4 Numerical integration: the trapezium rule Man functions can not be anti differentiated and the trapezium rule is a wa of estimating the area under the curve. Divide the area under = f () into n strips, each of width h. = f () Join the top of each strip with a straight line to form a trapezium. n Then the area under the curve sum of the areas of the trapezia 0 h h h a b b f ) d h ( + ) + h ( + ) + h ( + ) h ( + ) a b ( 0 n f ) d h ( ) a b ( 0 n f ) d h ( + + ( )) a ( 0 n n area under curve ½ width of each strip ( ends + middles ). n n C 4/04/0 SDB
24 Inde Area of triangle, 8 Binomial series, 4 binomial coefficients, 4 n C r or n, 4 r Circle Centre at origin, 0 General equation, 0 Tangent equation, Cosine rule, 8 Cubic equations, 6 Differentiation, 7 Equations a = b, 7 Eponential, 5 Factor theorem, 4 Factorials, 4 Factorising eamples, Functions decreasing, 7 increasing, 7 Integrals area under curve, definite, Logarithm, 5 change of base, 6 rules of logs, 5 Mid point, 0 Pascal s triangle, 4 Polnomials, long division, Radians arc length, 6 area of sector, 6 area of segment, 6 connection between radians and degrees, 6 Remainder theorem, Series Geometric, finite, Geometric, infinite, Geometric, proof of sum, Sine rule, 8 Stationar points gradients before and after, 9 maima and minima, 8 maimum and minimum problems, 0 second derivative, 8 Trapezium rule, Trig equations, 9 Trig functions basic results, 6 graphs, 7 identities, 8 Turning points, 8 C 4/04/0 SDB
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