Pure Core 2. Revision Notes

Transcription

1 Pure Core Revision Notes June 06

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3 Pure Core Algebra... Polynomials... Factorising... Standard results... Long division... Remainder theorem... 4 Factor theorem... 5 Choosing a suitable factor... 6 Cubic equations... 7 Trigonometry... 8 Radians... 8 Connection between radians and degrees... 8 Arc length, area of a sector and area of a segment... 8 Trigonometric functions... 9 Basic results... 9 Eact values for 0 o, 45 o and 60 o... 9 Sine and cosine rules and area of triangle... 9 Sine rule... 9 Ambiguous case... 0 Cosine rule... 0 Area of triangle... 0 Graphs of trigonometric functions... 0 Graphs of y = sin n, y = sin( ), y = sin( + n) etc.... Solving trigonometrical equations... Using identities... Coordinate Geometry... 5 Mid point... 5 Distance between two points... 5 Perpendicular lines... 5 Circle... 5 Centre at the origin... 5 General equation... 6 Equation of tangent... 7 Intersection of line and circle Sequences and series... 9 Geometric series... 9 Finite geometric series... 9 Infinite geometric series... 0 Proof of the formula for the sum of a geometric series... 9 Binomial series for positive integral inde... Pascal s triangle... Factorials... Binomial coefficients or n C r or nr...

4 5 Eponentials and logarithms... Graphs of eponentials and logarithms... Rules of logarithms... Changing the base of a logarithm... 4 A particular case... 4 Equations of the form a = b Differentiation... 5 Increasing and decreasing functions... 5 Stationary points and local maima and minima (turning points) Using second derivative... 6 Using gradients before and after... 7 Maimum and minimum problems Integration... 9 Definite integrals... 9 Area under curve... 9 Numerical integration: the trapezium rule Appendi... Binomial coefficients, n C r... Choosing r objects from n... n C r = n C n r... (a + b) n... Points of infleion... To find a point of infleion... Integration... Area under graph sum of rectangles... Integration as anti-differentiation... Inde... 4 C JUNE 06 SDB

5 Algebra Polynomials A polynomial is an epression of the form a n n + a n n +... a + a + a 0 where all the powers of are positive integers or 0. Addition, subtraction and multiplication of polynomials are easy, division must be done by long division. Factorising General eamples of factorising: ab + 6ac = a(b + c ) = ( )( ) 6 = ( 6) 6 0 = ( + )( 5) a by 6ay + b = a 6ay + b by = a( y) + b( y) = (a + b)( y) Standard results y = ( y)( + y), difference of two squares ( + y) = + y + y, ( y) = y + y Long division Eample: when is divided by +, the quotient is 5 + 9, and the remainder is + 6. C JUNE 06 SDB

6 Remainder theorem If 67 is divided by 6 the quotient is 04 and the remainder is. This can be written as 67 = In the same way, if a polynomial P() = a 0 + a + a +... a n n is divided by (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 P d ( c ) = 0 Q() + r d P( c ) = r. Theorem: If we put = d / c in the polynomial we obtain r, the remainder that we would have after dividing the polynomial by (c + d). Eample: The remainder when P() = + a + b + 9 is divided by ( ) is 6, and when P() is divided by ( + ) the remainder is. Find the values of a and b. Solution: ( ) = 0 when = /, dividing P() by ( ) gives a remainder P( ) = ( ) + a ( ) + b ( ) a + b = 9 I + 9 = 6 and ( + ) = 0 when =, dividing P() by ( + ) gives a remainder P( ) = ( ) + a ( ) + b ( ) + 9 = 4a b = 8 II I + II 7a = a = using I we get b = 0 4 C JUNE 06 SDB

7 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 ( ) by the factor theorem an equation is ( + )( ) = 0 5 = 0. or any multiple Eample: Show that ( ) is a factor of P() = and hence factorise the epression completely. Solution: Choose the value of which makes ( ) = 0, i.e. = remainder = P() = = = 0 ( ) is a factor by the factor theorem. We have started with a cubic and so we the other factor must be a quadratic, which can be found by long division or by common sense = ( )(6 7 ) = ( )( )( + ) which is now factorised completely. C JUNE 06 SDB 5

8 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 term and 6 has factors of,, and 6 the possible linear factors of are ( ± ), ( ± ), ( ± ), ( ± 6) ( ± ), ( ± ), ( ± ), ( ± 6) But ( ± ) = ( ± ) and ( ± 6) = ( ± ), so they 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 to give = ( )( + 5 ) = ( )( )( + ). 6 C JUNE 06 SDB

9 Cubic equations Factorise using the factor theorem then solve. N.B. The quadratic factor might not factorise in which case you 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 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 68 to D.P. C JUNE 06 SDB 7

10 Trigonometry Radians r radian r A radian is the angle subtended at the centre of a circle by an arc of length equal to the radius. r Connection between radians and degrees 80 o = π c Degrees Radians π π π / 6 π / 4 π / π / π / π / 4 5π / 6 π / Arc length, area of a sector and area of a segment r θ radians r s = rθ Arc length s = rθ Area of sector A = r θ. r θ radians r Area of segment = area sector area of triangle = r θ r sinθ. 8 C JUNE 06 SDB

11 Trigonometric functions Basic results sin A tan A= ; cos A sin( A) = sin A; cos( A) = cos A; tan( A) = tan A. Eact values for 0 o, 45 o and 60 o From the equilateral triangle of side we can see that 0 o sin 60 o = / sin 0 o = ½ cos 60 o = ½ cos 0 o = tan 60 o = tan 0 o = / / 60 o From the isosceles right angled triangle with sides,, we can see that sin 45 o = / cos 45 o = / tan 45 o = 45 o Sine and cosine rules and area of triangle C b a A c B Sine rule a b c = = sin A sin B sin C be careful the sine rule always gives you two answers for each angle, so if possible do not use the sine rule to find the largest angle as it might be obtuse; you may be able to use the cosine rule. C JUNE 06 SDB 9

12 Ambiguous case Eample: In a triangle PQR, PQ = 0, QPR = 40 o and QR = 8. Find PRQ. Q Solution: If we draw PQ and an angle of 40 o, there are two possible positions for R, giving two values of PRQ The sine rule gives sin 40 8 = sin R 0 sin R = P 40 o R R R = 5 5 o or = 6 5 o both answers are correct Cosine rule a = b + c bc cosa You will always have unique answers with the cosine rule. Area of triangle Area of a triangle = absin C = bcsin A = acsin B. Graphs of trigonometric functions y y=sin y y=cos y y=tan y = sin y = cos y = tan 0 C JUNE 06 SDB

13 Graphs of y = sin n, y = sin( ), y = sin( + n) etc. You should know the shapes of these graphs y = sin y y=sin y = sin is like y = sin but repeats itself times for 0 o 60 o, or 0 π c y y y=sin y=tan y=tan( )= tan y=sin( )= sin y = f () = sin y = f () = tan for a reflection in the y-ais, for a reflection in the y-ais, f( ) = sin( ) = sin, f( ) = tan( ) = tan, and for a reflection in the -ais, and for a reflection in the -ais, f () = sin f () = tan same graph for both reflections same graph for both reflections y = cos( ) and cos y = cos( ) is the same as the graph of y = f () = cos, since the graph of y = cos is symmetrical about the y-ais, and f ( ) = cos( ) = cos. y But y = cos = f () is a reflection of y = f () = cos in the -ais. y=cos y= cos y = sin( + 0) y = sin( + 0) is the graph of 0 y = sin translated through. 0 y y=sin y=sin(+0) C JUNE 06 SDB

14 Solving trigonometrical equations Eamples: Solve (a) sin = 0 45, (b) cos = 0 769, (c) sin = 0 876, (d) tan = 56, for 0 < 60 o. Solutions: (a) sin = 0 45 y y=sin = 6 9 y=0.45 using the graph we see that = = 6 9 or 5 = = (b) cos = y y=cos = 40 using the graph we see that = = 40 or 9 7 y= =40. = (c) sin = = 6 y y=sin using the graph we see that = or = 60 6 = 4 or y= = 6. =80+6. =60 6. (d) tan = 56 y = 57 using the graph we see that = y=.56 y=tan = 57 or 7 =57. = C JUNE 06 SDB

15 Eample: Solve sin ( π ) = 0.5 for 4 0c π c, giving your answers in radians in terms of π. y Solution: First put X = π 4 y=0.5 y=sinx sin X = 0 5 X = π 6 or π π 6 = 5π 6 X π/ π π/ π X = π 4 = π 6 or 5π = 5π or π. 6 X=π/6 X=π π/6 Eample: Solve cos = 0 47 for 0 o 60 o, giving your answers to the nearest degree. Solution: First put X = and find all solutions of cos X = 0 47 for 0 o X 70 o X = , or = 98 or = 4.77, or = 658 i.e. X = , 98, 4.77, 658 = X = o, 49 o, o, 9 o to the nearest degree. Using identities (i) using tan A sin A cos A Eample: Solve sin = 4 cos. Solution: First divide both sides by cos sin = 4 tan = 4 tan = 4 / cos = 5 o, or = o. C JUNE 06 SDB

16 (ii) using sin A + cos A = Eample: Given that cos A = 5 and that 70o < 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. y y=+/ y= / y=sin * Eample: Solve sin + sin cos = Solution: Rewriting cos in terms of sin will make life easier Using sin + cos cos = sin y sin + sin cos = y=/ y=sin sin + sin ( sin ) = sin + sin = 0 ( sin )(sin + ) = 0 y= sin = = 4 8 o, 8 o, or sin = = 70 o =4.8 = =70 N.B. If you are asked to give answers in radians, you are allowed to work in degrees as π above and then convert to radians by multiplying by So the answers in radians would be = 4 80 π / 80 = 0 70, or π / 80 = 4, or 70 π / 80 = π /. Under no circumstances should you use the 80 S A T C diagram. You need to understand the graphs and their symmetries, so get used to using them. 4 C JUNE 06 SDB

17 Coordinate Geometry Mid point The mid point, M, of the line joining P (a, b ) and Q (a, b ) is ( ( a a ), ( b b )) Distance between two points + +. Let P and Q be the points (a, b ) and (a, b ). Q (a, b ) Using Pythagoras s theorem M b b PQ = (a a ) + (b b ) P (a, b ) a a A Perpendicular lines Two lines with gradients m and m are perpendicular m m = Eample: Find the equation of the line through (, 5) which is perpendicular to the line with equation y = Solution: The gradient of y = is Gradient of perpendicular line is equation of perpendicular line is y 5 = ( ) using y y = m( ) + y + 9 = 0 Circles Centre at the origin Take any point, P, on a circle centre the origin and radius 5. Suppose that P has coordinates (, y) r y P (, y) Using Pythagoras Theorem we have + y = 5 + y = 5 which is the equation of the circle. and in general the equation of a circle centre (0, 0) and radius r is r r y r + y = r. r C JUNE 06 SDB 5

18 General equation In the circle shown the centre is C, (a, b), and the radius is r. CQ = a and PQ = y b and, using Pythagoras CQ + PQ = r ( a) + (y b) = r, y C P (, y) r y b a Q which is the general equation of a circle. Eample: Find the centre and radius of the circle whose equation is + y 4 + 6y = 0. Solution: First complete the square in both and y to give y + 6y + 9 = = 5 ( ) + (y + ) = 5 which is the equation of a circle with centre (, ) and radius 5. Eample: Find the equation of the circle which has diameter AB, where A is (, 5), and B is (8, 7). Solution: The centre is the mid point of AB is ( 8), (5 7) ) ( + = (5½, ) and the radius is AB = (8 ) + ( 7 5) = 6 5 equation is ( 5 5) + (y + ) = C JUNE 06 SDB

19 Equation of tangent Eample: Find the equation of the tangent to the circle + + y 4y = 0 at the point ( 4, 6). Solution: First complete the square in and in y 0 y y 4y + 4 = ( + ) + (y ) = 5. * 5 ²++y² 4y=0 Second find the gradient of the radius from the centre (, ) to the point ( 4, 6) gradient of radius = 6 4 = 4 gradient of the tangent at that point is 4, since the tangent is perpendicular to the radius and product of gradients of perpendicular lines is = 4 4 equation of the tangent is y 6 = ( 4) 4 4y + 6 = 0. 4y+6= * 5 Intersection of line and circle Eample: Find the intersection of the line y = + 4 with the circle + y = 5. Solution: Put y = + 4 in + y = 5 to give + ( + 4) = = = 0 (5 + )( + ) = 0 = or y = 0 4 or the line and the circle intersect at (, 0 4) and (, ) C JUNE 06 SDB 7

20 Showing a line is a tangent to a circle If the two points of intersection are the same point then the line is a tangent. Eample: Show that the line + 4y 0 = 0 is a tangent to the circle + + y + 6y = 5. Solution: Find the intersection of the line and circle + 4y 0 = 0 = 0 4y Substituting in the equation of the circle 0 4y + 0 4y + y + 6y = y + 6y y + 9y + 54y = 5 5y 50y + 5 = 0 y y + = 0 (y ) = 0 y = only, = line is a tangent at (, ), since there is only one point of intersection Note. You should know that the angle in a semi-circle is a right angle and that the perpendicular from the centre to a chord bisects the chord (cuts it eactly in half). 8 C JUNE 06 SDB

21 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 any non-zero real number. 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. Generally a geometric series can be written as S n = a + ar + ar + ar + ar ar n, 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 S n = a ( rn ) r = a (rn ) r. 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 multiply through by 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 ) S n = a ( rn ) r = a (rn ) r. For an infinite series, if < r < + r < then r n 0 as n, and S n S = a r. C JUNE 06 SDB 9

22 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 multiply by r times to go from the rd term to the 6 th term 6 = r r = 8 r = Now = r = r = ( ) = n th term, n = ar n = ( )n and the sum of the first terms is S = ( ) = S = 4 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 S n = a ( rn ). r Since r <, r n 0 as n and so S n S = a r Eample: Show that the geometric series S = converges to a limit and find its sum to infinity. Solution: Firstly the common ratio is the sum converges to a limit. The sum to infinity S = a 6 = r 6 = 4 4 which lies between and + therefore S = 64 0 C JUNE 06 SDB

23 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 C JUNE 06 SDB

24 Factorials Factorial n, written as n! = n (n ) (n ).... So 5! = 5 4 = 0 Binomial coefficients or n C r or n r If we think of row 6 in Pascal s triangle starting with the 0th 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 by r or n C r = n r = n! (n r)! r! n C r = n r = n(n )(n )(n ) r! up to r numbers This is particularly useful for calculating the numbers further down in Pascal s triangle. Eample: The fourth number in row 5 is 5 C 4 = 5 4 = 5! (5 4)!4! = 5!! 4! = = 65. You can also use n C r button on your calculator. Eample: Find the coefficient of in the epansion of ( ) 5. Solution: The term in is 5 C ( ) since is 0 9 ( 8 ) = 70 so the coefficient of is C = 0 For more ideas on using the binomial coefficients, see the appendi. C JUNE 06 SDB

25 5 Eponentials and logarithms Graphs of eponentials and logarithms y = is an eponential function and its inverse is the logarithm function y = log. Remember that the graph of an inverse function is the reflection of the original graph in y =. 4 y y= y= y = log Rules of logarithms log a = y = a y log a y = log a + log a y log a ( y) = log a log a y log a n = n log a log a = 0 log a a = Eample: Find log 8. Solution: Write log 8 = y 8 = y y = 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 = log a 5 log a (40 ) = log a 5 40 = 5 = 8 =. C JUNE 06 SDB

26 Eample: Solve log + log ( + 6) = + log ( + ). Solution: log (+6) (+) = (+6) (+) = = = = 0 ( 4)( + ) = 0 = 4 or But cannot be negative (you cannot have log when 0) = 4 only Changing the base of a logarithm log a b = log log c c b a Eample: Find log 4 9. Solution: log = = = 0 log4 9 4 log A particular case logb b log a b = log a 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 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 log 9 log = = = C JUNE 06 SDB

27 6 Differentiation Increasing and decreasing functions y is an increasing function if its gradient is positive, y is a decreasing function if its gradient is negative, dy > 0; d dy < 0 d Eample: For what values of is y = f () = + 7 an increasing function. Solution: y = f () = + 7 dy = f ' () = d For an increasing function we want values of for which f ' () = > 0 Find solutions of f ' () = = 0 ( + )( ) = 0 = / or so graph of meets -ais at / and and is above -ais for < / or > f ' () > 0 for < / or > f '() f '() =² / So y = + 7 is an increasing function for < / or >. Stationary points and local maima and minima (turning points). Any point where the gradient is zero is called a stationary point. Local maima and minima are called turning points. The gradient at a local maimum or minimum is 0. C JUNE 06 SDB 5

28 minimum Turning points maimum Stationary points of inflection Therefore to find ma and min first differentiate and find the values of which give gradient, dy, equal to zero: d second d y find second derivative and substitute value of found above d second derivative positive minimum, and second derivative negative maimum: N.B. If d y = 0, it does not help! In this case you will need to find the gradient d just before and just after the value of. Be careful: you might have a stationary point of inflection third substitute to find the value of y and give both coordinates in your answer. Using second derivative Eample: Find the local maima and minima of the curve with equation y = Solution: y = First find dy d = dy At maima and minima the gradient = d = = = 0 ( + 4) = 0 ( + 4)( ) = 0 = 4, 0 or. 6 C JUNE 06 SDB

29 d y Second find d = When = 4, d y = = 80, positive min at = 4 d d y When = 0, = 6, negative, ma at = 0 d d y When =, = = 0, positive, min at =. d Third find y values: when = 4, 0 or y = 5, 7 or 0 Maimum at (0, 7) and Minimums at ( 4, 5) and (, 0). N.B. If d y d inflection. = 0, it does not help! You can have any of ma, min or stationary point of Using gradients before and after Eample: Find the stationary points of y = Solution: y = dy = 4 + = 0 for stationary points d ( + ) = 0 ( ) = 0 = 0 or. d y 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. dy d = stationary point of inflection at (, ) N.B. We could have ma, min or stationary point of inflection when the second derivative is zero, so we must look at gradients before and after. C JUNE 06 SDB 7

30 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: cm 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 We have a problem here: A is a function not only of, but also of h. But the volume is 500 cm and the volume can also be written as V = π h cm π h = 500 h = 500 π and so I can be written A = π + π 500 A = π da = π d = 4π 000 = 4π For stationary values of A, the area, da π 000. d = 0 4π = π = 000 = = = π = 4 0 to S.F. h = 500 = π We do not know whether this value gives a maimum or a minimum value of A or a stationary point of inflection so we must find d A = 4π + d 000 = 4π Clearly 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. 8 C JUNE 06 SDB

31 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 y = 4. Solution: 0 4 d y y=² 4 0 since the area is below the ais 8 = = [ 8] ] [0 0] = 6 which is negative 4 required area is +6 4 Eample: Find the area between the -ais, =, = 4 and y =. Solution: First sketch the curve to see which bits are above (positive) and which bits are below (negative). y = = ( ) meets -ais at 0 and. y y= ² Area A, between and, is above ais: area A, between and 4, is below ais A A so we must find these areas separately. C JUNE 06 SDB 9

32 A = d = = [4 5] [ 6] = /. and 4 d = 4 = [ ] [4 5] = 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 = / ). Numerical integration: the trapezium rule Many functions can not be anti differentiated and the trapezium rule is a way of estimating the area under the curve. Divide the area under y = f () into n strips, each of width h. y y=f() Join the top of each strip with a straight line to form a trapezium. yn y0 Then the area under the curve sum of the areas of the trapezia a y y h h h b b f ) d h ( y + y ) + h ( y + y ) + h ( y + y ) h ( y + y ) a ( 0 n b f ( ) d h ( y 0 + y + y + y + y + y + y... + y y y n n n) a + + b f ) d h ( y + y + ( y + y + y y )) a ( 0 n n area under curve ½ width of each strip ( ends + middles ). n 0 C JUNE 06 SDB

33 8 Appendi Binomial coefficients, n C r Choosing r objects from n If we have n objects, the number of ways we can choose r of these objects is n C r. n C r = n C n r Every time r objects from n must, therefore, be the same as the number of ways of leaving n r behind. are chosen from n, there are n r objects left behind; the number of ways of choosing r objects n C r = n C n r. This can be proved algebraically. n C n r = n! n (n r)!(n r)! = n! (n n+r)! (n r)! = n! r! (n r)! = n C r (a + b) n In the epansion of (a + b) n = (a + b) (a + b) (a + b) (a + b) (a + b) (a + b) (a + b), where there are n brackets, we can think of forming the term a n r b r by choosing the r letter bs from the n brackets in n C r ways. Thus the coefficient of a n r b r is n C r. C JUNE 06 SDB

34 Points of infleion A point of infleion is a maimum or minimum of the gradient. When the gradient is also zero, in which case we have a stationary point of infleion, otherwise we have an oblique (sloping) point of infleion. - - minimum of gradient Oblique points of infleion maimum of gradient minimum of gradient maimum Stationary points of infleion of gradient To find a point of infleion. Find the value(s) of for which d y = 0, = α, β, d. Either show that d y 0 for these values of d or show that either = α d y d is +ve and = α + or = α d y d is ve and = α + d y d changes sign from = α to = α +. d y d d y d is ve is +ve Eample: Find the point(s) of infleion on the graph of y = Solution: y = dy d = d y d = 6 6 d y d = 0 6( ) = 6( + )( ) = 0 = or. d y d = 4 6 = d y d = 8 0, and = d y d = 8 0 C JUNE 06 SDB

35 points of infleion at A, (, ), 6 and B, (, ). Notice that dy = 0 when =, d but dy d = 0 when = 4 A, (, ), is an oblique point of 6 infleion, and B, (, ), is a stationary point of infleion. y=⁴ ³ ²+5+ 4 A * 4 y B * Integration Area under graph sum of rectangles In any continuous graph, y = f (), we can divide the area between = a and = b into n strips, each of width δ. The area under the graph (between the graph, the -ais and the lines = a and = b) is approimately the area of the n rectangles, as shown. the area under the graph n A y i δ i= b, and as δ 0, A = y a d y y=f() y y y y 0 n d d d a b Integration as anti-differentiation A = area under the curve from = a to δ A = increase in area from to + δ y y = f() δ A area of the rectangle shown δ A f () δ A f() δ A δa δ f() As δ 0 a δ + δ we have da d = f() to find the integral we anti-differentiate f (). C JUNE 06 SDB

36 Inde area of triangle, 9, 0 binomial coefficients n C r = n C n r, binomial coefficients, n C r, binomial series, binomial coefficients, n C r or n, r choosing r objects from n, circle centre at origin, 5 general equation, 6 intersection with line, 7 show line is tangent, 8 tangent equation, 7 cosine rule, 0 cubic equations, 7 differentiation, 5 distance between two points, 5 equations a = b, 4 eponential, factor theorem, 5 choosing a suitable factor, 6 factorials, factorising eamples, functions decreasing, 5 increasing, 5 geometric series finite, 9 infinite, 0 nth term, 9 proof of sum formula, 9 sum of infinite series, 0 sum of n terms, 9 integrals area under curve, 9 definite, 9 integration as anti-differentiation, sum of rectangles, logarithm, change of base, 4 rules of logs, mid point, 5 Pascal s triangle, perpendicular lines, 5 points of infleion, polynomials, long division, radians arc length, 8 area of sector, 8 area of segment, 8 connection between radians and degrees, 8 definition, 8 remainder theorem, 4 sine rule, 9 ambiguous case, 0 stationary points gradients before and after, 7 maima and minima, 5 maimum and minimum problems, 8 second derivative, 6 trapezium rule, 0 trig equations solving, trig functions basic results, 9 graphs, 0 identities, turning points, 5 4 C JUNE 06 SDB