Preparation work for A2 Mathematics [2017]

Transcription

1 Preparatin wrk fr A2 Mathematics [2017] The wrk studied in Y12 after the return frm study leave is frm the Cre 3 mdule f the A2 Mathematics curse. This wrk will nly be reviewed during Year 13, it will nt be retaught. This is t allw time t revise fully the Cre 3 and Cre 4 mdules befre the exams in June f Y13. A level Mathematics gets prgressively harder with each new tpic building n what has been learnt previusly. A significant amunt f Cre 3 and Cre 4 rests n the wrk studied thrughut year 12. It is essential that yu understand this wrk and can apply it t exam style questins; therwise yu will struggle t access large parts f the A2 curse. This piece f wrk is designed t review the key tpics that yu have cvered during year 12 and prepare yu fr the style f questins in the A2 exams. It cnsists f tw parts, Sectin A includes ntes, examples and exercises t help review and cnslidate the skills required t cmplete Sectin B. The answers fr sectin A are included t enable yu t assess yur wn understanding. Sectin B are exam style questins n the tpics met thrughut sectin A. This sectin will be submitted t yur class teacher fr marking at the start f year 13. It is t be cmpleted t a high standard with well-structured and clear slutins. Yu may als need t lk things up frm yur C1 and C2 ntes. T give yurself the best pssible start t the A2 curse I wuld encurage yu t devte the time and effrt needed t d this wrk well. Yu will be required t d crrectins t the questins that yu dn t get right. Sectin B will be due in at the start f Y13. Sectin A 1. Read thrugh the ntes and examples t review the tpics. 2. Cmplete the exercises at the end f each sectin. 3. Mark yur wrk using the slutins at the end. 1. Transfrmatins f Graphs Yu need t knw the basic graphs: y = x 2, y = (x b) 2 + c [ie cmpleted square frm], y = x 3, y = 1 /x, y = sin x, y = cs x, y = tan x and the transfrmatin results belw. Transfrming the graph f y = f(x) y = a f (x) r y /a = f(x) stretch in the y directin scale factr a y = f ( x /a) stretch in the x directin scale factr a y = f (x) + a r y a = f (x) translatin +a in the y directin y = f (x a) translatin +a in the x directin y = f (x) r y = f(x) reflectin in the line y = 0 (x axis) y = f( x) reflectin in the line x = 0 (y axis) Yu need t be able t sketch a transfrmed basic graph and label its intercepts and/r statinary pints Yu als need t be able t describe the transfrmatin that maps a basic graph t a transfrmed graph r vice versa.

2 Eg 1 Quadratic Translatin The riginal graph is y = x 2 It has been translated thrugh [ 2 3 ] y = (x - 2) Replace x with x 2 and y with y 3. The equatin f the transfrmed graph is y 3 = (x 2) 2 y = x 2 r y = (x 2) Eg 2 Trignmetric Translatin The riginal graph is y = sin x It has been translated thrugh [ 45 0 ] y = sin x Replace x with x ( 45) The equatin f the transfrmed graph is y = sin (x 45) y = sin (x + 45) y = sin (x + 45) Eg 3 Trignmetric Stretch [x directin] The riginal graph is y = sin x It has been stretched in the x directin scale factr ½ y = sin 2x Replace x with x ½ The equatin f the transfrmed graph is y = sin (x ½) y = sin x y = sin 2x Eg 4 Trignmetric Stretch [x directin] The riginal graph is y = sin x It has been stretched in the x directin scale factr 2 y = sin x Replace x with x /2 The equatin f the transfrmed graph is y = sin ½ x y = sin ½ x [Nte in this sketch 0 x 720]

3 Eg 5 Trignmetric Stretch [y directin] The riginal graph is y = sin x It has been stretched in the y directin scale factr 2 y = sin x Replace y with y /2 The equatin f the transfrmed graph is y /2 = sin x r y = 2 sin x y = 2sin x Eg 6 Reflectin The riginal graph is y = x 2 It has been reflected in the x axis [Line y = 0] y = x 2 Replace y with - y The equatin f the transfrmed graph is - y = x 2 r y = - x 2 y = - x 2 In Cre 2 when there are multiple transfrmatins they will be independent s the rder that yu d them in des nt matter. Fr example y = sin x translated thrugh [ 45 ] and stretched sf 2 in the y directin 0 Ding the translatin then the stretch gives: first y = sin (x + 45) then y /2 = sin (x + 45) r y = 2 sin (x + 45) Ding the stretch then the translatin gives: first y /2 = sin x r y = 2 sin x then y = 2 sin (x + 45) Hwever it is nt always the case that transfrmatins are independent and this will be lked at in Cre 3.

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6 2. Trignmetry [Equatins] The trignmetry f any angle can be related t the trignmetry f acute angles using the trignmetric graphs. The graph f y = sin x is nly shwn fr -360 t 360 but is defined fr all values f x. Yu can see frm the diagram all f the angles whse trignmetry relates t that f 30 0 including: sin (150) = sin (30) = ½ sin ( 30) = sin (30) = ½ Slving the equatin sin x = ½ fr -360 < x < 360 gives multiple slutins: x = 30 0, x = = x = = x = = The graph f y = cs x is nly shwn fr -360 t 360 but is defined fr all values f x. Yu can see frm the diagram all f the angles whse trignmetry relates t that f 60 0 including: cs (300) = cs (60) = ½ cs ( 120) = cs (60) = ½ Slving the equatin cs x = ½ fr -360 < x < 360 gives multiple slutins: x = 60 0, x = 60 0 x = = x = = The graph f y = tan x is nly shwn fr -360 t 360 but is defined fr all values f x except fr where cs x = 0. Yu can see frm the diagram all f the angles whse trignmetry relates t that f 45 0 including: tan (225) = tan (45) = 1 tan ( 45) = tan (45) = 1 Slving the equatin tan x = 1 fr -360 < x < 360 gives multiple slutins: x = 45 0, x = = x = = x = = Basic Trignmetric Equatins When slving trignmetric equatins fr a given interval the principal slutin is fund using yur calculatr and the ther slutins are fund using the trignmetric graphs as demnstrated in the wrk abve. Eg 1 Slve sin x = ½ fr 0 < x < 360 x = sin -1 (½) = 30 0 [fund by using yur calculatr] The ther slutin in the interval is fund by lking at the shape f the graph: x = = 150 0

7 Trignmetric Identities Yu need t learn the fllwing results (i) tan x = sin x / cs x (ii) sin 2 x + cs 2 x = 1 [nte: sin 2 x means (sin x) 2 t avid cnfusin with sin x 2 which means sin (x 2 )] and hence: sin 2 x = 1 cs 2 x and cs 2 x = 1 sin 2 x Als learn these techniques t use when trying t prve identities: Start with the mst difficult side T turn a sum int a prduct lk t pull ut a cmmn factr eg sin 3 x + sin x cs 2 x = sin x (sin 2 x + cs 2 x) Lk ut fr difference f tw squares 1 cs 2 x = (1 + cs x)(1 cs x) Lk ut fr ne f the brackets f a difference f tw squares in a fractin such as (1 + cs x) and create a DOTS by multiplying numeratr and denminatr by the ther bracket, (1 cs x) T split a single fractin int a sum f terms write each term in the numeratr ver the denminatr sinx+csx Eg = sinx + csx = tanx + 1 csx csx csx T turn a sum f terms int a single fractin write each term ver a cmmn denminatr Eg 1 + cs2 x = sin2 x + cs2 x = sin2 x+cs 2 x = 1 sin 2 x sin 2 x sin 2 x sin 2 x sin 2 x Yu can use 1 = sin 2 x + cs 2 x t replace numbers with trignmetry Multiply numeratr and denminatr by the same term t eliminate fractins within fractins Eg 2 Shw that sin 2 x 1 + cs x 1 cs x Start with ne side and keep manipulating until yu get the ther side. D nt try t d the same t bth sides like when slving equatins. LHS sin 2 x [Use sin 2 x + cs 2 x = 1 rearranged t replace sin 2 x] 1 cs x 1 cs 2 x [Factrise numeratr as DOTS] 1 cs x (1 + cs x) (1 cs x) [Cancel cmmn factr frm numeratr and denminatr] 1 cs x 1 + cs x RHS Eg 3 Shw that (sin x + cs x) 2-1 tan x 2cs 2 x LHS sin 2 x + 2 sin x cs x + cs 2 x 1 [Use sin 2 x + cs 2 x = 1] 2cs 2 x sin x cs x 1 [Simplify the numeratr by cancelling the 1s] 2 cs 2 x 2 sin x cs x [Cancel cmmn factrs frm numeratr and denminatr] 2 cs 2 x sin x [Use tan x = sin x / cs x] cs x tan x RHS

8 Quadratic Trignmetric Equatins The first slutin t an equatin is fund using yur calculatr and the ther slutins in the given interval are fund by lking at the shape f the graph. Quadratic Equatins may invlve difference f tw square and cmmn factrs as well as standard factrisatin. Eg 4 Slve 4 sin 2 x 1 = 0 fr 0 x < Factrise by difference f tw squares: 4 sin 2 x 1 = 0 (2 sin x 1) ( 2 sin x + 1) = 0 sin x = ½, sin x = - ½ sin x = ½ gives: x = sin -1 ( ½) = 30 0 and x = = sin x = - ½ gives: x = sin -1 (- ½) = which is utside the interval s x = = and x = = Eg 5 Slve 2sin x cs x + sin x = 0 fr 0 < x < D NOT divide by sin x as this causes yu t lse slutins! Instead factrise by cmmn factr: 2sin x cs x + sin x = 0 sin x (2 cs x + 1) = 0 sin x = 0, cs x = - ½ sin x = 0 gives: x = [Nte 0 0 and are NOT in the interval] cs x = - ½ gives: x = cs -1 (- ½) = and x = = T slve an equatin invlving tw different trignmetric functins which CANNOT be factrised by cmmn factr [like in example 5] yu ften need t use trignmetric identities Eg 6 Slve 2sin x 3cs x = 0 fr 0 < x < Give yur answers t 3sf. Rearrange and use the identity tan x = sin x / cs x: 2 sin x 3 cs x = 0 2 sin x = 3 cs x sin x = 3, hence: tan x = 3/2 cs x x = tan -1 ( 3 /2) gives: x = (3sf) and x = = (3sf) Eg 7 Slve sin 2 x = 2 (cs x 1) fr 0 x Substitute sin 2 x = 1 cs 2 x t create a quadratic equatin in cs x: 1 cs 2 x = 2 cs x 2 cs 2 x + 2 cs x 3 = 0 (cs x 1) (cs x + 3) = 0 cs x = 1, cs x = 3 cs x = 1 has slutins x = 0 and x = 360 [frm the graph - nte they ARE in the interval fr this questin]. cs x = 3 has n slutins as -1 cs x 1 fr all x.

9 Trignmetric Equatins invlving a substitutin and a change f interval These equatins cntain a multiple f x ie sin 3x = ½ r sin (2x 20) = 0.5 A basic equatin like sin x = ½ has tw slutins in the interval is 0 x 360. A questin invlving 2x usually has 4 slutins in this interval. A questin invlving 3x usually has 6 slutins in this interval. Learn the strategy fr slving these type f questins: sin (2x 20) = 0.5 fr 0 x (i) Let u = 2x 20 s the questin becmes sin u = 0.5 (ii) The interval fr x is 0 x s the interval fr u becmes 2(0) 20 u 2(360) Ie -20 u (iii) Draw the trig graph y = sin u fr -20 u (iv) Slve sin u = 0.5 using yur calculatr and find all f the ther slutins fr u frm yur graph (v) u = 2x 20 s x = (u + 20)/2 Use the substitutin t change all f yur slutins fr u int slutins fr x Eg 8 Slve sin (2x 20) = 0.5 fr 0 x Let u = 2x 20 Hence sin u = 0.5 fr -20 u sin u = 0.5, u = sin -1 (0.5) = 30 0 Frm the graph there are n slutins fr u in the interval -20 t 0 but there are 3 mre slutins in the interval 0 t 700: u = = u = = u = = Using the substitutin: u = 2x 20 we have x = (u + 20)/2 Hence: u = 30 u = 150 u = 390 u = 510 x = ( ) / 2 = 25 x = ( ) / 2 = 85 x = ( ) / 2 = 205 x = ( ) / 2 = 265 Final answers: x = 25, 85, 205 0, 265 Nte: if yur answers fr u are nt whle numbers yu shuld nt rund them as ptentially this can give yu slightly incrrect answers fr x.

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11 3. Differentiatin Calculating Gradients Tangent t the curve y = f(x) at the pint P P Nrmal t the curve y = f(x) at the pint P y = f(x) The gradient f a curve at a pint P n the curve is the gradient f the tangent t the curve at that pint. Differentiatin is the prcess f calculating the gradient functin which is a rule t find the gradient at any pint n the curve. The ntatin fr gradient is dy /dx. r f (x). The rules fr differentiating are: Fr Any cnstant differentiates t 0. n y kx dy n 1 nkx fr n being any ratinal number. dx The gradient at a specific pint is fund by substituting the x crdinate f the pint int the gradient functin. Yu always differentiate term by term s yu must expand any brackets r break up any fractins [by writing each term ver the denminatr] prir t differentiating Eg 1 ( x 4)2 y = 2 x Find the gradient f the curve at the pint where x = 4. ( x 4)2 y = 2 x = x 8 x+16 2x 1 = x x 1 2 dy = 1 dx 4 x x 3 2 = x ( x) 3 When x = 4 dy /dx = = 3 8 Yu culd be given a functin and the value f the gradient at a specific pint and wrk back twards the pints n the curve. Eg 2 Find the crdinates f the pint n the curve y = 1 x2 + 5 where the gradient = - ¼ y = x dy /dx = -2x -3 Hence -2x -3 = - ¼ 2 x 3 8 = x 3 2 = x = 1 4 When x = 2 y = 1 x2 + 5 = ¼ + 5. Hence the required crdinates are (2, 5¼)

12 Tangents and Nrmals If P = (a, b) is a pint n the curve y = f(x) then the gradient f the tangent at pint P is the same as the gradient f the curve at P i.e. f (a). The equatin f the tangent t the curve at the pint P is y b = f (a) (x a) The nrmal at P is perpendicular t the tangent s its gradient = -1 /f (a) The equatin f the nrmal t the curve at the pint P is y b = -1 /f (a) (x a) Eg 3 Find the equatin f the nrmal t the curve y = When x = 4, y = (2 4)2 4 = 1 s (a, b) = (4, 1) ( x 4)2 2 x at the pint where x = 4. Frm eg 1: the gradient f the curve y = = ( x 4)2 2 x at x = 4 was 3 /8 Hence the gradient f the nrmal = 8 /3 The equatin f the nrmal is: y 1 = 8 /3 (x 4) Statinary Pints Statinary [r turning] pints ccur where f (x) = 0. T find the x crdinates f statinary pints, slve the equatin f (x) = 0. Yu can find the crrespnding y crdinates using the equatin f the curve y = f(x). A statinary pint may be a [lcal] maximum, a [lcal] minimum r a pint f inflectin. The nature f a statinary pint can be determined by using the secnd derivative d2 y 2 f f (x) dx At statinary pint (a,b) ie where x = a If f (a) > 0 the pint is a minimum If f (a) < 0 the pint is a maximum If f (a) = 0 then abandn this methd and use the methd belw instead The nature f a statinary pint can als be determined by cnsidering the value f the gradient, dy /dx, either side f the statinary pint MAX MIN POINTS OF INFLECTION Eg 7 f(x) = 2(x+3). Find the crdinates f the statinary pint and determine its nature. x f(x) = 2x ½ + 6x ½ f (x) = x 1 2 3x 3 2 = 1 3 x x x At statinary pints f (x) = 0 hence Multiplying thrugh by x x gives: When x = 3, y = 12 = 12 3 = = 0 x x x The crdinates f the statinary pint are (3, 4 3) x 3 = 0 Hence x = 3 is the x crdinate f the statinary pint f (x) = 1 2 x x 5 2 = x x 2x 2 x When x = 3 f (x) = = = 1 > 0 Hence (3, 4 3) is a minimum pint

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14 Intrducing Integratin 4. Integratin Integratin is the reverse f differentiatin it can be used t find a functin frm its gradient functin. The diagram shws the curves y = x 2 and y = x These functins bth have the same gradient functin f (x) = 2x. There are infinitely many curves with gradient functin = 2x but they are all vertical translatins f the graph f y = x 2. This grup f functins can be described as y = x 2 + c where c is a cnstant. When integrating we need t indicate +c t shw all f the pssible functins. If we knw a pint n the curve then we can find the value f c and hence ne specific functin. If dy/dx = f (x) then we say y = T integrate: If f ( x) dx f ( x) c where c is the cnstant f integratin. n 1 n y kx then kx y dx c fr n being any ratinal number except -1. n 1 Yu always integrate term by term s yu must expand any brackets r break up any fractins [by writing each term ver the denminatr] prir t integrating Eg 1 If f (x) = x (5x 6) and (2, 3) lies n the curve find the equatin f the curve y = f(x) y = x(5x 6) dx y = 5x 3 2 6x 1 2 dx y = 5x x c 2 y = 2x 5 2 4x c When x = 2, y = 3 s... 3 = 2 ( 2) 5 4( 2) 3 + c 3 = c Hence y = 2x 5 2 4x = c hence c = 3 Definite Integratin & Area under a curve b Definite integratin = f ( x) dx = [g(x)] b = g(b) g(a) = a single value a a where g(x) is the functin yu get when f(x) is integrated. There is n cnstant, c, here. The area f the regin bunded by the curve y = f (x), the rdinates x = a and x= b and the x axis can be fund by definite integratin: Area = b f ( x) dx when it exists. a

15 Eg 2 Find the area bunded by the x axis, the curve y = 12 x 5 x2 and the rdinates x = 1 and x = x 1 2 5x 2 dx = [ 12x x 1 ] = [8x ] 2 x 1 R = [8( 2) ] [8( 1) ] = [ /2] [13] = ½ [square units] Integratin will prduce a negative answer fr areas belw the x axis. Eg 3 Find the area bunded by the x axis, the curve y = 1 4 x2 and the rdinates x = 1 and x = x 2 dx = [x 4x 1 ] R = [x + 4 x ] 2 1 = [2 + 2] [1 + 4] = 4 5 = -1 The minus sign indicates the area is belw the x axis The area is 1 square unit. Areas partly abve and belw the x axis shuld be integrated separately when integrating with respect t x. Eg 4 Find the area bunded by the x axis, the curve y = 1 4 x2 and the rdinates x = 1 and x = 5. T find the x intercepts slve y = 0: 0 = 1 4 s 4 = 1, x 2 x 2 x2 = 4 and x = ± 2 The graph is belw the x axis fr 1 x < 2 and abve fr 2 < x 5 5 Area abve x axis = 1 4x 2 dx 2 = [x + 4 x ] 5 2 = [ ] [2 + 2] = 14 5 Frm eg (3) Area belw the x axis = 1, hence the area f the regin is The area between tw graphs T find the area f the finite regin bunded by tw curves y = f(x) and y = g(x) yu first slve their equatins simultaneusly t find the limits f integratin. b Assuming y = f(x) is abve y = g(x) between the tw limits the area wuld be A = f(x)dx As the limits f the integrals are the same this can be cmbined int A = f(x) g(x)dx b a a b g(x)dx a Where ne f the functins is a straight line yu can find the area bunded by the line, the x axis and the x rdinates using the area f a standard shape such as a triangle r trapezium. Area f trapezium = ½ (a+b)h (a and b are the lengths f the parallel sides & h is the perpendicular height)

16 Eg 5 The diagram shws the curve y = 4x ½ and the line 3y = 4x + 8. Find the area f the finite regin bunded by the line and the curve [marked R]. Yu need t slve the equatins y = 4x ½ and 3y = 4x + 8 simultaneusly t find the limits. 3(4x ½ ) = 4x x ½ = 4x + 8 3x ½ = x + 2 This is a disguised quadratic.let p = x ½ 3p = p p 2 3p + 2 = 0 (p 2)(p 1) = 0 p = 2 and p = 1 As p = x ½, x = p 2 Hence x = 4 and x = 1 R Integrating the curve between x = 1 and x = 4 gives the area under the curve. Regin R is fund by finding the area f a trapezium subtracted frm the area under the curve. Fr the trapezium: Finding a: When x = 1 y = 4x ½ = 4 Finding b: When x = 4 y = 4x ½ = 8 Finding h: h = 4 1 = 3 Area f trapezium = ½ (4 + 8) x 3 = 18 Fr the area under the curve: 4 A = 4x ½ dx = [ 4x3 2 ] 4 1 = [8( x)3 ] = [64] 3 [8] Hence area f R = = 2 3 f a square unit = 56 = Alternative appraches t example 5: Finding the area f the trapezium by integratin 4 1 Area f the trapezium = 4 3 x dx= [2x2 3 +8x 3 ]4 1 =[ ] [ ]=64 10 =18 3 Area under the curve = 4x ½ dx = [ 4x3 2 ] 4 1 = [8( x)3 ] = [64] 3 [8] 1 3 Hence area f R = = 2 3 f a square unit 3 2 = 56 = Or by cmbining the integratins int ne [curve line as curve is abve the line] t give 4 1 Area f R = 4x ½ 4 3 x 8 3 dx = [4x x2 8x ] = [8( x) 3 = [ ] [8 2 8 ] = x2 3 8x 3 ] 4 1

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18 5. Expnentials & Lgarithms The expnential functin and the natural lgarithm functin. An expnential functin is ne f the frm y = a x where a is a cnstant. Expnential graphs all have the same shape and all pass thrugh (0,1). Any expnential functin is greater than 0 fr all x. y = e x y = In x y = e x is called the expnential functin that has gradient defined t be equal t 1 at the pint (0, 1). This prperty allws y = e x t be differentiated exactly. e = an irratinal number y = ln x = lge x is the natural lgarithm functin. It is the inverse f e x. The graph is a reflectin f the graph f y=e x in the line y=x Laws fr Lgarithms a c = b is identical and interchangeable with c = lga b lga xy = lga x + lga y lga(x/y) = lga x lga y lga x n = nlgax lgaa = 1 lga1 =0 Nte: Because lga x + lga y is equal t lga xy it means it cannt be equal t lga (x + y) Hence lga (x + y) des NOT equal lga x + lga y and yu CANNOT expand a lg like yu wuld a bracket. Expnentials and lgarithms as inverse functins e x and In x are inverse functins in the same way that x 2 and x are inverses. (x 2 ) and ( x) 2 cancel t give x. In the same way e In x and In [e x ] cancel t give x.

19 When yu apply a functin t an equatin yu apply it t whle sides nt individual terms. Eg y = x + 4 Square each side - nt each individual term [ y] 2 = [x + 4] 2 In x = In y + 3 e In x = e In y + 3 e Inx = e In y e 3 x = ye 3 In x = In y + 3 In x In y = 3 In [ x /y] = 3 e In [x/y] = e 3 x /y = e 3 x = ye 3 Apply e t each side nt each individual term Use laws fr indices t break up the right hand side Alternatively...Cllect the lgs n ne side Use laws fr lgs t cmbine int a single lg Apply e t each side Yu cannt cancel e and ln if there is smething in the way e 2In y desn t cancel as 2 is in the way! EITHER x = 2 In y Pull the 2 inside the ln y OR x = 2 In y divide by 2 at the start x = In y 2 then apply e t each side x /2 = In y then apply e t each side e x = y 2 y = e ½x y = [e x ] ½ = e ½x Finally expanding brackets wrks when yu multiply a bracket by a value. Applying a functin t a bracket is NOT multiplying the bracket by that functin s yu cannt multiply ut functins like yu wuld expand brackets... Eg e (a + b) is e a e b using laws fr lgs. Yu cannt expand the bracket t get e a + e b. In (a + b) is NOT In a + In b in the same way that (a + b) is NOT a + b! Slving expnential equatins Basic type 2 3 x = 7 x + 2 In [2 3 x ] = In [7 x + 2 ] Apply lgs t bth sides (3 x) In 2 = (x + 2) In 7 Use laws fr lgs t bring ut the pwer 3 In 2 x In 2 = x In In 7 Expand and islate the terms in x 3 In 2 2 In 7 = x In7 + x In 2 Take x ut as a cmmn factr 3 In 2 2 In 7 = x (In7 + In 2) Divide thrugh t find x. 3 In 2 2 In 7 = x (In7 + In 2) Quadratic type 2 2x+ 3 10(2 x ) + 3 = 0 Express 2 2x + 3 in terms f 2 x (2 x ) 2 x (2 x ) + 3 = 0 Let y = 2 x and write as a quadratic 8y 2 10y + 3 = 0 Factrise & slve fr y (4y 3)(2y 1) = 0 y = ¾, y = ½ Remember y = 2 x s slve fr x using lgs 2 x = ¾, 2 x = ½ r inspectin if pssible In 2 x = In [¾] x = -1 x In 2 = In [¾] x = In [¾] / In 2 Disguised quadratic 8(2 x ) + 3(2 -x ) = x = 1 /2 x then multiply thrugh t give the exact same quadratic as the example abve.

20 6. Trignmetry (Cre 3) Inverse Trignmetric Functins An inverse functin exists nly if the riginal functin is ne t ne. The range f a functin is the dmain f the inverse functin. The dmain f a functin is the range f the inverse functin. The definitins f the inverse trig functins arc sin, arc cs and arc tan written as sin -1, cs -1 and tan -1 as the inverse functins f sin,cs and tan respectively. The graph f y = f -1 (x) is a reflectin f the graph y = f(x) in the line y = x. The graphs f the inverse trig functins can be fund by reflecting the riginal trig graphs with a restricted dmain in the line y = x. Dmain and range f the inverse functins sin -1 x has dmain -1 x 1 and range π /2 sin -1 x π /2 y = arcsin x P = (1, π /2) Q = (-1, - π /2) cs -1 x has dmain -1 x 1 and range 0 cs -1 x π Q = (-1, π) y = arccs x P = (1, 0)

21 tan -1 x has dmain all real numbers and range π /2 tan -1 x π /2 y = π / 2 y = arctan x y =- π / 2 Reciprcal Trignmetric Functins The definitins f the reciprcal trig functins csecant, secant and ctangent as. 1 cs ec sin 1 sec cs ct 1 tan These are nt the same as the inverse functins! y = sec x = 1 / cs x Nte: When cs x = 1, sec x is 1/1 = 1. When cs x = 0, sec x is 1/0 which is undefined s yu get an asymptte. When cs x gets smaller, sec x gets bigger

22 y = csec x = 1 / sin x Nte: When sin x = 1, csec x is 1/1 = 1. When sin x = 0, csec x is 1/0 which is undefined s yu get an asymptte. When sin x gets smaller, csec x gets bigger y = ct x = 1 / tan x Nte: When tan x = 0, ct x is 1/0 which is undefined s yu get an asymptte. Where tan x has an asymptte, ct x = 0. When tan x gets bigger, ct x gets smaller Identities Revisin f C2 identities: tan x = sin x / cs x (*) sin 2 x + cs 2 x =1 (**) New identities ct x = cs x / sin x (reciprcal f *) 1 + ct 2 x = csec 2 x (divide ** by sin 2 x) tan 2 x + 1 = sec 2 x (divide ** by cs 2 x) Techniques fr identities [hints and tips]

23 Start with the mst difficult side T turn a sum int a prduct lk t pull ut a cmmn factr eg sin 3 x + sinxcs 2 x = sinx (sin 2 x + cs 2 x) Lk ut fr difference f tw squares 1 cs 2 x = (1 + cs x)(1 cs x) Lk ut fr ne f the brackets f a difference f tw squares in a fractin such as (1 + cs x) and create a DOTS by multiplying numeratr and denminatr by the ther bracket, (1 cs x) T split a single fractin int a sum f terms write each term in the numeratr ver the denminatr sinx+csx Eg = sinx + csx = 1 + ctx sin x sin x sinx T turn a sum f terms int a single fractin write each term ver a cmmn denminatr Eg 1 + cs2 x = sin2 x + cs2 x = sin2 x+cs 2 x = 1 = sin 2 x sin 2 x sin 2 x sin 2 x sin 2 x csec2 x Multiply numeratr and denminatr by the same term t eliminate fractins within fractins Mst C3 identities are easily dne using the 3 new identities. If everything yu ve tried hasn t wrked turn everything int sin and cs and use C2 identities Slving Equatins Yu need t knw the exact results fr the trignmetry f the angles π /6 (30 ), π /3 (60 ) and π /4 (45 0 ) Equatins are mst likely quadratic in tw different trig functins that require sme f the new trig functins t rearrange r are ging t be questins that invlve a change f interval. Eg: 2csec 2 x + ct x 3 = 0 0 < x < 2π 3 csec (2x + π /6) 2 = 0 0 < x < 2π 2 (1 + ct 2 x) + ct x 3 = 0 csec y = 2 / 3 π /6 < y < 4π + π /6 2ct 2 x + ct x 1 = 0 sin y = 3 /2 (2ct x + 1)(ct x 1) = 0 Find all slutins fr y in the interval ct x = 1 ct x = - ½ π /6 < y < 4π + π /6 using the extended graph f sin tan x = 1 tan x = -2 y = π /3, 2π /3, 7π /3, 8π /3 x = π /4 x = (utside interval) Then rearrange t find x = (y - π /6) / 2 x = 5π /4 x = 2.03 (3sf) x = π /12, π /4, 13π /12, 5π /4 x = 5.18 (3sf) [Nte: as 2x and the interval is 0 < x < 2π yu expect 4 answers] Lk ut fr related questins eg slve 2csec 2 2x + ct 2x 3 = 0 0 < x < This is the same equatin as the first equatin abve except that it has 2x in place f x and the interval has halved [s yu expect nly 2 answers] and is in degrees. The slutins t this equatin are the same as the slutin t the first equatin except yu qute yur answers in degrees.

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25 Questin 1: Transfrmatins (Cre 2) Sectin B: Assessment The graph abve shws a sketch f the curve with equatin y = f(x). The curve crsses the crdinate axes at the pints (-1,0), (0,-1) and (3,0). The minimum pint is (1, -3). On separate diagrams sketch the curve with equatin: a) y = f(x + 1) [3] b) y = f( 1 x) [3] 2 On each diagram, shw clearly the crdinates f the minimum pint, and each f pint at which the curve crsses the crdinate axes. Questin 2: Trignmetry (Cre 2) a) Shw that the equatin can be written as 3sin 2 x = 5 cs x + 1 3cs 2 x + 5cs x 2 = 0 [2] b) Hence slve, fr 0 x < 360, the equatin 3sin 2 x = 5 cs x + 1, giving yur answers t 1 decimal place where apprpriate. [5] Questin 3: Integratin (Cre 2) The line with equatin y = 6 x cuts the curve y = 4 + 2x x 2 at the pints P and Q, as shwn. a) Find the crdinates f P and Q. [5] b) Find the area f the shaded regin between the line and the curve as shwn in the diagram. [7]

26 Questin 4: Integratin and Differentiatin (Cre 1 & 2) The curve C with equatin y = f(x) is such that dy dx = 4x + 4 x, x > 0 a) Shw that, when x = 2, the exact value f dy is [3] dx The curve C passes thrugh the pint (4,50). b) Find f(x). [6] Questin 5: Differentiatin (Cre 1) The curve C has the equatin y = 6 1 2x x, x > 0. x2 a) Write x x in the frm x n, where n is a fractin. [1] b) Find dy dx [3] c) Find an equatin f the nrmal t the curve C at the pint n the curve where x = 1. [4] d) Find d2 y dx2. [2] e) Deduce that the curve C has n minimum pints. [2] Questin 6: Expnentials and Lgarithms (Cre 3) Slve each equatin, giving yur answers in exact frm. a) ln(4x + 1) = 2 [3] b) 3e x + 2e -x = 7 [5] Questin 7: Expnentials (Cre 3) a) Sketch the graph y = e ax + b, given a and b > 0. Mark the crdinates f the pint where the graph meets either the x-axis r the y-axis. [2] b) Given that when x = 0, y = 4, find the exact value f b. [2] Questin 8: Trignmetry (Cre 3) a) Sketch the graph f y = sec x fr x in the interval 0 x 2π. Shw n yur sketch the crdinates f any maximum and minimum pints and the equatins f any asympttes. [2] b) Find the x crdinates f the pints where the graph f y = 2 + sec (x π ) crsses the x-axis. [5] 6 Questin 9: Trignmetry (Cre 3) a) Express tan 2 x 1 in terms f sec x. [3] cs x b) Hence slve, fr 0< x < 360, the equatin tan 2 x 1 cs x = 4 [4]

27 Questin 10: Trignmetry (Cre 3) Fig. 1 Fig. 2 Fig. 3 Each diagram abve shws part f a curve, the equatin f which is ne f the fllwing: y = sin 1 x, y = cs 1 x, y = tan 1 x, y = sec x, y = csec x, y = ct x. State which equatin crrespnds t a) Fig. 1 [1] b) Fig. 2 [1] c) Fig. 3 [1] Ttal marks = 75

28 Sectin A: Answers 1. Transfrmatins

29 2. Trignmetry

30 3. Differentiatin 4. Integratin

31 5. Expnentials and Lgarithm 6. Trignmetry (Cre 3)

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