Preparatin wrk fr A Mathematics [018] The wrk studied in Y1 will frm the fundatins n which will build upn in Year 13. It will nly be reviewed during Year 13, it will nt be retaught. This is t allw time t revise fully 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 the Year 13 cntent rests n the wrk studied thrughut year 1. 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 Year 13 curse. This piece f wrk is designed t review the key tpics that yu have cvered during year 1 and prepare yu fr the style f questins in the final 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 Year 1 ntes. T give yurself the best pssible start t Year 13 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.. 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, y = (x b) + 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. Eg 1 Quadratic Translatin The riginal graph is y = x It has been translated thrugh [ 3 ] y = (x - ) + 3 Replace x with x and y with y 3. The equatin f the transfrmed graph is y 3 = (x ) y = x r y = (x ) + 3 Eg 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 x Replace x with x ½ The equatin f the transfrmed graph is y = sin (x ½) y = sin x y = sin x Eg 4 Trignmetric Stretch [x directin] The riginal graph is y = sin x It has been stretched in the x directin scale factr y = sin x Replace x with x / The equatin f the transfrmed graph is y = sin ½ x y = sin ½ x [Nte in this sketch 0 x 70] Eg 5 Trignmetric Stretch [y directin] The riginal graph is y = sin x It has been stretched in the y directin scale factr y = sin x Replace y with y / The equatin f the transfrmed graph is y / = sin x r y = sin x y = sin x Eg 6 Reflectin The riginal graph is y = x It has been reflected in the x axis [Line y = 0] y = x Replace y with - y The equatin f the transfrmed graph is - y = x r y = - x y = - x
In Cre 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 in the y directin 0 Ding the translatin then the stretch gives: first y = sin (x + 45) then y / = sin (x + 45) r y = sin (x + 45) Ding the stretch then the translatin gives: first y / = sin x r y = sin x then y = sin (x + 45) Hwever it is nt always the case that transfrmatins are independent and this will be lked at in Cre 3.
. 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 = 180 30 = 10 0 x = 180 30 = 150 0 x = 360 + 30 = 330 0 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 ( 10) = cs (60) = ½ Slving the equatin cs x = ½ fr -360 < x < 360 gives multiple slutins: x = 60 0, x = 60 0 x = 360 60 = 300 0 x = 360 + 60 = 300 0 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 (5) = 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 = 45 180 = 135 0 x = 45 + 180 = 5 0 x = 135 180 = 315 0 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 = 180 30 = 150 0
Trignmetric Identities Yu need t learn the fllwing results (i) tan x = sin x / cs x (ii) sin x + cs x = 1 [nte: sin x means (sin x) t avid cnfusin with sin x which means sin (x )] and hence: sin x = 1 cs x and cs x = 1 sin 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 x = sin x (sin x + cs x) Lk ut fr difference f tw squares 1 cs 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 + cs x = sin x + cs x = sin x+cs x = 1 sin x sin x sin x sin x sin x Yu can use 1 = sin x + cs x t replace numbers with trignmetry Multiply numeratr and denminatr by the same term t eliminate fractins within fractins Eg Shw that sin 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 x [Use sin x + cs x = 1 rearranged t replace sin x] 1 cs x 1 cs 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) - 1 tan x cs x LHS sin x + sin x cs x + cs x 1 [Use sin x + cs x = 1] cs x 1 + sin x cs x 1 [Simplify the numeratr by cancelling the 1s] cs x sin x cs x [Cancel cmmn factrs frm numeratr and denminatr] cs x sin x [Use tan x = sin x / cs x] cs x tan x RHS
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 x 1 = 0 fr 0 x < 360 0 Factrise by difference f tw squares: 4 sin x 1 = 0 ( sin x 1) ( sin x + 1) = 0 sin x = ½, sin x = - ½ sin x = ½ gives: x = sin -1 ( ½) = 30 0 and x = 180 30 = 150 0 sin x = - ½ gives: x = sin -1 (- ½) = -30 0 which is utside the interval s x = 180 + 30 = 10 0 and x = 360 30 = 330 0 Eg 5 Slve sin x cs x + sin x = 0 fr 0 < x < 360 0 D NOT divide by sin x as this causes yu t lse slutins! Instead factrise by cmmn factr: sin x cs x + sin x = 0 sin x ( cs x + 1) = 0 sin x = 0, cs x = - ½ sin x = 0 gives: x = 180 0 [Nte 0 0 and 360 0 are NOT in the interval] cs x = - ½ gives: x = cs -1 (- ½) = 10 0 and x = 360 10 = 40 0 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 sin x 3cs x = 0 fr 0 < x < 360 0 Give yur answers t 3sf. Rearrange and use the identity tan x = sin x / cs x: sin x 3 cs x = 0 sin x = 3 cs x sin x = 3, hence: tan x = 3/ cs x 180 0 x = tan -1 ( 3 /) gives: x = 56.3 0 (3sf) and x = 180 + 56.3099 = 36 0 (3sf) Eg 7 Slve sin x = (cs x 1) fr 0 x 360 0 Substitute sin x = 1 cs x t create a quadratic equatin in cs x: 1 cs x = cs x cs x + 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.
Trignmetric Equatins invlving a substitutin and a change f interval These equatins cntain a multiple f x ie sin 3x = ½ r sin (x 0) = 0.5 A basic equatin like sin x = ½ has tw slutins in the interval is 0 x 360. A questin invlving x 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 (x 0) = 0.5 fr 0 x 360 0 (i) Let u = x 0 s the questin becmes sin u = 0.5 (ii) The interval fr x is 0 x 360 0 s the interval fr u becmes (0) 0 u (360) - 0 0 Ie -0 u 700 0 (iii) Draw the trig graph y = sin u fr -0 u 700 0 (iv) Slve sin u = 0.5 using yur calculatr and find all f the ther slutins fr u frm yur graph (v) u = x 0 s x = (u + 0)/ Use the substitutin t change all f yur slutins fr u int slutins fr x Eg 8 Slve sin (x 0) = 0.5 fr 0 x 360 0 Let u = x 0 Hence sin u = 0.5 fr -0 u 700 0 540 70 sin u = 0.5, u = sin -1 (0.5) = 30 0 Frm the graph there are n slutins fr u in the interval -0 t 0 but there are 3 mre slutins in the interval 0 t 700: u = 180 30 = 150 0 u = 360 + 30 = 390 0 u = 540 30 = 510 0 Using the substitutin: u = x 0 we have x = (u + 0)/ Hence: u = 30 u = 150 u = 390 u = 510 x = (30 + 0) / = 5 x = (150 + 0) / = 85 x = (390 + 0) / = 05 x = (510 + 0) / = 65 Final answers: x = 5, 85, 05 0, 65 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.
180 degrees 360 30 45 360 60
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 dx nkx n 1 fr n being any ratinal number. 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) y = x Find the gradient f the curve at the pint where x = 4. ( x 4) y = x = x 8 x+16 x 1 = 1 x1 4 + 8x 1 dy = 1 dx 4 x 1 + 0 4x 3 = 1 4 4 x ( x) 3 When x = 4 dy /dx = 1 8 4 8 = 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 Find the crdinates f the pint n the curve y = 1 x + 5 where the gradient = - ¼ y = x - + 5 dy /dx = -x -3 Hence -x -3 = - ¼ x 3 8 = x 3 = x = 1 4 When x = y = 1 x + 5 = ¼ + 5. Hence the required crdinates are (, 5¼)
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 = ( 4) 4 = 1 s (a, b) = (4, 1) ( x 4) x at the pint where x = 4. Frm eg 1: the gradient f the curve y = = ( x 4) 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 d y 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) = (x+3). Find the crdinates f the statinary pint and determine its nature. x f(x) = x ½ + 6x ½ f (x) = x 1 3x 3 = 1 3 x x x At statinary pints f (x) = 0 hence Multiplying thrugh by x x gives: When x = 3, y = 1 = 1 3 = 4 3 3 3 1 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 x 3 + 9 x 5 = 1 + 9 x x x x When x = 3 f (x) = 1 + 9 = 1 + 3 = 1 > 0 Hence (3, 4 3) is a minimum pint 6 3 18 3 6 3 6 3 3 3
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 and y = x + 3. These functins bth have the same gradient functin f (x) = x. There are infinitely many curves with gradient functin = x but they are all vertical translatins f the graph f y = x. This grup f functins can be described as y = x + 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 (, 3) lies n the curve find the equatin f the curve y = f(x) y = x(5x 6) dx y = 5x 3 6x 1 dx y = 5x5 5 6x3 3 + c y = x 5 4x 3 + c When x =, y = 3 s... 3 = ( ) 5 4( ) 3 + c 3 = 4 4 + c Hence y = x 5 4x 3 + 3 3 = 8 8 + 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
Eg Find the area bunded by the x axis, the curve y = 1 x 5 x and the rdinates x = 1 and x =. 1 1x 1 5x dx = [ 1x3 3 5x 1 ] 1 1 = [8x 3 + 5 ] x 1 R = [8( ) 3 + 5 ] [8( 1)3 + 5 1 ] = [16 + 5 /] [13] = 16 10½ [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 x and the rdinates x = 1 and x =. 1 4x dx = [x 4x 1 ] 1 1 1 R = [x + 4 x ] 1 = [ + ] [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 x and the rdinates x = 1 and x = 5. T find the x intercepts slve y = 0: 0 = 1 4 s 4 = 1, x x x = 4 and x = ± The graph is belw the x axis fr 1 x < and abve fr < x 5 5 Area abve x axis = 1 4x dx = [x + 4 x ] 5 = [5 + 4 5 ] [ + ] = 14 5 Frm eg (3) Area belw the x axis = 1, hence the area f the regin is 4 5 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)
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 + 8 1x ½ = 4x + 8 3x ½ = x + This is a disguised quadratic.let p = x ½ 3p = p + p 3p + = 0 (p )(p 1) = 0 p = and p = 1 As p = x ½, x = p 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 ] 4 1 = [8( x)3 ] 4 3 1 = [64] 3 [8] 1 3 Hence area f R = 18 3 18 = 3 f a square unit 3 = 56 = 18 3 3 Alternative appraches t example 5: Finding the area f the trapezium by integratin 4 1 Area f the trapezium = 4 3 x+8 3 4 dx= [x 3 +8x 3 ]4 1 =[3 3 +3 3 ] [ 3 +8 3 ]=64 10 =18 3 Area under the curve = 4x ½ dx = [ 4x3 ] 4 1 = [8( x)3 ] 4 3 1 = [64] 3 [8] 1 3 Hence area f R = 18 3 18 = 3 f a square unit 3 = 56 = 18 3 3 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 = [4x3 3 x 8x ] 4 3 3 1 = [8( x) 3 = [ 64 3 3 ] 3 3 3 [8 8 ] = 3 3 3 3 3 x 3 8x 3 ] 4 1
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.7188188.. 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 and x are inverses. (x ) and ( x) cancel t give x. In the same way e In x and In [e x ] cancel t give x.
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] = [x + 4] 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 In y desn t cancel as is in the way! EITHER x = In y Pull the inside the ln y OR x = In y divide by at the start x = In y then apply e t each side x / = In y then apply e t each side e x = y 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 3 x = 7 x + In [ 3 x ] = In [7 x + ] Apply lgs t bth sides (3 x) In = (x + ) In 7 Use laws fr lgs t bring ut the pwer 3 In x In = x In 7 + In 7 Expand and islate the terms in x 3 In In 7 = x In7 + x In Take x ut as a cmmn factr 3 In In 7 = x (In7 + In ) Divide thrugh t find x. 3 In In 7 = x (In7 + In ) Quadratic type x+ 3 10( x ) + 3 = 0 Express x + 3 in terms f x ( x ) x 3 10( x ) + 3 = 0 Let y = x and write as a quadratic 8y 10y + 3 = 0 Factrise & slve fr y (4y 3)(y 1) = 0 y = ¾, y = ½ Remember y = x s slve fr x using lgs x = ¾, x = ½ r inspectin if pssible In x = In [¾] x = -1 x In = In [¾] x = In [¾] / In Disguised quadratic 8( x ) + 3( -x ) = 10 -x = 1 / x then multiply thrugh t give the exact same quadratic as the example abve.
6. Functins Intrductin t functins and mappings. A functin maps an element f the dmain (x) t an element f the range (y) s y = f(x). A functin can be 1 t 1...such as f(x) = x + 3 which maps each x value t ne y value It can als be many t 1...such as y = x where different x values can map t the same y value. A functin cannt be ne t many...where an x value maps t mre than 1 y value. Dmain and range The dmain is the set f x values that the functin maps t y values. The dmain is expressed in terms f x. The range is the set f y values that the functin maps t. The range is expressed in terms f y r f(x). Eg f(x) = (x + 1) - 3 range dmain The dmain f the functin is x є IR [x is any element f the real numbers] The range is f(x) є IR, f(x) -3 When the dmain fr a functin is restricted the range des nt necessarily cme frm using just the end pints f its dmain. Eg: f(x) = (x + 1) 3 is defined fr - < x < 1 Using the end pints... f(-) = - and f(1) = 1. This might make yu think the range is - < f(x) < 1 but this is wrng! If yu lk at the graph (abve) and take int accunt the minimum pint yu can see that the range is actually -3 f(x) < 1. When cnsidering the range, draw a diagram and cnsider statinary pints fr a quadratic yu ften cmplete the square t find the min / max pint. When writing a functin yu must state the functin f(x) = and the dmain x. Egs y = e x x є IR y = In x x > 0, x є IR [In is nt defined fr x = 0 r x < 0] y = 3 /x x 0, x є IR [Yu can t divide by 0 hence x 0] y = tan x x (n +1) π /, x є IR [tan x = sinx / csx is nt defined wherever csx = 0 ie the dd numbered multiples f π/] Cmpsite functins Mst functins are a cmbinatin f mre basic functins. These are called cmpsite functins. gh(x) means g[h(x)] - ie apply h first then g Generally gh(x) is nt the same as hg(x)
Eg g(x) = x 4 x є IR and h(x) = 5 /x x 0, x є IR gh(x) = g[h(x)] hg(x) = h[g(x)] = g[ 5 /x] = h[x 4] = [ 5 /x] 4 = 5 /[x 4] = 5 /x 4 Cnsider the dmain f the cmpsite functins: the dmain f gh(x) is the same as the dmain f h. Hwever the dmain f hg(x) is nt the same as the dmain f g. The dmain f hg(x) must be x ±, x є IR. In this case the dmain f g has t be restricted t avid g mapping t 0 which is nt in the dmain f h. [Nte this means -5/4 will nt be in the range f hg(x)]. Yu culd be asked t find a cmpsite functin [eg find hg(x)], the result f a cmpsite functin fr a specific value [eg find hg(3)] r t slve an equatin invlving cmpsite functins [eg slve hg(x) = 3]. Inverse functins Inverse functins need t be ne t ne mappings. A many t ne functin can nly have an inverse if its dmain is restricted t make it ne t ne. Eg y = (x + 1) 3 x є IR is many t ne s des nt have an inverse Hwever if yu restrict its dmain t y = (x + 1) 3 x > -1, x є IR then it becmes ne t ne [as it s minimum pint is (-1, - 3)] This part f the curve is n lnger part f the functin y = f -1 (x) y = x y = f(x) with restricted dmain The relatinship between a functin and its inverse can be shwn graphically each is a reflectin f the ther in the line y = x є IR The range f f(x) is the dmain f f -1 (x) and the range f f -1 (x) is the dmain f f(x) The methd fr finding the inverse functin: y = f(x) s x = f -1 (y), hence make x the subject f the frmula... y = (x + 1) 3 Nte: y + 3 = (x + 1) (1) yu need t determine if + r square rt using the dmain f f(x). ± (y + 3) = x + 1 As the dmain f f(x) is x > -1 yu need the psitive square rt] -1 + (y + 3) = x () The dmain f the inverse functin must be stated it is the same as the -1 + (y + 3) = f -1 (y) range f f(x) Using a change f variable... f -1 (x) = -1 + (x + 3), x > -3 x є IR Sme functins such as f(x) = 1 /x are self inverses.
7 Ratinal Functins (Algebraic Fractins) Simplifying fractins by cancelling cmmn factrs Factrise everything first! Yu can nly cancel when everything is multiplied tgether - yu usually have t factrise t ensure this is the case. Lk ut fr difference f tw squares. x 4x 3 ( x 1)( x 3) ( x 1) x 18 ( x 3)( x 3) ( x 3) Multiplicatin & Divisin f fractins Factrise everything first! Fr multiplicatin yu multiply the numeratrs tgether and multiply the denminatrs tgether and cancel in yur wrking. Fr divisin yu must multiply the first fractin by the reciprcal f the secnd fractin. (ie.turn the secnd fractin upside dwn and then multiply by the first fractin.) x x x 1 3x 6 6x 1 x( x 1) ( x 1)( x 1) 3( x ) 6( x ) x( x 1) 6( x ) 3( x ) ( x 1)( x 1) x x 1
Additin & Subtractin f fractins Factrise everything first! Put the fractins ver the lwest cmmn denminatr using equivalent fractins and add / subtract the numeratrs Be careful t use brackets fr questins invlving subtractin s that yu subtract all the terms yu shuld subtract and nt just the first f thse terms. DO NOT USE THE METHOD OF CROSS MULTIPLICATION AS IT CAN MAKE THESE QUESTIONS VERY DIFFICULT. x 3x 4 x 4 x 16 x 3x 4 x 4 ( x 4)( x 4) ( x )( x 4) 3x 4 ( x 4)( x 4) ( x 4)( x 4) Factrise and then see that the lwest cmmn denminatr will be (x 4)(x + 4) Use equivalent fractins t write ver the cmmn denminatr is (x 4)(x + 4) ( x x 8) (3x 4) ( x 4)( x 4) Subtract bth the 3x and the 4. Frgetting this is the mst cmmn errr in fractins! x x 8 3x 4 ( x 4)( x 4) x x 1 ( x 4)( x 4) ( x 4)( x 3) ( x 4)( x 4) x x 3 4 Dn t expand the denminatr. Instead try t factrise the numeratr and see if yu can simplify the fractin. 8. Plynmial Divisin Plynmials & Algebraic Divisin A plynmial is an expressin f the frm: anx n +... + a3 x 3 + a x + a1 x + a0 where the ai are cnstants The degree f a plynmial is the highest pwer f the variable in the plynmial An imprper fractin is a fractin whse denminatr has degree equal t r greater than the degree f its denminatr. Algebraic divisin can be used t turn an imprper fractin int a mixed plynmial. x 3 x 4x 5 x is an imprper fractin This example shws hw t divide x³ x² 4x 5 by x - The qutient is x² + x The remainder is 9 x 3 x 4x 5 x is the same as x² + x x Yu can als write: x³ x² 4x 5 = (x )( x² + x ) - 9 9 x x 3 ( x 3 x x x x ( x ) x 4x 4x x) 5 x 5 ( x 4) 9 Cmparing cefficients An alternative t algebraic divisin is the methd f cmparing cefficients: If yu are dividing a cubic by a linear expressin yu will get a quadratic expressin with remainder.
x³ x² 4x 5 (x )(ax + bx + c) + r Cmpare cefficient f x 3 : 1 a x : 1 b a hence 1 b s b 1 x: 4 c b hence 4 c s c x 0 : 5 c + r hence 5 4 + r s r 9 Hence x³ x² 4x 5 = (x )( x² + x ) - 9 Factr Therems If f(x) is a plynmial, if fr x = a we have f (a) = 0 then (x a) is a factr f f (x). If f(x) is a plynmial, if fr x = a /b we have f ( a /b) = 0 then (bx a) is a factr f f (x). Questin Types Express an imprper fractin as a mixed plynmial expressin Shw (x 1) is a factr f P(x) Slve a cubic equatin and sketch a cubic graph. Hence slve a cubic inequality If P(x) is a cubic with tw unknwn cefficients, a and b, use infrmatin abut factrs and remainders t frm and slve tw simultaneus equatins t find a and b Examples [1] Shw (x - 1) is a factr f f(x) = x³ 3x² 3x +. Hence write f(x) as a prduct f three linear factrs f(½) = (½) 3 3(½) 3(½) + = 1 /4 3 /4 3 / + = 0 Hence by the factr therem as f( 1) = 0 then (x + 1) is a factr f f(x) Using algebraic divisin r cmparing cefficients yu can shw that the quadratic factr is x x Hence f(x) = (x 1)(x x ) f(x) = (x 1)(x )(x + 1) as a prduct f 3 linear factrs [] Sketch y = x³ 3x² 3x +. Hence slve x³ 3x² 3x + > 0. When x = 0... y = This is the same cubic as in [] hence y = (x + 1)(x 1)(x ) When y = 0... x = -1, x = ½ and x = Frm the sketch yu can slve the inequality x³ 3x² 3x + > 0. -1 ½ This is where the graph is abve the x axis.
[3] Shw that the nly slutin t x³ x² + 4x - 3 = 0 is x = 1. If x = 1 is a slutin then (x 1) is a factr. Using algebraic divisin r cmparing cefficients yu can shw that x³ x² x + 3 = 0 is the same as (x 1)(x x + 3) = 0 x = 1 frm the first bracket is clearly a slutin. Use the discriminant: b 4ac = (-1) 4(1)(3) = -11 < 0 hence x x 3 = 0 has n slutins. INSERT SKILLS EXTRA 1A cre 4
Questin 1: Transfrmatins 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] On each diagram, shw clearly the crdinates f the minimum pint, and each f pint at which the curve crsses the crdinate axes. Questin : Trignmetry a) Shw that the equatin can be written as 3sin x = 5 cs x + 1 3cs x + 5cs x = 0 [] b) Hence slve, fr 0 x < 360, the equatin 3sin x = 5 cs x + 1, giving yur answers t 1 decimal place where apprpriate. [5] Questin 3: Integratin The line with equatin y = 6 x cuts the curve y = 4 + x x 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]
Questin 4: Integratin and Differentiatin The curve C with equatin y = f(x) is such that dy dx = 4x + 4 x, x > 0 a) Shw that, when x =, the exact value f dy is 8 +. [3] dx The curve C passes thrugh the pint (4,50). b) Find f(x). [6] Questin 5: Differentiatin The curve C has the equatin y = 6 1 x x, x > 0. x 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 d y dx. [] e) Deduce that the curve C has n minimum pints. [] Questin 6: Expnentials and Lgarithms Slve each equatin, giving yur answers in exact frm. a) ln(4x + 1) = [3] b) 3e x + e -x = 7 [5] Questin 7: Expnentials 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. [] b) Given that when x = 0, y = 4, find the exact value f b. [] Questin 8: Functins The functin f is defined by f: x 3 x x 5 x R, x 5 a) Find f -1 (x) [3]
The functin g has dmain -1 x 8 and is linear frm (-1,-9) t (,0) and frm (,0) t (8,4), The abve diagram shws a sketch f the graph f y=g(x) b) Write dwn the range f g. [1] c) Find gg() [] d) Find fg(8) [] e) State the dmain f the inverse functin g -1. [1] Questin 9: Factr Therem f(x) = x 3 7x 10x + 4 a) Use the factr therem t shw that (x + ) is a factr f f(x). [] b) Factrise f(x) cmpletely. [4] Questin 10: Ratinal Functins Express (3x + ) 9x 4 3x + 1 as a single fractin in its simplest frm. [4] Ttal marks = 77
Sectin A: Answers 1. Transfrmatins
. Trignmetry 75, 105 10, 70, 130 60 188.63, 351.37 0, 10, 360
3. Differentiatin 4. Integratin
5. Expnentials and Lgarithm 6. Functins
7.