2013 Revision Lecture 2 Function (Solution)

Pioneer Junior College Mathematics Department H2JC2 -2013 Revision Lecture 2 : Function Basic Q12012 JC2 MI Prelim P2/2 The function f is defined by 2: 1xf x e , xe . (i)Show that1f exists and state its range.[2] (ii)Find1f and write down its domain.[3] (iii)Sketch the graph of1f , giving the equations of any asymptotes and the exact coordinates of any points where the curves cross the axes.[3] (iv)Hence, find the value(s) of x that satisfies the equation 21xe x = + .[3] Q22012 JC2 CJC Prelim P2/3(i), (ii), (iii) The function f is defined as follows. 11: f2+xxfor9 e x . (i)Sketch the graph of y = f(x).[1] (ii)If the domain of f is restricted tok x s , state with a reason the greatest value of k for whichthe function) ( f1x exists.[2] (iii)With the restricted domain in (ii), find an expression for) ( f1x and the real value of a such that) f( ) ( f1a a =.[4] Q32012 JC2 JJC Prelim P1/9 The functions f andgare defined byf : x, x < 0 , g : x sin x, . (i)Explain why f 1 exists.[1] (ii)Define f 1 in a similar form.[3] (iii)Sketch,onthesamediagram,thegraphsof,and, giving the coordinates of any points where the curves cross the x- and y- axes.[3] (iv)Show that the composite function fg does not exist.[1] (v)The function h is defined byh : x sin x, 2 x < < . Define fh and state the range of fh.[3] xx10 2 x s sf( ) y x =1f ( ) y x=1f f( ) y x=Q42012 JC2 DHS Prelim P1/10 The functions f and g are defined by f : x 22 4,1 x x x + + s andg : x( )1ln 2 1 , 2x x > . (i)Find an expression for( )1f xand state the domain of 1f . [3] (ii)Sketch the graphs for( ) f y x = and( )1f y x= on the same diagram. What can you say about the solution of the equation1f ( ) = f ( )? x x State your reason clearly.[3] (iii)Determine if the composite function gf exists. If so, find gf(x) and the exact range of gf. [5] Average Q52012 JC2 RV Prelim P1/6 (modified) The function f is defined by 2f : 2 3, . x x x x a + s (i)Explain why f 1 does not exist when1 a = .[1] (ii)State the largest value ofasuch that f 1 exists. Find f 1, stating its domain.[3] (iii)Find the exact solution of the equation( )2f x x =, using the value ofafound inpart (ii).[2] The function g has domain (0, 3) and its graph passes through the point with coordinates(1.2, 1). The graph of g is given below. For the rest of the question, take1 a = . (iv)Give a reason why fg does not exist, where f is the function given above.[1] (v)State the largest domain of g such that fg exists. Hence, find the range of fg, showingclearly your working.[3] x 0 y 23 (1.2, 1) Q62012 JC2 YJC Prelim P2/3 The functions f and g are defined by 2f : 4 4 , ,where is a constant, x x x x e( ) g: ln 1 , 1. x x x < (i)Explain why function f does not have an inverse.[1] (ii)The function f has an inverse if its domain is restricted tox k < . Find the largest value ofk in terms of , and define 1f corresponding to this domain of f.[4] (iii)If 12 > , show that the composite function 1gf exists and find the range of 1gf in terms of .[3] Q72012 JC2 RI Prelim P1/11(b) Functions f and g are defined by ( )2f : 2 3 x a x + ,xewhere a is a positive constant; g :5xxx,, 5. x x e = (i)Sketch the graph off ( ) y x = , indicating the coordinates of the stationary point and intersections with the axes if any.[2] (ii)The function k, a restriction of the function f is defined by ( )2k : 2 3, , 7. x a x x x + e >Given that the composite function gk exists, find the range of values for a. [3] (iii)Given that 12a = , solve the inequalityf ( ) g( ) x x > . Hence write down thesolution of ( )2 12 32 5xxx+ +>+.[3] Q82010 JC2 HCI MYE (a)Given a one-one function f, explain why the composite functionalways exists. [1] (b)The functions g and h are defined by , where a is a positive real constant, . (i)Sketch, on separate diagrams, the graphs of g and h. Define the function in similar form.[4] 1ff2 2g: ,,0 x a x x x e s2h : 3e ,,0xx x x e >1g(ii)Find the value of x such that.[2] (iii)State the range of values of x for which.[1] (iv)Show that hg does not exist.Find the maximal domain of g such that hg exists.[2] Advanced Q92012 JC2 SAJC Prelim P1/7 The functions f, g and f1h are given by2212 3f :, , 1 where is a constant,,1g :( 1) 0.25, , 0 3.5,f h : ,1, 0.xx axx x x a axx x x xx e x x+ e > =+ e>< (iv)0, 0.797 2)(iii) xxx f =1) (1,1 0 x s (iii) 1 132 (v)( |3, 0 6)(ii) 1 2 2f : 2 4 4 , 4 4 x x x + < +(iii)( ) ( )ln 1 2 , + 7)(ii) 225a >(iii)5 or5.59 x x < > ,5 or5.59 (3 s.f.) x x > < 8)(b)(i) 1 2 2: , g x a x x a s (ii)0.716 9) 1 a > (i) 51[ 3, )7(ii) 22 3 3h : , , 01x xxe ex x xe e >+ 10)(ii) 4(iii)( )2g( ) 1 x x = 11)(iv)( | | |fgR 2, 1 0, 64 = 12)(iii)563

Pioneer Junior College Mathematics Department H2JC2 -2013 Revision Lecture 2 : Function (Solution) Q1 (i) Every horizontal line y=k intersects the curve y=f(x) at most once,f is a one-one function, hence f-1 exists. ( ) 1,fR = (ii) ( )( )( )21121,ln 1 2ln 12( ) ln 1 , 1xy e xy xyxf x x x= e+ =+== + > (iii) (iv) 22110, 0.797xxe xe xx x= + == = y = -1 0 y = k y = f y = x 0 x = 1 y = -1 y = f

Q2 (i) (ii) For inverse to exist, function f must be one-one and thus, largest k is 0. (iii) 112+ =xy12 = + y y xyyx =12 yyx =1 Since0 x s, 1 yxy = xxx f =1) (1,1 0 x s < Solving( ) ( ) x x f f-1=is equivalent to solving( ) x x = f : xx=+112 0 13= + + x xSolving by GC and taking the real root: a = 0.682 (3sf) Q3 (i) y 0 x f( ) y x = y=k Since any horizontal line y = k, k ewill cut the graph of f exactly once, hence f is one-one. Thus, f -1 exists. (ii)Let xx y1 =21 0 x yx = 42122+ = yyx But x < 0, 2142 2yx y = +Hence, f -1 : x 2142 2xx + , x e (- , ) (iii)yy x =

f( ) y x =

(-1,0)0 1f ( ) y x=x (0,-1) 1f f( ) y x=

(iv)Df : (-,0) and Rg : [-1 , 1].Since Rg . Df , therefore fg does not exist. (v) fh(x) = f(sin x) = xxsin1sin Hence, fh : x xxsin1sin , t < x < 2t Range of fh =[0, ) Q4 (i) ( )( )( )2221Let2 4 1 31 31 33 1 ( 1)f 3 1y x x xx yx yx y xx x= + + = + ++ = = = s = Note : ( )1f f x x=and( )1ff x x= .However, the domain is generally not the same. 1ff fD D=while 1 1ff fD D =| ) 1ffD R 3, = = (ii) Since there is no intersection between the 2 graphs, there is no real value of x for 1f( ) = f ( ). x x (iii) | )f g1R 3, , D ,2| |= = |\ . Since f gR D , _gf exists. ( )( )( )( )( )( )222gf( ) g 1 3ln 2 1 3 1ln 2 1 5 , 1x xxx x= + + (= + + = + + s From graph of gf(x) for x 1 ,

| )gfR ln5, = Alternatively ( | | ) | )f g, 1 3, ln5, | )gf R ln5,= Q5 (i) 2f : 2 3, 1. x x x x + s

(-1,3) y y = x O 3- | -1 y = f(x) -1-

(3,-1) | 3 ln5 y x O | -1 Note : Rgf is equal toRg , with the domain of g restricted toRf. x 3 1 (1,4) y

A horizontal line y = k where4 0 k < scuts the graph of y = f(x) more than once, thus f is not one-to-one. Therefore f 1 does not exist. (ii) For f 1 to exist, largest domain is( , 1]. Largest value of1 a = . 221Let2 3.2 3 02 4 4( 3 )22 4 1621 4Since1, 1 4f ( ) 1 4, 4y x xx x yyxyyx x yx x x= + + = = +== +s = + = + > (iii) ( )2f x x =( ) ( )1f f x x= y = f(x), y = f 1(x) and y = x intersect at the same point.22f ( )2 33 01 1321 13Since1, .2x xx x xx xxx x=+ =+ = = s = (iv) Rg = (0,2) Df =( ,1] Rg . Df Thus, fg does not exist. (v) For fg to exist, Rg _ Df. Let Rg = (0,1] Thus Dg = [1.2,3) g f[1.2, 3) (0,1] ( 3, 0] Note : In this case, the function g is not given. It is impossible to find the range of fg from the graph of fg.Rfg is equal toRf , with the domain of f restricted toRg. Note : Concept here is if()1f y x=, then( ) f y x = ( |3, 0fgR= Q6 (i)( )2f 4 4 x x x =

( ) ( )( )2 2222 2 42 4 4xx (= + + = + + + Since there exists a horizontal line y = k cutting the graph( ) f y x = more than once, f is not a one-one function. f does not have an inverse. (ii)Largest value of k =2 Let( )222 4 4 y x = + + + 22 4 4 x y = +Since2 x < , 22 4 4 x y = + 1 2 2f : 2 4 4 , 4 4 x x x + < + (iii)( ) 1f, 2 R = ( )g, 1 D = Given 12 > ,( ) ( ) 1gf, 2 , 1 R D = c = 1gfexists ( ) ( ) 1gfln 1 2 , R = + Q7 (i) ( )22 3 y a x = +xy(2,3) (0, 4a+3)

Note : 122 12 1> > < (ii) In order for gk to exist,R Dk g_Domain of g is( ) ( ) , , 5 5 , ( )2k( ) 2 3 x a x = + ,7 > xRange of k is| ) + , 3 25aSo,5 3 25 > + a08 . 0 or252> a(iii) Whenf ( ) g( ) x x > ,( )2 12 32 5xxx +> 5 or5.59 (3 s.f.) x x < > ( )2 12 32 5xxx+ +>+ ( )( )2 12 32 5xxx +>( ( )2 12 32 5xxx +> Replace with , 5 or5.59 (3 s.f.) x x x x > < Q8 (a)Since f is one one, 1f exists. As ,always exists. (b)(i) 2 222since 0,y a xx a yx x a y= = s =

(b)(ii) To find x such that, we may find x such that. 1ffR D =1ff1 2 2g : ,x a x x a s1h ( ) h( ) x x= h( ) x x =23exx=-a a2 O y x g(x) . 3 O y x h(x) .

23 0xe x = Using GC, x = 0.716 (3 s.f.) (b)(iii) For, we find x such that11hh ( ) , 0 3h h( ) , 0x x xx x x= < s= >

Hence we obtain. b(iv) Q9 22( 1)(4 ) (2 3)( 1)dy x x a x axdx x+ + + =+ 22 220,( 1)(4 ) (2 3) 04 4 2 3 02 4 ( 3) 0dydxx x a x axx x ax a x axx x a=+ + + =+ + + + =+ + + = For stationary points to exist:24 4(2)( 3) 01(since1)aa a + >< = For 1fto exist, there should not be any stationary point in the given domain, hence1 a > . Therefore, the set of values: { : 1}. a a e > (i) 22 3 3( )1x xf xx =+ 51: (0, 3.5) :[ 0.25, 6) :[ 3, )7g g fgD R R

1 1hh ( ) h h( ) x x =1h hhD D R= =0 3 x < s2g hg h2gg( , ][0, )Since , hg does not exist.For hg to exist,[0, ].Hence maximal[ , 0].R a DR DR aD a= = .== y =g(x) 1 6 3.5 y x 14 -16 y=f(x) y x 517 Note : For turning point to exist, there must be at least one solution to the quadratic equation that resulted from dy/dx =0 Note : Rfg is equal toRf , with the domain of f restricted toRg. Note : ( )1f f x x=and( )1ff x x= .However, the domain is generally not the same. 1ff fD D=while ( )1ff x x= -a a2 O y x g(x) . (ii) 1f h( )xx e= 2h( ) f(e )2 3 31xx xxxe ee= =+ 22 3 3h : , , 01x xxe ex x xe e >+ Q10 (i) Let

,, 3 x x e = Since rule and domain of( ) f xand( )1f xare the same, (ii)( )2 1f ( ) ff ( ) ff x x x x= = = 2 2 2f ( ) f f ( ) x x x = =Similarly,( )20f x x = ( ) ( ) ( )( )( )21 20213 7f ff f33 5 7f 5 45 3xx x xx = = == =

(iii) 22: 4 , , 33xgf x x xx | |e = |\ . 3 73xyx =( 3) 3 7 y x x = 3 3 7 xy y x = 3 3 7 xy x y = ( 3) 3 7 x y y = 3 73yxy =13 7( )3xf xx=1f f =-3 Note : Concept here is if()1f y x=, then( ) f y x = ( )( )12 22 22g gff ( )3 7 3 74 2 / 33 31 24 /3 31x xx xx xxx xx= ( ( | | | |= ||(( \ . \ . | | | |= || \ . \ .=

Q11 (i) Since there exists a line y = k cutting f more than once, f is not a one-one function and so f-1 does not exist. (ii) x y 1 2 2 1 y = f(x) 1010 8 81

x y 8 2 2

y = f(x) y = x | )(( ) 10, 0 1 2 2,10 or 10, 0 1 2 2,10(( + + (iii) Solving( ) ( )1f f x x=is the same as solving( ) f x x = and so finding the number of intersection points between the graph of f and the liney = x in the domain in (ii) give rise to the number of real solution. So from the graph above, there will be 2 distinct real roots. (iv) | || |gfR 1, 9D 10.10 = = Since g fR D _ so fg(x) exist. ( ) | | | |22 , 10, 1 1,10fg :1, 1 1x xxx x e < < | | | | ( | | |g f10,10 1, 9 2, 1 0, 64 ( | | |fgR 2, 1 0, 64 = Q12 (i) f(4) =f(2) = f(2) = 1. (ii) y x 1 1 1 Note :( ) ( 6) f x f x = +and( ) ( ) f x f x = - 4 4 3 69 1 3 6 (iii) 75f ( ) d x x} =( )32114 4 1 32x dx ( + ( } =331(3 )4 23x ( ( +( ( (

= 48 [0 8]3 = 563 Lesson Guide for Revision Lecture 2 : Function Time Content/Learning Objectives/Skills involved Suggested Guideline Level of Difficulty 10 min Question 3 - explain whyf-1 exists and findf-1 - solve( ) f x x =Band 1, 2 skip Band 3, 4 go through Basic 10 min Question 2 - restrict domain such that inverse exists. -find inverse -solve 1f ( ) = f ( ) x x (concept ( ) f x x = ) Band 1 , 2 skip Band 3,4 go through Basic 15 min Question 3 - restrict domain such that inverse exists. -find inverse - sketch f, f-1 , ff-1,-explain why fg does not exists -find fh and range of fh Band 1, 2- skip Band 3, 4 go through Basic 15 min Question 4 - find inverse - sketch f and f-1 - solve 1f ( ) = f ( ) x x(i.e( ) f x x = ) -Determine if gf exist, find gf and its range Band 1 - skip Band 2, 3, 4 go through Basic 15 min Question 5 - explain why inverse exists - restrict domain such that inverse exists. -find inverse - solve( )2f x x = (i.e. using inverse, leading to 1f ( ) = f ( ) x x Band 1, 2 go through Band 3, 4 go through (i), (ii), (iii) Average and hence( ) f x x = ) - explain why fg does not exist -maximal domain of g such that fg exists. - find range of fg , where rule of g is not given. (i.erange of fg = range of f with domain restricted to range of g) 10 min Question 6 - explain why f (f contains unknown) does not exist - restrict domain such that inverse exists. -find inverse - show gf-1 exists and find its range Band 1,4 skip Band 2 go through Band 3 may go through if still have time Average 10min Question 7 - Given that gk exists, find range of value of a, where a is an unknown constant in k -Given a, solvef ( ) g( ) x x > , hence solvef ( ) g( ) x x > Band 1, 2 go through Band 3, 4 - skip Average 15 min Question 8 - explain why exists - sketch g , f. - find h-1. - solve 1h ( ) = h ( ) x x(i.eh ( ) =x x ) - range of x where (i.e consider domain) -explain why hg does not exists.- find maximal domain for g such that hg exits. Band 1, 2 go through Band 3 , 4 go through (a), (b)(i) and (b)(ii) Average 10 min Question 9 - Given that f has stationary point, find unknown constant a in the function and set of values for a for inverse to exist (i.e use discriminant) - find range of fg Band 1, 2 go through Band 3, 4 - skip Advanced 1ff1 1hh ( ) h h( ) x x = Band 1 85min Band 2 80 min Band 3 70 min Band 4 60 min + TYS questions - find h given f-1h (i.eif()1f y x=, then( ) f y x = ) 5 minQuestion 10 - show f = f-1 and hence find 21f (5)(i.e every f2 = x , hence f21 = f ) - given gf and f, find g Band 1, 2- go through (ii), (iii) Band 3, 4 - skip Advanced 15 min Question 11 - explain why inverse does not exists for a piecewise function - maximal domain for inverse to exist. - solve( ) ( )1f f x x=- why fg exists, find fg and its range, where g is also a piecewise function. Band 1 go through Band 2, 3,4 - skip Advanced 15 min Question 12 -Given piecewise and recurrence function and( ) ( 6) f x f x = + , ( ) ( ) f x f x = -sketch( ) f x Band 1 go through Band 2, 3,4 - skip Advanced