DHS Functions (9758) (Revision Solutions)

Y5 Topical Revision Package 1 3. Functions (solutions) 1 3. Functions (solutions) 1(i) (ii) (iii) 1 11Let 2 2 1 2 1Hence g : , 2 2 y xy x xxy xx x − = +  − =  = − → − fR =  gD \ 0= fgRD Hence gf does not exist 2 2 1f 2 1 1121 120 using GC, 2.62, 1, or 0.382 x xx xx x  +=    + − =        + + = =− − −

Y5 Topical Revision Package 1 3. Functions (solutions) 2 2(b) g: x 3x2 + 2, x  -, Rg = (2, )  (0, ) = fD Hence fg exists. fg(x) = 2(3 2)xe−+ , Dfg = Dg = ( - , 0) or − fg: x 2(3 2)xe−+ , x < 0. 2( ,0) (2, ) (0, )gf e−− ⎯⎯ →  ⎯⎯ → Rfg = 2(0, )e− 2(a) f: x e-x, x  + , Since the graph of f(x) is strictly decreasing, f is one-to-one Hence f-1 exists. Let y = f(x) y = e-x  x = - ln y  f -1(x) = - ln x f -1 : x - ln x, 0 < x < 1 ( 1 ffDR− = =(0,1)) Range of f-1 = + ff -1(x) = x, where 0 < x < 1 since -1 -1ff fD =D . x y y= ff -1(x) y= f -1(x) y= f(x) 1 1 y=g(x)

Y5 Topical Revision Package 1 3. Functions (solutions) 3 3(i) (ii) (iii) (iv) (v) ( ) 2 1f 2 0 for 1 2 x x x x  = +     f is strictly increasing. Since f is strictly increasing, its minimum and maximum values correspond to the minimum and maximum x values. Thus Rf = 171 1,4 0, 22    − − =     . ( ) ( ) 1ff xx −=  ( )f xx=  2 1xx x−=  32 10xx− − =  x = 1.47. Since Rg =  )1, 2 = Df, fg exists. Since Rf = 7 π0, 0,22        = Dg, gf does not exist. fg(x) = f( sin 1x+ ) = ( ) 2 1sin 1 sin 1x x+− + . Dfg = Dg = π0, 2   . fg : x → ( ) 2 1sin 1 sin 1x x+− + , π, 0 2xx   .  ) gfπ70, 1, 2 0,22    ⎯⎯ → ⎯⎯ →    Rfg = 70, 2   .

Y5 Topical Revision Package 1 3. Functions (solutions) 4 5(i) 2 2 21f ( ) 2 11 x xx x x + + −= + − − = 2 2 2 21 xx xxx + + − =+−+ . ( ) 32f ( ) f f ( ) f ( )x x x== , and ( ) 4 3 2f ( ) f f ( ) f ( )x x x x= = =  2012f ( ) xx= . 5(ii) g [ 1, 1)R =− and f \{ 1 }D = Since gfRD , fg exists. 4(i) ( ) ( ) ( ) ( ) 2 2 22 2 2 11h( ) f( ) tan 1 sech '( ) tan 1 since sec 0, tan 1 0 for 0 < < 2 sech '( ) 0 for 0 < < 2tan 1 h is a decreasing function. h is a one -one function. x xx xx x x x x xxx x x   == + =− +  +  =−  +  (ii) 1 11 11Let tan 1tan 1 1h ( ) tan 1 yx xy x x − −− =  = − +  =−  (iii) f g f gR (1, ), D (0, ) R D and gf exists.=  =    1 tan 2gf( ) g( tan +1) 1 or tan + 1 tan + 1 xxx xx += = + = Thus 1gf : 1 ,0 tan + 1 2xx x +   . ( ) ( )fgπ0, 1, 1, 22  ⎯⎯ →  ⎯⎯ → gfR (1,2)= y x 1 O

Y5 Topical Revision Package 1 3. Functions (solutions) 5 5(iii) cos 2fg : cos 1 xx x + − , 02 x  . 5(iv) f g fg [ 1, 1)RR= − ⎯⎯ → From the graphs of g and f, fg 1, 2R = − −  . y x O

Y5 Topical Revision Package 1 3. Functions (solutions) 6 ( ) ( ) 1 1 f f -1 ff -1 , , ,ff exists. ff ( ) , 1 R D RD x x x − − = −  = −  = = 6(i) 1 2 2 Let f ( ) ln( 1) e1 e 1 since 1 f ( ) e 1, y y x yx yx x xx xx −= =− =+ = +   = +  (ii) (iii) x y x = 1 1f ( )yx −= ( )1ff 1,R − = (1, 1) x y

Y5 Topical Revision Package 1 3. Functions (solutions) 7 7 f(x) = 3 – 2x – x2 = 4 – (x + 1)2 For 1f− to exist, f must be one–one.  k = −1 To find 1f− : Let y = 4 – (x + 1)2  yx −−= 41 Since x  −1, yx −−−= 41 Thus, 1f− : x x−−− 41 , x  (−, 4] (i) Range of f = (−, 4], domain of g = [0, 4] Since range of f / domain of g, gf does not exist. (iii) Let 1Xx=+ 1 1 g g( 1) 1 g g( ) xx XX − − + = + = 1 g( )g g( )D D 0 4 0 1 4 1 3XX X x x− =      +  −   1 ( 1)g g xD − + = [ −1, 3] ----- (1) 11 2 2 2 g( )gg ( ) g ( )D D R 1 e 1 1 e 0 e 1 XXX X x x−− = =      +     − 1( 1)gg xD − + = [0, e2 –1] ----- (2) Taking the intersection of (1) & (2), set of values of x = [0, 3] (−1, 4) −3 1

Y5 Topical Revision Package 1 3. Functions (solutions) 8 8(i) ( ) 2 2 2 2 2 4 24 24 2 4 since 2 y x x yx xy x y x       =−  = − −  − = +  = + +  1 2 2g : 2 4 , 4x x x   − + + − 8(ii) For gf to exist, ( ) ( )fg1, R D 2 , 121 2    =  =      8(iii) When 1 =− , ( ) 2g 4 , 2x x x x= + − ( ) ( ) g f gfR 1, 5, R=  →  = 9(i) f ( )yx= is concave upwards when ( )f0 x  ( ) 2f ( ) ln 1xx=+  2 2f ( ) 1 xx x  = + and ( ) ( ) ( ) ( ) 2 2 2222 2 1 2 (2 ) 22f 11 x x x xx xx +− − == ++ ( ) 2 22 22 0 1 x x −  + Since ( ) 22 1x + is always positive, 22(1 ) 0x− ( )( )1 1 0 11 xx x − +  −   O x y – 1 2 y = f(x) ( ) 1f , 2 1 1xx x= −  −+ -1 -2 (-2,1) -2 1 5 x y y = g(x)

Y5 Topical Revision Package 1 3. Functions (solutions) 9 9 (ii) (iii) 2f ( ) ln( 1)y x x= = + , x 1f− exist if xk , where 0.k  (iv) ( ) 2g : ln 1 , 1.x x x + Let 2 2 ln( 1) 1e e1 y y yx x x =+ += = −  But 1x , therefore e1yx=− g g(1) ln 2 R [ln 2, ) = = 1g ( ) e 1 xx− =− , x [ln 2, ) or 1g : e 1 xx− − , x [ln 2, ) 10(i) (ii) (iii) (iv) Every horizontal line y = k, -2 ≤ k ≤ 2, cuts the graph of f at exactly 1 point. Therefore f is one-one and f-1 exists. set of values of x for which f(x) = f-1(x) is [0,2]. f-1(x) = 3  x = f (3) = -3/4 3 2 f ( ) dxx = ( ) 3 2 2 1 2d4 x x x− = 1 3− 3 4 2 1f ( ) dxx− −  = ( )3 1 1 473 2 24 3 2 12   − +  = 11(i) The horizontal line y = a cuts the graph ( )fyx= at most once. f is an one-to-one function and thus 1f− exists. y = f-1(x) y = f(x) (-2,4) (4,-2) 2 2 y = x y = k O x y y = f(x) ( )2,8 ya= ( )2, 1−−

Y5 Topical Revision Package 1 3. Functions (solutions) 10 (ii) For 20 x−   , 2 44 4 (since 2 0) xy x y x y x =−  = − =− − −   1 3 4 , 1 0,f: , 0 8. xxx xx − − − −    (iii)  )g 1,0R =− and  )h 2,0D =− . Since gh ,RD hg exists. (iv) ( ) ( ) 2 gh hg 4 xxx= =− Since  )gh h 2,0DD= = − and  )hg g 1,0DD= = − , solution for ( ) ( )gh hgxx= is the intersection of the two domains, i.e. 10 x−   . 12i) ( ) ( ) ( ) 2 2 2 Let 2 1 12 21 12 12 y x x yx yx xy xy = − − = − − + = − − = + =  + Since 11 x−   , 12xy= − + 1f : 1 2, 2 2x x x− − + −   (ii) ( ) 2 9 3 , 0 3, g: 3 , 3 6, xx x xx −   −   and that ( ) ( )g g 6xx=+ for all real values of x. y x O 9 3 6 (8, 3) (-2, 1) For 02 x , 3 3y x x y=  = y x O -1 (1, -2) (-1, 2) -1 (-2, 1) (2, -1)