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 +   . ( )