NJC 03_2016_-_2017_H2_Mathematics_Functions_Assignment_Solutions_(LMS)

National Junior College Mathematics Department 2016 Functions Page 1 of 2 National Junior College 2016 – 2017 H2 Mathematics Functions Assignment Solutions 1(i)  g 0,R      f\ 3 = fg does not exist.D   fg \ 0 gf exists.RD    2 1gf : , , 3. ( 3)x x x x    Take note of the correct use of set notations. Domain is required whenever a function is defined. 1(ii) \ 3{ } ® \ 0{ } ® 0,¥( ) Therefore we have  gfR 0, . The mapping method is shown here. An alternative approach is through a sketch of the graph of the function gf over its domain. 1(iii) Every horizontal line y = k,  \0k cuts the graph of f ( )yx exactly once. Hence f is one - one, and so 1f  exists. 1 1 1 333y x xx y y         1 1f : 3 , \ 0 .xx x    The inverse function has to be defined in a similar form as stated in the question.  1 ffD R \ 0  1(iv) The origin is excluded because  11 fff fD D R \ 0    . y x 3  fyx yk y x O  gyx

National Junior College Mathematics Department 2016 Functions Page 2 of 2 2(i) fg ( , ) ( ,1) = .RD     Therefore largest value of 1. 2(ii) Need to show clearly the symmetrical relationship between the function and its inverse about the line y = x. You are to use the same scale on both axes to illustrate this. The coordinates of the end points of each curve must be clearly stated. 2(iii) From the graph, the solution to 1f( ) f ( )xx  is the same as that of f( ) .xx 2 2 1 ( 2) 5 3 0 5 13 2 5 13 5 13 (reject 2).22 xx xx x xx                From the graphs in 2(ii), solving 1f ( ) f ( )xx  is equivalent to s olving f(x) = x . Hence it is not necessary to find the inverse function to solve the given equation. 3(i)   2 2 2 2 2 4 3 f( ) 6 5 . f ( ) f 5 25 30 .                Note that f 2 is a composite function and not the square of the function. 3(ii) 22f( ) ( 3 ) 9 .xx    Therefore 3.k  When 3,x  2 2 2 2 2 2 12 ( 3 ) 9 ( 3 ) 9 39 3 9 . f ( ) 3 9 . y x x y xy xy xx                          