Chapter 5 : Functions and Graphs

Chapter 5 : Functions
& Graph

Step By Step Solution

LIM HWEE CHENG

KOLEJ MATRIKULASI JOHOR KPM

Chapter 5 : Functions & Graph Mathematics
SM015

Q1 : 2003 / 2004 Q2 : 2004 / 2005

Given that f (x) = 2x +1 and h(x) = 2x2 + 4x +1, Given g(x) = 3 x and h(x) = 1 . Find f (x)
x3
find a function g such that ( f  g)(x) = h(x) .
such that ( f  g  h)(x) = x .
Write g(x) in the form a(x + b)2 + c , where a ,b
and c are constants. x +1

Given ( f  g)(x) = h(x) ( °ℎ)( ) = [ℎ( )]
= ( 13)
[ ( )] = 2x2 + 4x +1
= 3√ 13
2[ ( )] + 1 = 2x2 + 4x +1
2[ ( )] = 2 2 + 4 =1
( ) = 2 + 2

Completing the square
Given ( f  g  h)(x) = x
( ) = ( + 1)2 − 1
x +1
{ a(x + b)2 + c form }
[ °ℎ( )] =

+1

(1) =

+1

Let = 1  = 1



1

( ) =

1 +1

1

=
1+


=1

1+

Thus, ( ) = 1
1+

1

Chapter 5 : Functions & Graph Mathematics
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Q3: 2005/2006

Function f and g are defined as f (x) = e2x ,

g(x) = 1− x , x  . Find f −1(x) and hence

( )obtain g  f −1 (x)

To find the f −1(x) y =k

 Let f  f −1(x) = x

e 2 f −1( x) = x

2 −1( ) = Since y = k intersects f(x) at 2 different points,
Therefore f(x) is not one to one function.

−1( ) = 1

2

( )Therefore, = [1 ]
g  f −1 (x) Q5: 2007/2008
2 Given h(x) = 3x .

= 1 − 1 x−3

2 Defining h2 (x) = (h  h)(x) , determine the

Q4: 2007/2008 function h2 (x) and hence deduce the inverse of
h(x) . Evaluate h13(9) .
A function f is defined by f (x) = x2 − 2x − 3 for
0  x  5 . State the range of f and determine h2 (x) = (h  h)(x)
whether f is one to one.
= 3ℎ( )
Given Domain = [0,5] ℎ( )−3
f (x) = x2 − 2x − 3
= 3( 3− 3)
= [( − 1)2 − 1] − 3
= ( − 1)2 − 4 3− 3−3

9

= −3
3 −3( −3)
−3

9

= −3
3 −3 +9
−3

= 9 =

9

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Chapter 5 : Functions & Graph Mathematics
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To find the ℎ−1( ) To find −1
Let ( −1( )) =
Let ℎ(ℎ−1( )) = −1( ) − 3 =

Thus, ℎ−1( ) = ℎ( ) = 3 −1( ) = + 3

−3 To find ( ° )( )
( ° )( ) = [ ( )]
*** Because ℎ2( ) = (ℎ ∘ ℎ)( ) =
= [2 2 + 1]
ℎ13(9) = ℎ12(ℎ(9)) = 2 2 + 1 − 3
= 2 2 − 2
= ℎ12 (3(9))
To find the ( ° )−1( )
9−3 Let ( ° )( ) = ( )
Thus ( ° )−1( ) = −1( )
= ℎ2 (ℎ2 (ℎ2 … (9))) Let ( −1( )) =

2 2( −1( ))2 − 2 =
( −1( ))2 = +2
= ℎ2 (9)
2
2
−1( ) = ±√ +2
=9
2
2
Since ≫ 0 , −1( ) = √ +2
Q6 : 2007/2008
Given f (x) = 2x2 + 1, x  0 and g(x) = x − 3 , 2
find
(a) The inverses of f and g and verify that
(g  f )−1 = f −1  g −1

(b) The function of h if (g  f )−1  h(x) = 1
x

(c) The value of x for which f  g = g  f

(a) To find the −1 Thus, ( ° )−1( ) = √ +2
Let ( −1( )) =
2[ −1( )]2 + 1 = 2
2[ −1( )]2 = − 1
[ −1( )]2 = −1 To find the −1( −1( ))
−1( −1( )) = −1( + 3)
2 = √ +3−1

−1( ) = ±√ −1 2

2 = √ +2

Since ≫ 0 , thus −1 = [0, +∞) 2

−1( ) = √ −1 Therefore, (g  f )−1 = f −1  g −1 shown.

2 ( +ve only )

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Chapter 5 : Functions & Graph Mathematics
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(b) Given (g  f )−1  h(x) = 1 To find the −1
x Let ( −1( )) =

√ℎ( )+2 = 1 10−2 −1( ) =

2

ℎ( )+2 = 1 10 − 2 −1( ) =
2 2 −1( ) = 10− = 5 −

ℎ( ) = 2 − 2 22
2
−1( 2) = 5 − 2
= 2− 2
2 2

(c) Given f  g = g  f To find g ( )
° ( ) = ( − 3)
= 2( − 3)2 + 1 -------- (1) 2

° ( ) = (2 2 + 1) 2
= (2 2 + 1) − 3 g (2) = 5 − 2 (2)
= 2 2 − 2 ------------ (2)
= 5 − 2 ( 2)

4

2
g (2) = 5 − 2

Given f −1(x2 ) = g  x 
 2 

Thus Eq(1) = Eq (2) 5 − 2 = 5 − 2
2( − 3)2 + 1 = 2 2 − 2
2( 2 − 6 + 9) + 1 = 2 2 − 2 22

2 2 − 12 + 18 + 1 − 2 2 + 2 = 0 2 = 2
21 − 12 = 0
= 21 = 7 22

12 4 = 1

22

= 1

Q7: 2008/2009 ( )f −1 o g (0). = −1[ (0)]

Given that f (x) = 10 − 2x and g(x) = 5 − 2x2 . Find = −1[5 − 2(0)2]
k = −1(5)
= 5 − 1 (5)
the value of k so that f −1 ( x2 ) = g  x  .
 2  2

=5

2

( )Hence, find f −1 o g (0).

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Chapter 5 : Functions & Graph Mathematics
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Q8: 2008 /2009 6 + 3 ≫ 1
Let f (x) = 4x −1 and g(x) = x + 2 ={ 4
(a) Find the interval of x which f (x)  g(x) −2 + 5
(b) If h(x) = f (x) + 2g(x),express h(x) as a > 1
4
piecewise functions.
Q9: 2008/2009
(a) For f (x)  g(x) Let f (ax) = a3x2 + a2x + 3a where a is non-zero.

|4 − 1| < + 2 {LESS AND} (a) Find a if f (0) = 6 .

4 − 1 < + 2 and 4 − 1 > −( + 2) (b) Determine f (x) .

3 < 3 5 > −1 (c) Determine the domain and range
< 1  > −1 of f (x) . Hence, state the
interval in which f is one to one
5

−1 1 (a) Given f (0) = 6 , ≠ 0
5  = 0
= 0
Solution ： { : −1 < < 1}
5 Therefore, ( (0)) = 6
3(0)2 + 2(0) + 3 = 6
4 − 1 , ≫ 1
( ) = |4 − 1| = { 4 3 = 6
(b) −(4 − 1), = 2
<1
4 (b) (2 ) = 8 2 + 4 + 6
Let = 2
ℎ( ) = ( ) + 2 ( ) =

4 − 1 + 2( + 2) ≫ 1 2
={ 4
−(4 − 1) + 2( + 2) Thus, ( ) = 8 ( )2 + 4 ( ) + 6
> 1
4 22

= 2 2 + 2 + 6

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Chapter 5 : Functions & Graph Mathematics
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∴ ( ) = 2 2 + 2 + 6 −1( ) = 1 + 1
2
(c) ( ) = 2 2 + 2 + 6
= 2( 2 + + 3) = 1+ 2
12 1 2
= 2 (( + 2) − 4 + 3)
1 2 11 Given that x > 1,
= 2 ( + 2) + 2
= −1 = (1, +∞) = −1 = (0, +∞)

= , = [11 , +∞]

2

1

The intervals of f for which is one to one

function are (−∞, − ] or [− , , +∞)



Q10: 2009/2010

A function g is defined by Q11: 2009/2010
g(x) = 1 , x  1.Find g −1(x) and state its
(a) Show that y − y 2 + 1  0 for all real
x −1 values of y.
domain and range.
(b) Let f be a function defined by
To find the −1, f (x) = ex − e−x . Find f −1(x) .
2
Let [ −1( )] =
(a) Since 2 < 2 + 1
1 = √ 2 < √ 2 + 1
√ −1( )−1 < √ 2 + 1
− √ 2 + 1 <0 shown
√ −1( ) − 1 = 1
6


−1( ) − 1 = 1
2

Chapter 5 : Functions & Graph Mathematics
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(b) To find the f −1(x) Q12: 2010/2011
Given the function f and g as follows:
Let [ −1( )] =

= −1( )− − −1( ) f (x) = 2 − x2 , g(x) = x + 2

2 (a) Find f  g and g  f

−1( ) − − −1( )= 2

−1( ) − 1 = 2 (b) Find (g  f )−1
−1 ( )

Let = −1( ) (c) Determine the value of x such that

− 1 = 2 f  g(x) = g  f (x)

(a) f  g = [ ( )]
= [ + 2]
2−1 = 2 = 2 − ( + 2)2
= −( 2 + 4 + 2)

g  f = [ ( )]
2 − 2 − 1 = 0 = [2 − 2]
= 2 − 2 + 2
By using the quadratic formula = 4 − 2

= −(−2 )±√(−2 )2−4(1)(−1) (b) To find the (g  f )−1

2(1) Let [ ° ]( ° )−1( ) =
4 − [( ° )−1( )]2 =
= 2 ±√4 2+4
[( ° )−1( )]2 = 4 −
2
( ° )−1( ) = ±√4 −
= 2 ±2√ 2+1
(c) Given f  g(x) = g  f (x)
2 −( 2 + 4 + 2) = 4 − 2
− 2 − 4 − 2 = 4 − 2
= ± √ 2 + 1 −4 − 2 = 4
−4 = 6
∴ −1( ) = ± √ 2 + 1
−1( ) = ln( ± √ 2 + 1)

From part (a), − √ 2 + 1 < 0
−1( ) = ln( + √ 2 + 1)

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Chapter 5 : Functions & Graph Mathematics
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3
= − 2

Q13:2011/2012 Q14 : 2011/2012

The functions f and g are defined as: (a) Given ( ) = − and ( ) = 2.
(i) Find the domain and range of f and g.
f (x) = x −1, x 1 (ii) Show that ( ° )( ) = −2 .
g(x) = x2, x0
e −2x , −  x0
(b) Given h( x) =  , x0
 x +1
Find the inverse function, f −1(x) and determine
(i) Find h−1(x) .
its range. Then, evaluate ( f  g)(−2)

(ii) Sketch the graph for h(x) and h−1(x)

To find −1, Let ( −1( )) = (a) (i) = , = (0, +∞)

√ −1( ) − 1 = = , = [0 , +∞)
−1( ) − 1 = 2
(ii) ( ° )( ) = ⌈ ( )⌉
−1( ) = 1 + 2 = [ − ]
= −1 = [1, +∞) = [ − ]2
= −1 = [0, +∞) = −2 shown.

( f  g)(−2) = [ (−2)] (b) (i) For ℎ( ) = −2

= [(−2)2] ℎ = (−∞, 0] , ℎ = [1, +∞)
= (4) Thus, ℎ[ℎ−1( )] =

= √4 − 1 −2ℎ−1( ) =
= √3 −2ℎ−1(x) =
ℎ−1( ) = −1

2

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Chapter 5 : Functions & Graph Mathematics
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For ℎ( ) = + 1 Q15 : 2011 / 2012

ℎ = [0, +∞) , ℎ = [1, +∞) (a) A function f(x) is defined by
Thus, ℎ[ℎ−1( )] = f (x) = 3x for x  6 . Show that f(x) is
ℎ−1( ) + 1 = x−6
ℎ−1( ) = − 1 a one–to–one function. Find the values of

Therefore, x such that ( f  f )(x) = 0

ℎ−1( ) = { − 1 , ≫ 1 (b) Given that f (x) = 1− 3x and
2

− 1 , ≫ 1  .

g(x) = x −1. Find f g −1  − 7 
2  2 

(ii) y =x
y

y =h(x) (a) Algebraic Approach

y =h-1(x) Let ( 1) = ( 2)

x 3 1 = 3 2

1−6 2−6

1( 2 − 6) = 2( 1 − 6)
1 2 − 6 1 = 1 2 − 6 2

−6 1 = −6 2
1 = 2

∴ f(x) is one to one function

Given ( f  f )(x) = 0

[ 3 ] = 0

−6

= 03( 3− 6)

3− 6−6

9 ∙ −6 = 0
−6 3 −6( −6)

9 = 0

36−3

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Chapter 5 : Functions & Graph Mathematics
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9 = 0 3 −1( )+4 =
= 0 −1( )−2

(b) To find the −1( ) 3 −1( ) + 4 = ( −1( ) − 2)
Let ( −1( )) = −1( ) − 3 −1( ) = 4 + 2
( −1( )) − 1 =
−1( )[ − 3] = 4 + 2
2
( −1( )) −1( ) = 4+2

= + 1 −3

2 To find −1( )

−1( ) = 2( + 1) ( )Let g  g −1(x) = x
Thus, −1 (−7) = 2(− 7 + 1)
3 − −1( ) =
22 −1( ) = 3 −

= 2 (−5) ( )  (b) f  g −1 (3) = f g −1(3)

2 = [3 − 3]

= −5 = (0)

( −1 (− 27)) = (−5) = 3(0)+4
= √1 − 3(−5)
= √16 0−2
=4
= −2
Q16: 2012/2013
( )(c) g  f −1 (k) = 2
The functions f and g are defined as 3
f (x) = 3x + 4 , x  2 and g(x) = 3 − x .
[ −1( )] = 2
x−2
3
(a) Find f −1(x) and g −1(x)
[4+2( )] = 2
( )(b) Evaluate f  g −1 (3).
−3 3
( )(c) If g  f −1 (k) = 2 , find the value of k.
3 3 − 4+2( ) = 2

(a) To find f −1(x) −3 3

( )Let f  f −1(x) = x 3( −3)−(4+2 ) = 2

−3 3

3 −9−4−2 = 2
−3 3

−13 = 2

−3 3

3( − 13) = 2( − 3)

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Chapter 5 : Functions & Graph Mathematics
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= 39 − 6 =x

k = 33 (g  f )(x) = g(ln(2x + 3))

Q17: 2013/2014 = eln(2x+3) − 3
Given f (x) = ln(2x + 3) and g(x) = ex − 3 . 2

2 = 2x + 3 − 3
2
(a) Show that f(x) is a one-to-one function
algebraically. =x
Since ( f  g)(x) = (g  f )(x) = x ,
(b) Find ( f  g)(x) and (g  f )(x) . Hence, state thus f and g are inverse for each other
(c)
the conclusion about the results.

(c) Sketch the graph of f(x) and g(x) on the
same axes. Hence, state the domain and
range of f(x).

(a) Algebraic Approach

Let ( 1) = ( 2)
(2 1 + 3 ) = (2x2 + 3 )
(2 1 + 3 ) = (2 2 + 3 )
2 1 = 2 2
1 = 2

∴ ( ) is one to one function

(b) =  − 3 , 
 2
Df , R f = (− , )

 e x − 3 
2
( f  g)(x) = f

2 e x − 3  
2 3
= ln + 

= ln(e x − 3 + 3)

= ln e x

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Chapter 5 : Functions & Graph Mathematics
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Q18 : 2013 / 2014 [ +8] =

Consider the function f (x) = 1+ ln( x), x  1. 4 −5
Determine f −1(x) and state its range. Hence,
evaluate f −1(3). = ( 4 −+58)+8

Given f (x) = 1+ ln( x), x  1 4( 4 −+58)−5
Let ( −1( )) =
( +8)+8(4 −5) =x
1 + [ −1( )] =
[ −1( )] = − 1 4 −5
−1( ) = −1 4( +8)−5(4 −5)
−1 = = [1, +∞)
4 −5
−1(3) = 3−1 = 2
2 +8 +32 −40 =
Q19 : 2013 / 2014
Given g(x) = kx + 8 , x  5 where k is a constant. 4 +32−20 +25

4x − 5 4 2 + 8 + 32 − 40 = 4 2 + 32 − 20 2 + 25

(a) Find the value of k if (g  g)(x) = x. 20 2 − 25 − 40 = 4 2 − 2 − 8

(b) Find the value of k so that g(x) is not a one- Comparing coefficients
to-one function.
x2 : 4k = 20
(a) Given (g  g)(x) = x. = 5

[ ( )] = x : k2 = 25
k = 5 ( coefficient of x2 is 5 )

constant : 8k = 40
k =5

Thus, k = 5

(b) Algebraic Approach
Let ( 1) = ( 2)

1+8 = 2+8

4 1−5 4 2−5

( 1 + 8)(4 2 − 5) = ( 2 + 8)(4 1 − 5)
4 1 2 − 5 1 + 32 2 − 40 =

4 1 2 − 5 2 + 32 1 − 40
5 1 − 5 2 + 32 1 − 32 2 = 0
5k( 1 − 2) + 32( 1 − 2) = 0

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Chapter 5 : Functions & Graph Mathematics
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( 1 − 2)(5 + 32) = 0

If ( ) is not one to one function, 1 ≠ 2 (b)
∴ 1 − 2 ≠ 0
= ( ) x=4
Therefore, for ( 1 − 2)(5 + 32) = 0
∴ 5 + 32 = 0 y=x

5 = −32 y=4

= −32 y=x

5 = −1( )

Thus, when = −32 , g(x) is not a one-to-one

5

function.

= −1 = (4 , +∞)

Q20 : 2013 / 2014 = (4 , +∞) −1 =
Given f (x) = e3x + 4 , x  R.
Q21 : 2014/ 2015
(a) Find f −1(x).
(a) Given f (x) = 2x +1and g(x) = x2 + 2x −1.
(b) On the same axes, sketch the graphs of f (x)
and f −1(x) . State the domain of f (x) and (i) Find ( f − g)(x).
(ii) Evaluate (3g − 2 f )(1).
f −1(x) .
(c) Given f (x) = 2x + 1 . State the domain
(a) Let ° −1( ) = 2
3 −1( ) + 4 =
3 −1( ) = − 4 and range of f (x) . Hence, on the same
3 −1( ) = ( − 4) axes, sketch the graph of f (x) and f −1(x) .
−1( ) = 1 ( − 4)

3

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Chapter 5 : Functions & Graph Mathematics
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(a) (i) ( − )( ) = ( ) − ( ) Q22 : 2014/ 2015
= (2 + 1) − ( 2 + 2 − 1) (a)Determine whether f (x) = 1 and
= 2 − 2
x−4
(ii) ( − )( ) g(x) = 4x +1 are inverse function of each
= 3g(1) − 2f(1)
= 3[12 + 2(1) − 1] − 2[2(1) + 1] x
= 3(2) − 2(3)
=0 other by computing their composite functions.

(b) Given f (x) = ln(1− 3x).

(b) f (x) = 2x + 1 (i)Determine the domain and range of f (x) .
2 Then sketch the graph of f (x) .

For f(x) to be defined, 2 + 1 ≫ 0 (ii) Find f −1(x) , if it exists. Hence, state the
domain and range of f −1(x) .
2
2 ≫ − 1
2
≫ − 1
4 (a) [ ( )] = [4 +1]

= [− 1 , +∞)

4 =1

4 + 1−4

= [ 0 , +∞ ) =1
4 +1−4


y=x

= 1

1



y=f

=

y =f -1 [ ( )] = [ 1 ]

-−41 −4
-−41
= 4( −14)+1
1
−4

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Chapter 5 : Functions & Graph Mathematics
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4+ −4 3 −1( ) = 1 −

= −4 −1( ) = 1 (1 = )
1 3
−4
−1 = = ( −∞ , +∞ )
= ∙ −4
1
−4 1 −1 = (−∞ , 3 )

=

Since ° ( ) = ° ( ) =

Therefore f(x) and g(x) are inverse for each other.

(b) Given ( ) = ( 1 − 3 ). Q23: 2015 /2016

(i) For f(x) to be defined, Given a function f (x) = 3 − 2x.

1 − 3 > 0 (a) Show that f is a one to one function.

−3 > −1 (b) Find the domain and range of f
< 1
(c) Determine the inverse functions of f and
3 state its domain and range.

= (−∞ ,1 ) , = (−∞, +∞ ) = (d) Sketch the graphs of f and f −1 on the same
= 1 axis.
3
3

y = ln(1-3x)

1 (a) Algebraic Approach
3 Let ( 1) = ( 2)
√3 − 2 1 = √3 − 2 2
(iii) Since f(x) is one to one function, therefore 3 − 2 1 = 3 − 2 2
f-1(x) exists. −2 1 = −2 2
Let [ −1( )] = 1 = 2
(1 − 3 −1( )) =
1 − 3 −1( ) = ∴ is one to one function

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Chapter 5 : Functions & Graph Mathematics
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(b) f(x) is defined when 3 − 2 ≫ 0 Q24 : 2015 / 2016

−2 ≫ −3 (a) The function f is given as

≪ 3 ( ) = + , ≠ .
−
2
If ( f  f )(x) = x , find the value of a.

= (−∞ ,3 ] , = [ 0 , +∞)

2 (b) Let f (x) = ln 3x + 2 and g(x) = e−x + 2 be

two functions. Evaluate (g  f )−1(3).

(c) Let [ −1( )] = (a)

√3 − 2 −1( ) = Given ( f  f )(x) = x

(3 −+42) =

3 − 2 −1( ) = 2 = ( −+ )+2

3( −+ )−4

2 −1( ) = 3 − 2 ( + 2) + 2(3 − 4)

3( 3 − 4 − 4) =
+ 2) − 4(3
−1( ) = 1 (3 − 2)
3 − 4
2

−1 = = [0 , +∞ ) 2 + 2 + 6 − 8
3 + 6 − 12 + 16 =

−1 = = (−∞ , 3] 2 + 2 + 6 − 8 = 3 2 − 12 2 + 22

2

(d) 12 2 − 16 − 8 = 3 2 − 2 − 2

y=x

= ( ) Comparing :
2 ∶ 3 = 12

= 4

∶ 2 = 16

y = −1( ) = 4 ( + )

Coefficient of 2 4

∶ 2a = 8
= 4 D

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Chapter 5 : Functions & Graph Mathematics
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(a) ( ° )( ) = [ln(3 + 2)] (c) Assume that the domain for g(x) is x  0 ,
= − ln(3 +2) + 2 determine g −1(x) and state its domain
= ln (3 +2)−1 + 2 and range.
1
= 3 + 2 + 2 (a) Let °[ −1( )] =

To find the ( ° )−1( ) [ −1( )]2+1 =

Let ( ° )°[( ° )−1( )] = x 5

1 + 2 = [ −1( )]2 + 1 = 5
3( ° )−1( )+2

[ −1( )]2 = 5 − 1

1 −1( ) = ±√5 − 1
3( ° )−1( ) + 2 = − 2

3( ° )−1( ) + 2 = 1 Since −1 = = [0 , +∞ )

−2

1 −1( ) = √5 − 1
− 2 − 2
3( ° )−1( ) = [ ( )] = 1 ( 2(3 −1) + 1)

Given 5 ,

( ° )−1( ) = 11 ⌈ ( )⌉2+1 = 1 ( 2(3 −1) + 1)
3 ( − 2 − 2)
55

⌈ ( )⌉2 + 1 = 2(3 −1) + 1

Therefore, ⌈ ( )⌉2 = ( − )

( ∘ )− ( ) = 1 ( 1 − 2) = −1
3 3−2 3 ⌈ ( )⌉2 = [ 3 −1]2

∴ ( ) = 3 −1 shown.

Q25 : 2016/ 2017 (b) [ (2)] = [22+1]

Given f (x) = x2 +1, x  0 5
5
= (1)
(a) Determine f −1(x) . Hence, if
= 3(1)−1
( )f g(x) = 1 e2(3x−1) +1 , show that
5 = 2
g(x) = e3x−1 = 7.389

(b) Evaluate g f (2) correct to three decimal

places.

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Chapter 5 : Functions & Graph Mathematics
SM015

(c) Given = [ 0 , + ∞) , thus g is one to one ( ° )( ) = [ ( )]
and g-1 exists.
= (| − 3|)
Let ° −1( ) =
= | −3|
3 −1( )−1 =
3 −1( ) − 1 = = { − ( − −33), , ≥ 3
3 −1( ) = 1 + < 3
−1( ) = 1+
(b) Graph Sketching
3

−1 = = [ −1, +∞)

−1 = = [0 , +∞)

= −( −3) = −3

Q26: 2016 /2017

Given f (x) = ex and g(x) = x − 3

(a) Show that ( ∘ )( ) = { − ( − −33) , ≥ 3
, < 3

(b) Sketch the graph of y = ( f  g)(x) . Hence, ( ° )−1 ( ) exists when (−∞ , 3] or
state the interval in which ( f  g)−1(x) exists. [3 , +∞)

(c) Determine ( f  g)−1(x) , for x  3 (c) To determine the ( f  g)−1(x) for

(d) Find the function h(x) for x  1 , given that [3 , +∞)
3 Let ( ° )[( ° )−1( )] =

(h  f )(x) = 2e x . Hence show that h(x) is one ( ° )−1( ))−3 =
( ° )−1( )) − 3 =
1 − 3e x ( ° )−1( ) = 3 +

to one function.

(a) | − 3| = {− ( − −3,3) , ≫3
< 3

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Chapter 5 : Functions & Graph Mathematics
SM015

(d) Given (h  f )(x) = 2e x −1 = = (2 , +∞)

1 − 3e x −1 = = (1 , +∞)

ℎ( ) = 2 Let ( −1( )) =
1−3 ln( −1( ) + 1) =
−1( ) + 1 =
To find h(x) , let = −1( ) = − 1
= ln
−1 = = ( 0 , +∞)
2 ln −1 = = (0 , +∞)
ℎ( ) = 1 − 3 ln

= 2

1−3

Thus, ℎ( ) = 2

1−3

Q27: 2017 / 2018 (b) (g  f )(x) = ( ( ))

Consider functions of f (x) = (x − 2)2 +1 , x  2 = [( − 2)2 + 1]
and g(x) = ln( x +1) , x  0 = (( − 2)2 + 1 + 1)
= (( − 2)2 + 2)
(a) Find f −1(x) and g −1(x) , and state the ( ∘ )( ) = ((2 − 2)2 + 2)
domain and range for each of the inverse = ln 2
function.
19
(b) Obtain (g  f )(x) . Hence, evaluate (g  f )(2) .

(a) Let ( −1( )) =

[ −1( ) − 2]2 + 1 =

[ −1( ) − 2]2 = − 1

−1( ) = 2 ± √ − 1

Since −1 = = (2 , +∞)

Thus, −1( ) = 2 + √ − 1

Chapter 5 : Functions & Graph Mathematics
SM015

Q28: 2017 / 2018 To find g −1(x) ,
Given the function g(x) = 1
Let °( −1( )) =
2x − 5
(a) Find the domain and range of g(x) . 1 =
(b) Show that g(x) is one –to-one function. 2 −1( )−5

Hence, find g −1(x) . 2 −1( ) − 5 = 1
(c) On the same axis, sketch the graph of g(x) a

g −1(x)
(d) Show that g  g −1(x) = x . 2 −1( ) = 5 + 1



−1( ) = 1 (5 +1)

2

= 5 +1

2

(c)
=

(a) g(x) = 1 is defined when
2x − 5

2 − 5 ≠ 0 = −1( )
= ( )
≠ 5 = −1( )

2

= ( −∞ , 5 ) ∪ (5 , +∞)

2 2

= ( −∞ ,0 ) ∪ (0, +∞)

= ( )

(b) Algebraic Approach (d) ° −1( ) = [ −1( )]
Let ( 1) = ( 2) = (1+5 )
1= 1
2
2 1−5 2 2−5
1
2 2 − 5 = 2 1 − 5 = 2 (1 +2 5 ) − 5
2 1 = 2 1
1 = 2 =1
1+5 −5
Thus, g(x) is one to one function

= = shown.

1

20

Chapter 5 : Functions & Graph Mathematics
SM015

Q29 : 2018 /2019 (a) (ii) Given ( f h) (x) = x

(a) Given f (x) = 5x +1 and g(x) = x +1 − 2 . (ℎ( )) =
4x x2 − 4 5ℎ( )+1 =

Find 4ℎ( )

(i) the domain of g(x) 5ℎ( ) + 1 = 4 ℎ( )
4 ℎ( ) − 5ℎ( ) = 1
(ii) h(x), if ( f h) (x) = x ℎ( ) = 1

(b) Given p(x) = ln(3x + 6) and q(x) = ex − 2 . 4 −5
3
(b) ° ( ) = ( − 2)
Show that p(x) and q(x) are inverse each
3
other.
= (3 ( − 2) + 6)
(a) (i)
For g(x) to be defined, 3
+ 1 ≫ 0 ∩ 2 − 4 ≠ 0
= ( − 6 + 6)
≫ −1 ∩ 2 ≠ 4 =
=
≫ −1 ∩ ≠ ±2 ° ( ) = (ln (3 + 6)
= − 2 ln (3 +6)
(−∞, −2) ∪ (−2,2) ∪ (2 , +∞ )
3
−2 − 1 2
= 3 +6 − 2
= [ −1 , 2 ) ∪ ( 2 , +∞ )
3

= 3 +6−6

3

= 3

3

=

Since ° ( ) = ° ( ) = ,
therefore p and q are inverse of each other.

21

Chapter 5 : Functions & Graph Mathematics
SM015

Q30 : 2019 / 2020 (b) From ( ) = 1
1+1+1 1
Given f ( x) = 1 .
1
1+ 1 Case 1 ∶ 1 + 1
1+


x

(a) Simplify f (x) and evaluate f  1  . Thus, ≠ 0
 2 

(b) The domain of f (x) is a set of real Case 2 ∶ 1 +
number except three numbers.
Determine the numbers. +1

Thus, + 1 ≠ 0
≠ −1

(a) ( ) = 1 +1
1+1+1 1 Case 3 ∶ 2 +1

= 1 Thus, 2 + 1 ≠ 0
≠ − 1
1+ 1
+1 2



= 1
1+ + 1

=1 The three numbers are :
+1+ ≠ 0 , ≠ −1 , ≠ − 1
+1
2

= +1

2 +1

f 1 = 21+1
 2  2(12)+1

3

=2

2

=3

4

22

Chapter 5 : Functions & Graph Mathematics
SM015

Q31 : 2019 / 2020 Domain

Given a function f (x) = ln(2x +1) −1 = = (−∞ , +∞)
(a) State the domain and range of f (x) .
(b) Find the inverse function of f (x) and state Range

its domain and range. Hence, find the value −1 = = (− 1 , +∞)
of x for which f −1(x) = 0 . 2
(c) Sketch the graphs of f (x) and f −1(x) on the
same coordinate axes. When f −1(x) = 0
1 ( − 1) = 0
(a) Given f (x) = ln(2x +1)
2

− 1 = 0
= 1
= ln 1 = 0

Domain

f(x) is defined when 2 + 1 > 0 (c)

> −1

2

= ( −1 , +∞ )

2

Range

= (−∞ , + ∞)

(b) Let ( −1( )) =
(2 −1( ) + 1 ) =
2 −1( ) + 1 =
2 −1( ) = − 1
−1( ) = 1 ( − 1)

2

23