function

Notes of Optional Mathematics for SEE competitors

Chapter-1
Algebra

1.2 Composite Function
Composite Function: Let : → and : → are any two functions, then the new function defined

from → is called composite function of f and g (not g and f). It is denoted by gof and read as g oh f or
g(f). Composite function is also known as product function or function of a function.

Suppose : → and : → are defined by the following function:

Note: in gof(x), function f(x) act as the main function followed by g(x) function whereas in fog(x), function
g(x) act as the main function followed by f(x) function.
For finding gof: From above figure we can write the value of f and g in ordered pair as shown below.
f = {(1,2), (3,4), (5,6)}
g= {(2,5), (4,6), (6,8)}
Now, gof (1) = g [f (1)] = g [2] =5 gof (3) = g [f (3)] = g [4] =6 gof (5) = g [f (5)] = g [6] =8
Therefore, gof = {(1,5), (3,6), (5,8).

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Suppose : → and : → are defined by following function where f= {(0,0), (1,1), (2,4), (3,9)} and
g= {(1,0), (2,1), (3,2), (4,3)}. Find fog using arrow diagram.

Using arrow diagram,

fog = {(1,0), (2,1), (3,4), (4,9)}.

Without using arrow diagram,

(1) = [ (1)] = [0] = 0 (2) = [ (2)] = [1] = 1

(3) = [ (3)] = [2] = 4 (4) = [ (4)] = [3] = 9

Therefore, fog = {(1,0), (2,1), (3,4), (4,9)}.

Alternative way (Changing given order pair into function and then finding fog)
f= {(0,0), (1,1), (2,4), (3,9)}. This order pair is defined by function ( ) = 2

g= {(1,0), (2,1), (3,2), (4,3)}. This order pair is defined by function ( ) = − 1
Now, fog(x) = f [ g(x)] = f [ − 1] = ( − 1)2
When x = 1, (1) = (1 − 1)2 = 0
When x = 2, (2) = (2 − 1)2 = 1
When x = 3, (3) = (3 − 1)2 = 4
When x = 4, (4) = (4 − 1)2 = 9

Therefore, fog = {(1,0), (2,1), (3,4), (4,9)}.

Combination of Function: The sum, difference, product and quotient of the function is called

combination of the function.

Sum: ( + ) ( ) = ( ) + ( ) Difference: ( − ) ( ) = ( ) − ( )

Product: ( × ) ( ) = ( ) × ( ) Quotient: ( / ) ( ) = ( )/ ( ); ( ) ≠ 0

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Example: Let f: R→ and g: R→ be defined by ( ) = 2 − 1 and ( ) = 3 2 − . Find

a) ( − 3)( ) b) ( 3)( )
c) (3 + 4 )( ) d) ( )( ) ( ≠ 0, 1)

3

Solutions:

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

b) ( 3)( ) = ( × × )( ) = ( ) × ( ) × ( ) = (2 − 1)(2 − 1)(2 − 1) = (2 − 1)3

c) (3 + 4 )( ) = 3 ( ) + 4 ( ) = 3(2 − 1) + 4(3 2 − ) = 12 2 + 2 − 3

d) ( ) ( ) = ( ) = 2 −1
( ) 3 2−


Try Yourself: Let f: R→ and g: R→ be defined by ( ) = 4 3 − 1 and ( ) = 3 3. Find

a) (2 − 5)( ) b) ( 2)( ) c) (3 − 4 )( )

d) ( )( ) e) ( )( ) ( g(x) ≠ 0)



Decomposition of Function: The process of decomposing of composite function into its component

function is called decomposition of functions.

For the decomposition of composite functions, we have to look for an inner function and an outer function.

Example: Decompose the composite function 1 .( There are many ways to decompose composite
( −2)2

function so we have to approach any one of them.)

Solution:

Let us consider the composite function be fog(x)

Then fog(x)= ( −12)2, here we have to find f(x) and g(x) separately to decompose given composite function.

In order to find f(x) and g(x), we have to find an inner function and outer function.

Now,

1

fog(x)= ( −2)2

Here g(x) function act as the main function so inner function must be ( ) = − 2, g(x) is followed by f(x).

Therefore, f(x) = 1 2.

Checking of answer:

1

( ) = [ ( )] = ( − 2) = ( −2)2

Example: Decompose the composite function √ 2 − 9.

Solution:

Let fog(x)= √ 2 − 9
Here g(x) act as main function so inner function must be ( ) = 2 − 9.

Therefore, f(x) = √ .

Checking of answer:

fog(x)=f [ g(x)] = f ( 2 − 9)= √ 2 − 9

Try yourself: Decompose the composite functions: a) (1 − )2 b) √5 −

c) ( +2)2

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Properties of Composite Function 2. ( ) = ( )( ) = [ ( )]
4. 3( ) = ( ) = [ ( ( ))]
1. ( ) = ( )( ) = [ ( )]
3. 2( ) = ( ) = [ ( )] 6. ( ± ) = ( ) ± , where k is constant.
5. ℎ ( ) = (ℎ )
7. ( ) ( ) = ( )

Domain and Range of a Function

A function is defined from set A to set B i.e., f: A→ then the set A is known as domain of function denoted
by Dom(f) and defined by
Dom(f) = {x: x A}
The set of values y=f(x) B for every x is known as range of the function. Range is also defined as the
image of domain. The set B is the co-domain of the function and the range is the subset of co-domain of a
function. It is denoted by R(f) and defined by
R(f)= {y: y ,y=f(x) for all x }

Domain and Range of a composite function

Domain of fog function is equal to domain of g function and range of fog is equal to range of f function.
Domain of gof function is equal to domain of f function and range of gof is equal to range of g function.

In the given figure, A= {1,2,3,4} is the domain of the function g and B= {0,1,2,3} is the range of function g.
Similarly, B= {0,1,2,3} is the domain of the function f and C= {0,1,4,9} is the range of function f.
Also, A= {1,2,3,4} is the domain of the composite function fog(x) and C= {0,1,4,9} is the range of fog(x).,
Example: If f= {(m, n), (a, b)} and g= {(n, a), (b, c)} then what is the range of fog.
Solution: Range of fog = range of f function = {n, b}.
Give a try: Find out the domain and range of function gof.

Inverse Function

Let f: A→B be one to one onto, then a function can be defined from B to A such that every element of B
associate with a unique element of A, then the function defined from B to A is known as the inverse function f
and is denoted by −1.
An inverse of a function f: A→ exists only when the function is one to one onto (bijective).
Example: If f = {(a, x), (b, y), (c, z)} then −1= {(x, a), (y, b), (z, c)}

Note: Relation between identity function and its inverse is given by [ −1( )] = −1[ ( )] =

One to one onto (bijective): A function that is one to one (injective) and onto (surjective) is called a bijective
function.

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Very Short Questions.

➢ If f: A→ and h: → be two functions then what will denote the composite function from → ?
 hof(x) will denote the composite function from → .
➢ Which range of the functions between f and g is equal to the range of the functions gof(x)?
 Range of function g is equal to the range of function gof(x).
➢ Which range of the functions between f and g is equal to the range of the functions fog(x)?
 Range of function f is equal to the range of function fog(x).
➢ Which domain of the functions between f and g is equal to the domain of the functions gof(x)?
 Domain of function f is equal to the domain of function gof(x).
➢ Which domain of the functions between f and g is equal to the domain of the functions fog(x)?
 Domain of function g is equal to the domain of function fog(x).
➢ What is the necessary condition to find the composite function of two functions?
 The necessary condition to find the composite function of two function is that range of the inner function

should be the subset of the domain of the outer function.

➢ In what condition, fog(x) and gof(x) are equal?
 fog(x) and gof(x) are equal when both of them are identity function i.e., fog(x)=gof(x)=x. In other word

when function f(x) and g(x) is inverse to each other fog(x) and gof(x) are equal.

➢ If h = {(1, 2), (3,5), (4,1)} and g= {(2,3), (5,1), (1,3)}, then find goh (3).

 Solution: goh (3) = g [h (3)] = g (5) =1

➢ If f = {(2,5)} and g = {(5,7), what is the order pair of gof?

 Solution: gof (2) = g [ f (2)] = g (5) = 7. Therefore, gof = {(2,7)}.

➢ If f(x) = x + 2, then find fof(x).

 Solution: fof(x) = f(x+2) = (x +2) + 2 = x + 4

➢ If f(x) = 2x and g(x) = 2x + 2, then find fog(x).

 Solution: fog(x) = f(2x+2) = 2(2x+2) = 4x + 4

➢ If f = {(m, n), (a, b)} and g= {(n, a), (b, c)}, then find the range and domain of gof.

 Solution: Range of gof = Range of g = {a, c} Domain of gof = Domain of f = {m, a}

➢ If f(x) = x and g(x) = x + b, then find gof(x) and fog(x).

 Solution: gof(x) = g(x) = x + b fog(x)=f (x+ b) = x + b.

➢ Under what condition, the inverse of a function is possible?

 The inverse of a function is possible only when the function is one to one onto function.

➢ What is the relation between a function and its inverse?

 [ −1( )] = −1[ ( )] =

➢ If −1= {(a, x), (b, y), (c, z)} then find the function f.

 Solution: f = {(x, a), (y, b), (z, c)}

➢ If f: → and g: →R are inverse to each other then for each a , what is the value of gof(a)?

 gof(a)=a

➢ If f: → and g: →R are inverse to each other then for each a , what is the value of fog(a)?

 fog(a)=a

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Short Questions
Type-1
➢ If f = {(3, 4), (4, 5), (5, 6)} and g = {(2, 3), (3, 4), (4, 5)}, then show the composite function fog in an arrow

diagram and find it in ordered pair.

 Solution:

Therefore, fog = {(2, 4), (3, 5), (4, 6)}

➢ If f = {(0, 0), (1, 1), (2, 4), (3, 9)} and g = {(1, 0), (2, 1), (3, 2), (4, 3)} then show the composite function

fog in an arrow diagram and find it in ordered pair.

 Solution:

Therefore, fog = {(1, 0), (2, 1), (3, 4), (4,9)}

➢ If f = {(1, 2), (2, 3), (3, 4)} and g = {(2, a), (4, c), (3, b)} then find the composite function gof in ordered

pair.

 Solution:

gof (1) = g [f (1)] = g (2) = a
gof (2) = g [f (2)] = g (3) = b
gof (3) = g [f (3)] = g (4) = c
Therefore, gof = {(1, a), (2, b), (3, c)}

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Type-2
➢ If ( ) = + 1 ( ) = 2 + 1,find the value of gof(x).
 Solution:

Given, f(x) = x + 1 & g(x)=2x+1. To find gof(x)=?

Now,

gof(x)= g[f(x)] = g(x+1) = 2(x+1) +1 = 2x+2+1= 2x+3

∴ gof(x) = 2x+3

➢ If g(x) = +2 ℎ( ) = 3 − 2, then find goh(x).
5

 Solution:

Given, h(x) = 3x-2 & g(x) = +2. To find: goh(x)=?

5

Now,

goh(x)= g[h(x)] = g(3x-2) = (3 −2)+2 = 3 −2+2 = 3
5 5 5

∴ goh(x) = 3
5

➢ If h(x) = 3x + 4 then find the value of hoh (3).

 Solution:

Given, h (x) = 3x + 4, To find: hoh (3) =?

Now,

h (3) = 3 × 3 + 4 = 13

∴ hoh (3) = h [h (3)] = h (13) = 3 × 13 + 4 = 43

Alternative way,

hoh(x) = h [h(x)] = h(3x+4) = 3(3x+4) + 4 = 9x + 12 + 4 = 9x + 16

∴ hoh (3) = 9 × 3 + 16 = 27 + 16 = 43

➢ If f(x) = x – 2 and g(x) = 3x + 1, then find gf(x) and gf (2)
 Solution:

Given, f(x) = x – 2 & g(x) = 3x + 1, To find gf(x)=? & gf (2) =?

Now,
gf (x) = g(x-2) = 3(x-2) + 1 = 3x – 6 + 1 = 3x – 5

gf (2) = 3×2 – 5 = 1

➢ If f(x) = 3x + 15 and h(x) = 3x + 2, find the value of foh(−2).
 Solution:

f(x) = 3x + 15 & h(x) = 3x + 2, To find: foh(−2) =?

Now,

h(−2) = 3(−2) + 2 = −6 + 2 = −4

∴ foh (−2) = f (−4) = 3(−4) + 15 = −12 + 15 = 3

➢ If f(x) = 2x + 5 and g(x) = −25,prove that fog(x) is an identity function.
 Solution:

fog(x) = f[g(x)] = ( −5 ) = 2( −5 )+5 = x – 5 + 5 = x. Proved
2 2

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Type-3

➢ If f(x) = 3x + b and ff (2) = 12, find the value of b.

 Solution:

ff (2) = 12 ⇒ f [ 3(2) + b] = 12 ⇒ f (6 + b) = 12 ⇒ 3(6 + b) +b = 12 ⇒ 18 + 4b = 12 ⇒ 4b = – 6 ⇒ b= −3
2

Alternative way,

ff(x) = f(3x+b) = 3(3x+b) + b = 9x + 4b

ff (2) = 9(2) + 4b = 18 + 4b

According to question,

ff (2) = 12

∴ 18 + 4b = 12 ⇒ 4b = 12 – 18 ⇒ 4b = -6 ⇒ b = −3
2

➢ If f(x)= 4x and g(x) = x + 1 and fog(x) = 20, find the value of x.

 Solution:

Given, fog(x)= 20

⇒ f[g(x)] = 20

⇒ f (x+1) = 20

⇒ 4(x+1) = 20

⇒ 4x + 4 = 20

⇒ 4x = 16

⇒x=4

∴ The required value of x is 4.

➢ 2

If f(x) = +3 and g(x) = then find the domain in 4 fog(x) = 1.

 Solution:

2 2 2
2+3
fog(x) = f ( ) = =

2 +3

According to question,

4 fog(x) = 1

⇒ 4 2 = 1 ⇒ 8 = 1 ⇒ 8 = 2 + 3x ⇒ 6 = 3x ⇒ x = 2
2+3 2+3

∴ 2 is the required domain.

➢ If f(x) = 2x + 5a -1 and fof (3) = 4. Find the value of a.

 Solution:

f (3) = 2(3) + 5a -1 = 6 + 5a -1 = 5a + 5

we have,

fof (3) = 4

⇒ f(5a+5) = 4
⇒2(5a+5) + 5a – 1 = 4
⇒10a + 10 + 5a – 1 = 4

⇒15a = - 5

⇒ a = − 1
3

∴ The value of a is − 1 .
3

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Type-4

➢ If f(x) = 4x + 5 and fog(x) = 8x + 13. Find g(x).
 Solution:

We have, fog(x) = 8x + 13
⇒4g(x) + 5 = 8x + 13
⇒4g(x) = 8x + 13 – 5
⇒4g(x) = 8x + 8
⇒g(x) = 2x + 2

∴ ( ) = 2 + 2

➢ If f: R→R: f(x) = −3 and fog: R→R: (fog(x)) = x. Then find g(x).
2

 Solution:

We have, fog(x) = x

⇒ ( )−3 = x
2

⇒g(x) – 3 = 2x

⇒g(x) = 2x + 3

∴ ( ) = 2 + 3

➢ If f: N→R: f(x) = 2and g: R→R (fog(x)) = 2 − 2 + 1 then find g(x).
 Solution:

We have, fog(x) = 2 − 2 + 1
⇒ ( ( ))2 = ( − 1)2

⇒ ( ) = − 1

∴ ( ) = − 1

➢ If fog(x) = 2x + 13 and g(x) = +1, find f (1).

2

 Solution:

We have, fog(x) = 2x + 13

⇒ ( +21) = 2x + 13

⇒ ( +21) = 4 (2 ) + 13

⇒ ( +21) = 4 ( +12−1) + 13

⇒ ( +2 1) = 4 ( +2 1) − 4 + 13
2

⇒ ( +21) = 4 ( +21) – 2 + 13

⇒ ( +21) = 4 ( +21) + 11

⇒ ( ) = 4 + 11

∴ f (1) = 4 × 1 + 11 = 15.

Give a try:

➢ If g(x) = 4 – x and fog(x) = 11 – 2x then find f(x).
➢ If f(x) = 3x – 2 and fog(x) = 6x – 2, find g(x).

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Type-5

➢ If h(x) = (2x – 3 )2 and h(x) = fog(x) then finds any two possible formula of f(x) and g(x).
 Solution:

Given, h(x) = fog(x) = (2x – 3) 2

Here g(x) act as the inner function and f(x) act as outer function.
Let g(x) = (2x – 3)
Then, f(2x-3) = (2x – 3) 2
∴ f(x) = x2
∴ one possible formula of f(x) is x2 and g(x) is (2x – 3).

Another possible formula:
Let g(x) = (2x – 3) 2
Then, f [(2x – 3) 2] = (2x – 3) 2
∴ f(x) = x
∴ Another possible formula of f(x) is x and g(x) are (2x – 3) 2

: ℎ ( ) ( ).

➢ If h(x) = 1 where x≠ -3 and h(x) = fog(x) then finds the possible formulas of f(x) and g(x).
( +3)2

 Solution:

Given, h(x) = fog(x) = 1 , x≠ -3
( +3)2

Here g(x) act as inner function and f(x) act as outer function.
Let g(x) = ( + 3)2

Then, f (( + 3)2 = 1
( +3)2

∴ f(x) = 1


∴ One possible formula of f(x) is 1 and g(x) is ( + 3)2


Another possible formula:

Let g(x) = x + 3

Then, f (x + 3) = 1
( +3)2

∴ f(x) = 1
2

∴ Another possible formula of f(x) is 1 and g(x) is x + 3.
2

Give a try:

➢ If h(x) = (3 + 2)3 and h(x) = fog(x), find the possible formulas of f(x) and g(x).

➢ If p(x) = 1 where x≠ -2 and p(x) = fog(x), find the possible formulas of f(x) and g(x).
( +2)3

➢ Decompose the function √5 − 2 as the composition of two function.

➢ 4

Decompose the function 3−√(4+ 2) as the composition of two function.

Hint: let fog(x) be two composite function of f and g. Let g(x) = 4 + 2. Then find f(x).

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Long Questions
Type-1

➢ If f(x) = 2x + 1 and g(x) = x2 – 2, then find a) fof(x) b) gof(x) c) fog(x) d) gog(x)
 Solution:

Given, f(x) = 2x + 1 & g(x) = x2 – 2

a) fof(x) = f[f(x)] = f (2x + 1) = 2(2x + 1) + 1 = 4x + 3
b) gof(x) = g[f(x)] = g (2x + 1) = (2x + 1)2 –2 = 4x2 + 4x + 1 – 2 = 4x2 + 4x – 1
c) fog(x) = f[g(x)] = f (x2 – 2) = 2(x2 – 2) + 1 = 2x2 – 4 + 1 = 2x2 – 3
d) gog(x) = g[g(x)] = g (x2 – 2) = (x2 – 2)2 – 2 = x4 – 4x2 + 4 – 2 = x4 – 4x2 + 2

➢ If f(x) = 3x + 2 and g(x) = 2x2 – 1 then find a) fog(4) b) gof (2) c) fof(1) d) gog(0)
 Solution:

Given, f(x) = 3x + 2 & g(x) = 2x2 – 1
a) fog(4) = f [g (4)] = f[2(4)2 – 1] = f(31) = 3(31) + 2 = 93 + 2 = 95
b) gof(2) = g[f(2)] = g[3(2) + 2] = g(8) = 2(8)2 – 1 = 127

c) fof(1) = f[f(1)] = f[3(1)+2] = f(5) = 3(5)+2 = 17
d) gog(0) = g[g(0)] = g[2(0)2 – 1] = g(-1) = 2(-1)2 – 1 = 2 – 1 = 1

Another way: To find fog(4), first find fog(x) and then put x =4. Same goes for other questions.

➢ If f(x) = 3x+4 and g(x) = 2(x+1) then prove that fog(x) = gof(x).
 Solution: g(x) = 2x + 2

L.H.S

fog(x) = f[g(x)] = f(2x+2) = 3(2x+2) + 4 = 6x+6+4 = 6x+10

R.H.S

gof(x) = g[f(x)] = g(3x+4) = 2(3x+4) +2 = 6x + 8 + 2 = 6x+10

∴ L.H.S = R.H.S Proved.

➢ If f(x) = 3x2 + 5x and g(x) = 4x2 – 5 then prove that fog(4) = 3f(4)g(4) – 1298.
 Solution:

L.H.S = fog(4) = f[g(4)] = f[4(4) 2 – 5] = f(59) = 3(59) 2 + 5(59) = 10738
R.H.S = 3f(4)g(4) – 1298 = 3{3(4) 2+5(4)}{4(4) 2 – 5} - 1298 = 3(68)(59) – 1298 = 10738

∴ L.H.S = R.H.S Proved.

➢ Show that fog(x) = gof(x) = x.

 Solution:

a) f(x) = 2x – 6 and 2g(x) = x+ 6 b) f(x) = 4 – 3x and 3g(x) = 4 – x

L.H.S. = fog(x) = f( +26)= 2( +26) – 6 = x L.H.S. = fog(x) = f(4−3 ) = 4 - 3(4−3 )= x

R.H.S. = gof(x) = g(2x – 6) = (2 −6)+6 = x R.H.S. = gof(x) = g(4 – 3x) = 4−(4−3 ) = x
2 3

∴ fog(x) = gof(x) = x Proved. ∴ fog(x) = gof(x) = x Proved.

c) f(x) = ax + b and a g(x) = x – b d) f(x) = 3x – 4 and 3 g(x)= 4 + x

L.H.S. = fog(x) =f( − ) = a( − ) +b = x L.H.S.= fog(x) = f(4+3 )=3(4+3 ) – 4 = x

( + )− R.H.S. =gof(x) = g(3x – 4) = 4+(3 −4) = x
3
R.H.S. = gof(x) = g(ax + b) = = x
∴ fog(x) = gof(x) = x Proved. ∴ fog(x) = gof(x) = x Proved.

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Type-2

➢ 6 ( x ≠ 2, ∈ R), g(x) = ax2 – 1 and gof(5) = 7 then find the value of a.

If f(x) = −2

 Solution:

6 ∴f(5) 6 = 2

We have, f(x) = −2 =5−2

Also, gof(5) = 7 ⇒ g(2) = 7 ⇒ a (2)2 – 1 = 7 ⇒ 4a – 1 = 7 ⇒ 4a = 8 ⇒ a = 2

∴ The value of a is 2.

➢ If f(x) = 2x3 – 3x2 + 4x – 1 and g(x) = 2, then find the value of x such that fog(x) + gof(x) = x + 3.
 Solution:

fog(x) = f[g(x)] = f(2) = 2(2)3 – 3(2)2 + 4(2) – 1 = 16 – 12 + 8 – 1 = 11

gof(x) = g[f(x)] = 2 [ since g(x) is a constant function]

We have,

fog(x) + gof(x) = x + 3

⇒ 11 + 2 = x + 3
⇒13 – 3 = x

⇒ x = 10

∴ The value of x is 10.

➢ If f(x) = ax + 5, fog(x) = 8x + 13 and gof(5) = 8a then find the value of a.
 Solution:

We have, Now, f(5) = a(5) + b = 5a + 5 ⇒ 40a + 48 = 8a2
fog(x) = 8x + 13 Given, ⇒ 8a2 – 40a - 48 = 0
⇒ f[g(x)] = 8x + 13 gof(5) = 8a ⇒ a2 – 5a – 6 = 0
⇒ a g(x) + 5 = 8x + 13 ⇒ g(5a +5) = 8a ⇒ a2 – 6a + a – 6 = 0

⇒ a g(x) = 8x + 8 ⇒ 8(5 +5)+8 = 8a ⇒(a – 6) (a + 1) = 0
Either a = 6 or a = -1

⇒ g(x) = 8 +8 ⇒ 40a + 40 + 8 = 8a2


∴ The value of a is 6 or -1.

➢ If f(x) = , g(x) = 11 – bx2, fof(5) = 4 and fog(2) = 1 , find the values of a and b.
3− 5 2

 Solution:

∴ f(5) = = −
3−5 2
We have, f(x) = 3−

Similarly, g(x) =11 – bx2 ∴ g(2) = 11 – b(2)2 = 11 – 4b

Now, Also,

4 1

fof(5) = 5 fog(2) = 2

⇒ f (−2 ) = 4 ⇒ f(11 – 4b) = 1
5 2

⇒ =45 ⇒ = 1
3−(11−4 ) 2
3− −
2

⇒ 5a = 4(6 + ) ⇒ 2a = - 8 + 4b
2 ⇒ 2(4) + 8 = 4b

⇒ 5a = 12 + 2a

⇒ 3a = 12 ∴ a = 4 ⇒ 4b = 16 ∴ b = 4

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Type-3

➢ If f(x) = 3x – 2 and fog(x) = 6x – 2 then find the value of x such that gof(x) = 8.
 Solution:

We have, Also,
fog(x) = 6x – 2
⇒f[g(x)] = 6x – 2 gof(x) = 8
⇒3g(x) – 2 = 6x – 2 ⇒g(3x-2) = 8
⇒ 2(3x-2) = 8
⇒3g(x) = 6x ⇒ 6x – 4 = 8

⇒g(x) = 2x ⇒ 6x = 12

∴x=2

➢ If f(x) = 4x + 2, fog(x) = 8x + 14 and gof(x) = 18 then find the value of x.
 Solution:

We have, Also, gof(x) = 18

fog(x) = 8x + 14 ⇒ g(4x+2) = 18

⇒4g(x) + 2 = 8x + 14 ⇒ 2(4x+2) +3 = 18

⇒ 4g(x) = 8x + 12 ⇒ 8x +4 +3 = 18

⇒ g(x) = 2x + 3 ⇒ x = 11
8

∴ 11

x= 8

➢ If f(x) = 2x +1, fog(x) = 2x + 11 and gof(x) = 8 then find the value of x.

 Solution:

We have, Also, gof(x) = 8

fog(x) = 2x + 11 ⇒ g(2x+1) = 8

⇒ 2g(x) + 1 = 2x + 11 ⇒ (2x+1) +5 = 8

⇒ 2g(x) = 2x + 10 ⇒ 2x +1 +5 = 8
⇒ g(x) = x + 5 ⇒2 x = 8 – 6

∴ x=1

➢ If f(x) = 3x + 5 and fog(x+2) = 12x + 17, find the value of x such that gof(x) = 88.
 Solution:

We have, Also, gof(x) = 88

fog(x+2) = 12x + 17 ⇒g(3x+5) = 88
⇒ 3g(x+2) +5 = 12 x + 17 ⇒ 4(3x+5) – 4 = 88

⇒3g(x+2) = 12x + 12 ⇒12x + 20 = 92

⇒3g(x+2) = 12x + 24 – 12 ⇒ 12x = 72

⇒3g(x+2) = 12(x +2) – 12 ⇒x=6

⇒g(x) = 12 −12 = 4x – 4 ∴ x=6
3

Give a try:

➢ If f(x) = 4x + 5 and fog(x) = 8x + 13, find the value of x such that gof(x) = 28.

➢ If f(x) = 3x – 2 and fog(x+1) = 6x + 4, find the value of 2x such that gof(x) = 8.

➢ If g(x) = 2x and fog(x) = 6x – 2 then find the value of x such that gof(x) = 10.

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

Some Extra Questions

➢ Function f(x-1) = 3x -4 is given. If f[g(x)+5] = 3x + 2, find g(8).
 Solution:

We have, f(x – 1) = 3x – 4 = 3(x – 1) – 1
∴ f(x) = 3x – 1

Now, f[g(x)+5] = 3x + 2

⇒ 3[g(x) + 5] – 1 = 3x + 2

⇒ 3g(x) + 15 – 1 = 3x + 2
⇒ 3g(x) = 3x – 12

⇒ g(x) = x – 4

∴ g(8) = 8 – 4 = 4

➢ Try: Function f(x+2) = 3x + 5 is given. If f[g(x) + 2] = x + 3, find g(8).
➢ If f(x)= 3x2 – 7 and g(x) = 2x + a, find value of ‘a’ so that graph of fog(x) crosses the y- axis at 68.
 Solution:

fog(x)= f(2x+a) = 3(2x + a)2 – 7

When fog(x) crosses y-axis, then x = 0.

So, fog(0) = 68

⇒ 3[2(0) + a]2 – 7 = 68

⇒3a2 = 75

⇒ a2 = 25

⇒ a2 = (± 5)2

∴a=±5

➢ Try: If f(x)= 2x2 + 5 and g(x) = 3x + a, find value of ‘a’ so that graph of fog(x) crosses the y- axis at 23.

➢ If f(x) = kx + 3, g(x)= 24 ++ 2 , fof(5) = 57 and gof(4) = 2 then find the value of m and k.
 Solution:

We have, f(x) = kx + 3 ∴ f(5) = 5k + 3 ∴ f(4) = 4k + 3

4 +2

g(x) = 2 +

Now, fof(5) =57 Also, gof(4) = 2

⇒ f(5k + 3) = 57 ⇒ g(4k +3) = 2
4(4 +3)+2
⇒ k (5k + 3) + 3 = 57
⇒ 5k2 +3k – 54 = 0 ⇒ 2(4 +3)+ = 2
16 +12+2
This is quadratic in k.
⇒ 8 +6+ = 2
−3±√32−4(5)(−54)
⇒ 16k + 14 = 16k = 12 + 2m
k = 2(5) ⇒ 2m = 2
⇒m=1
⇒k = −3±√9+1080 ∴m=1
10

⇒k = −3±√(33)2

10

⇒ k = −3±33
10

Taking +ve sign Taking – ve sign

∴k=3 ∴ k = - 3.6

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

➢ Try: If f(x) = 3x + k, g(x)= 42 ++ 1 , fof(2) = 30 and gof(1) = 2 then find the value of m and k.
➢ If f, g: → defined by, f(x) = x3 + 2 and g(x) = 4x – 1, find fog(x) and gof(x), and show that the composite

functions are not commutative.

 Solution:

fog(x) = f(4x-1) = (4x-1) 3 + 2 = 64x3 – 48x2 + 12x + 1
gof(x) = g(x3 + 2) = 4(x3 + 2) – 1 = 4x3 + 7
fog(x) ≠ gof(x). So, the composite functions are not commutative.

➢ If f(x) = 2x + 3, gof(x) = 2x – 1 then find fog(2) and gof(2).
 Solution:

We have, gof(x) = 2x – 1
⇒g(2x+3) = (2x + 3) – 4
⇒g(x) = x – 4

Now,
g(2) = 2 – 4 = -2

f(2) = 2(2) + 3 = 7

∴ fog(2) = f(-2) = 2(-2) + 3 = -1

∴ gof(2) = g(7) = 7 – 4 = 3

➢ Try: If f(x) = x2, gof(x) = x2 – 3 then find fog(2) and gof(2).
➢ If a function f(x) = 5 then find f(4) and fof(-5).
 Solution:

Given f(x) = 5

∴ f(4) = 5

∴ fof(-5) = f(5) = 5 [ since f(-5) = 5]

➢ If f(x) = 3 and g(x) = 2x – 1 then prove that fog(x) is a constant function.
 Solution:

fog(x) = f(2x-1) = 3

∴ fog(x) is a constant function.

➢ Try: If f(x) = 4x + 5 and g(x) = 15 then prove that fog(x) is a constant function.

➢ If f(a) = , g(x) = 5x – 7 and f[g(x)] = 15x – 21 then find the value of m.
 Solution:

∴ f(x) =

f(a) =

We have, f[g(x)] = 15x – 21

⇒ f(5x – 7) = 15x – 21

⇒ 5 +7 = 3(5x+7)


⇒ m = 1
3

➢ If f(x) = √ − ; x ∈ N then prove that fof(x) = x.

 Solution:

fof(x) = f( √ − ) = √ − ( √ − ) = √ − + = √ = x. Proved

Prepared by: Prabin Poudel

Notes of Optional Mathematics for SEE competitors

➢ If f(x) = 1+ then find fof(tan ).
1−

 Solution:

fof(tan ) = f(11+− ) = 1+11+− = 1− +1+ =−2 2 = -cot
1−11−+ 1− −1−

➢ Try: f(x) = 1+ , find the value of fof(sin )

−1

➢ If 2f(x)+ f ( 1 ) = 6x 3 then find the value of fof(2).

+

 Solution:

Given, 2f(x)+ f ( 1 ) = 6x + 3 ...................(i)


11

Replacing x by and by x we get

2f( 1 ) + f(x) = 6 + 3x ...................................(ii)


Multiplying equation (i) by 2 we get

4f(x)+ 2f ( 1 ) = 12x + 6 ..............................(iii)


Solving equation (ii) and (iii) we get

4f(x)+ 2f ( 1 ) = 12x + 6


2f( 1 ) + f(x) = 6 + 3x


-- --

3f(x) = 9x

∴ f(x) = 3x

∴f(2) = 3(2) = 6

∴fof(2) = f(6) = 3(6) = 18

➢ Try: If 2f(x) - f ( 1 ) = 6x - 3 then find the value of fof(2).


➢

If k(x) = and g(x) = 2x + 5 then what value of r makes 3[kg(x)] = 2x + 5?

 Solution:

We have, 3[kg(x)] = 2x + 5

⇒3[k(2x+5)] = 2x + 5

⇒ 3[ 2 +5 ] = 2x + 5


⇒r=3

➢ Try: If f(x) = ax + 3 and fof(x) = a2x then find the value of a.

➢ −1 −1

If f(x) = +1 then show that fof(x)=

➢ If f(x) = 3 and g(x) = 2 then find
+2
2

a) fog(4) b) gof(2) c) fof(1) d gog(0)

Prepared by: Prabin Poudel