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Functions - [A]
FUNCTIONS Section - 1

1.1 Introduction to Functions

Functions provide us with a convenient way to handle the relationship between the values of one variable
quantity that depends on the values of another variable quantity. For example, let us imagine a spherical
rubber balloon into which air is being pumped. The radius of the balloon (r) is changing with time t. In
mathematics, we say that r is a function of time t and symbolically, it may be written,
r = f (t) : r is a function of time t.

Similarly the volume of the balloon also depends on time t. Hence we can write
V = g (t) : V is function of time t.

Different letters f and g are used because they represent different mathematical relations.
In general, if the values of a variable y depend on the values of another variable x, we write
y = f (x) i.e., y is a function of x.
Explanation : Let y = f (x) be a real valued function.
Here f (x) can be understood as an operator between x and y. ‘x’ is feeded as input and ‘y’ is the corresponding
output.

Consider y = f (x) = x + 1

i.e. x (input) 0 –3 1
y (output) 1 –2 2

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Ordered Pair : The combination of an input and output is called an ordered pair.
Representation of Ordered Pair : (x , y)
 
input output
So in above example orderd pairs are (0, 1), (–3, –2), (1, 2) etc., which satisfy the function:

1.2 Graph of a Function

It is the pictorial representation of a function. It is formed by plotting ordered
pairs that satisfy function.

Let y  f (x)  x  1. Graph of y  f (x) is shown :

y = f (x) is a linear polynomial whose graph is a straight line.

Note : A unique line passes through two given points. So to draw the graph of linear polynomials we needed to plot
only two ordered pairs and join them.

Illustrating the Concept :

1. Suppose it is given that y is the square of x. Then we may write,
y = f (x) = x2
2. Velocity of uniformly accelerating particle starting from rest depends on time t.
If acceleration is 2 m/s2, we can write, velocity (v) as a function of t.
i.e v (t) = 0 + 2t { Using v = u + at }

1.3 Intervals and Notations

To express values a variable can take, we use following notations.
(i) Open interval :
If x can take values which lie strictly between a and b then we can write
a<x <b or x  (a, b)
(ii) Closed interval :
If x can take values which lie strictly between a and b or x can be equal to a or x can be equal to b,
then we can write
a< x< b or x  [a, b]
(iii) Half-open interval :
If only one end point is included for values of x, then the interval is called as half-open interval.
a< x  b or x  (a, b]

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a < x<b or x  [a, b)

(iv) Infinite intervals :
If x can take all real values, than we can write as
–<x<  x  (– , ) or x  R
(v) Other Notations :
a<x  x  (a, )
a<x  x  [a, )
x< b  x  (– , b)
x<b  x  (– , b]

(vi) If x can take specific values, say x = a, x = b and x = c, then we can write x  {a, b, c}

1.4 Domain
While defining real-valued functions, we have to observe some restrictions. One such restriction is that we
can never divide by zero (0). Hence in the function
1
y = f (x) = x  1 , x cannot be equal to 1.

Domain of y = f (x) is collection of all inputs that operator can take so that output of operator exists
OR
The set of values of x for which y takes real values (so that the function is well defined) is known as
the domain for that function.
Hence the domain of y = f (x) in above examples is x  (–, 1)  (1, ) or x  R – {1}.
Illustrating the Concept :

Find the domain of the following functions.

1
(i) y = 1  x2 (ii) y = 2 sinx (iii) y
x2

1 1
(iv) 3 x  (v) x x (iv) 1  log10 x
log10 x
(i) Square root of a negative number is not defined
 1 – x2  0  x2  1  –1  x  1

Hence the domain of x is the interval [– 1, 1].

(ii) y = 2 sin x
Trigonometric function sin x is defined for all value of x
 x  R i.e. x  (– ,)

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(iii) As denominator cannot be zero, x can not be 2

 x  (– , 2)  (2, ) i.e. x  R – {2}
(iv) Square root of a negative number is not defined
 3  x is defined if 3 – x  0
 x 3 . . . . . .(i)
log10x is defined if x > 0 and also denominator cannot be 0
 log10x cannot be 0  x  1
1
Hence log x is defined if x > 0 and x  1
10

 x  (0, ) – {1} . . . . . . (ii)

Combining (i) and (ii), domain is :
x  (0, 1)  (1, 3]
(v) f (x) is defined if : x + | x |  0

  x if x  0  
 x  –x  x>0  Using | x |   
   x if x  0  
Hence domain is x  (0, )
(vi) f (x) is defined if
1 – log10 x  0 and x>0
 log10 x  1 and x>0
 x  10 and x > 0
 domain is x  (0, 10]

1.5. Range
Range of y = f (x) is collection of all outputs {f (x)} corresponding to each real input in the
domain.
OR
The set of values which y can take is known as the set of Range for that function.

NATURE OF A FUNCTION Section - 2

2.1 Even Function
If a function y = f (x) satisfies f   x   f  x  for all values of x in its domin, then y = f (x) is called an even
function.

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Note : As an even function satisfies, f   x   f  x  , f (x) possesses same value for values of x which are equal in
magnitude and opposite in signs.
For example : f  1  f 1 , f  2   f  2  , f  3   f  3  . . . . .
Therefore graph of an even function is symmetrical about y-axis i.e. left half is mirror image of right half and
right half is mirror image of left half, considering y-axis as mirror.

Illustrating the concept :

(i) Consider f  x   x 2

f   x     x 2  x 2
 f  x  f  x
Hence f  x   x 2 is an even function.

Note : Graph of f  x   x 2 is symmetrical about y-axis.

(ii) Consider f (x) = cos x
f   x   cos   x   cos x [using cos     cos ]
 f  x  f  x 

Hence f (x) = cos x is an even function.

Note : Graph of f (x) = cos x is symmetrical about y-axis.

2.2 Odd function

If a function y = f (x) satisfies f   x    f  x  for all values of x in its domain, then y = f (x) is called an odd
function.
Note : As an odd function satisfies, f   x    f  x  , f (x) possesses values equal in magnitude but opposite in sign
for all values of x which are also equal in magnitude but with opposite signs.

For example : f  1   f 1 , f  2    f  2  , f  3   f  3 , . . . . .

Therefore graph of an odd function is symmetrical about origin i.e. if we rotate the graph in right half about
origin through 180 , then we get graph in left half.

Illustrating the concept :

(i) Consider f  x   x3
3
f   x     x    x3

 f (x) satisfies, f   x    f  x 
Hence f  x   x3 is an odd function.

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(ii) Consider f  x   sin x

f   x   sin   x    sin x [Using sin      sin  ]

 f  x  satisfies, f   x    f  x 

Hence f  x   sin x is an odd function.

Note : As graph is symmetrical about origin, left half of the graph can also be drawn by taking reflection of right half
in both x-axis as well as y-axis :

2.3 Properties of Even and Odd Function

(a) Sum : (i) even + even = even
(ii) even + odd = neither even nor odd
(iii) odd + even = neither even nor odd
(iv) odd + odd = odd
(b) Difference : (i) even – even = even
(ii) even – odd = neither even nor odd
(iii) odd – even = nether even nor odd
(iv) odd – odd = odd
(c) Product : (i) even × even = even
(ii) even × odd = odd
(iii) odd × even = odd
(iv) odd × odd = even
(d) Division : (i) even  even = even
(ii) even  odd = odd
(iii) odd  even = odd

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(iv) odd  odd = even

(e) (i) if f (x) + f (– x) = 0  f is odd function
(ii) if f (x) – f (– x) = 0  f is even function
(f) The square of an even or an odd function is always an even function.

2.4 Periodic function :

A function f (x) is said to be periodic function of x, if there exists a positive real number T such that f (x+ T)
= f (x).
The smallest value of T is called the period of the function.

Note : The positive T should be independent of x for f (x) to be periodic. In case T is not independent of x, f (x)
is not a periodic function.

Definition (Graphically)
A function is said to be periodic if its graph repeats itself after a fixed interval and the width of that interval is
called its period.
For example :
 Graph of f  x  = A sin x repeats after an interview of 2 .
Thus, f (x) = A sin x is periodic with period 2 .

Standard Results on Periodic Functions

Function Period
1. sin n x, cos n x , If n is even
secn x, cosecn x 2, If n is odd or fraction
2. tan n x , cot n x , n is even or odd
3. | sinx |, | cos x | 
| tan x |, | cot x | 
| sec x |, | cosec x | 
4. x   x    x 1

5. x , x2 , x3  5 etc. These function are not periodic

Properties of Periodic Function

(i) If f (x) has period T , then
(a) cf (x) has also period T
(b) f (x + c) is periodic with period T

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(c) f (x)  c is periodic with period T

(d) If constant is added, subtracted, multiplied or divided in periodic function, period remains
same.
(e) Every constant function is always periodic, with no fundamental period.
(f) Inverse of a periodic function does not exist. But in case of trigonometric function since do-
main and range one restricted and defined, hence inverse exists.
(ii) If f (x) is periodic with period T, then
T
k f (cx + d) has period c , hence period is affected by coefficient of x only .
(iii) If f (x) and g(x) are two functions with period T1 and T2 respectively and h  x   a f  x   b g  x 
then h (x) has period = LCM of {T1, T2}

Note : There are some exceptions to above result :

For Example : Period of {f (x) = | sin x | + | cos x |} = /2 instead of 
Period of {f (x) = sin4x + cos4 x} = /2 instead of 
Period of {f (x) = | tanx | + | cot x |} = /2 instead of 
(v) If f (x) is a periodic function with period T and g(x) is any function such that domain of f is a proper
subset of domain of g, then g (f (x)) is periodic with period T.
For Example : sin (x – [x]) = sin ({x}) is periodic with period 1 as x –[x] is periodic with period 1 .
HOW TO TAKE LCM
p r t LCM of  p ,r, t 
1. LCM of  q , s , u  
  HCF of  q , s , u 
2. (a) LCM of rational with rational is possible
(b) LCM of irrational with irrational is possible but LCM of two irrational number of different
kind (for example 2 3 and 3 2 ) does not exist.
(c) LCM of rational with irrational is not possible
For Example : LCM of (2, 2, 6) is not possible as 2, 6 Irrational and 2  rational.

CLASSIFICATION OF FUNCTION Section - 3

Basic functions can be categorised into the following categories.

1. Algebraic functions :
Algebraic functions can be of the following types :
(a) Monomial function (b) Polynomial function
(c) Rational function (d) Irrational function

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2. Transcendental functions :
Trancendental functions can be of the following types :
(a) Logarithmic function (b) Exponential function
(c) Trigonometric function (d) Inverse Trigonometric function
3. Piecewise defined functions :
Piecewise defined functions can be of the following types :
(a) Modulus function (b) Greatest Integer function
(c) Fractional Part function (d) Least Integer function
(e) Signum function
Following chart shows relationship between these functions.

3.1 Algebraic functions

(a) Monomial function :
Any function of the form f  x   kxn , where k is constant and n  N is known as monomial function.
For example :
f  x   3x 4 is a monomial function of degree 4.

f  x   2 x3 is a monomial function of degree 3.

(b) Polynomial function :
A function f (x) = a0xn + a1xn  1 +… + an , (a0  0)
where a0, a1, a2, . . ., an are real constants and n is non-negative integer, is called a polynomial
function, where n is the degree of polynomial function.

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For example :
f (x) = x1920 + 5x1919 + 6x (polynomial of degree 1920)

g (x) = x2 + 3x + 3 (polynomial of degree 2)

h (x) = 7 = 7x0 (polynomial of degree 0)

(i) Constant function : If degree of a polynomial function is 0, then polynomial function is called
as constant function.
(ii) Identity function : If polynomial function takes the form y = f (x) = x for all x  R , then it is
called as identity function on R.
p ( x)
(c) Rational algebraic function : A function of the form f (x) = , where p(x) and q(x) are
q ( x)
polynomials and q(x)  0, is called a rational function.
The domain of a rational function is the set of all real numbers except values where q(x) = 0.
(d) Irrational function : An algebraic function or rational function containing one or more radicals (non-
integral rational powers of x) is called an irrational function.

x3  x
e.g., , 3x3 – x3/2 + 9x – 1
2 x2  9

3.2 Transcendental function

(a) Logarithmic Function
Logarithmic function is represented as
y = loga x where x > 0, a  (0, 1) (1, )
If a > 1, y increases as x increases (as seen from graph).
If 0 < a < 1, y decreases as x increases.

Continuity :
The graph of f (x) = loga x is continuous (i.e. no break in the curve) in the respective domain.

Domain and Range :

The Domain of the function f (x) is x > 0 and Range is y  R.

Properties of Logarithmic Function :

(i) loga a = 1

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(ii) loga 1 = 0
(iii) loga (mn) = loga m + loga n
(iv) loga (m/n) = loga m – loga n
(v) loga xm = m loga x
log mb
(vi) log a b  m  1, m > 0
log m a

(vii) a log a x  x (viii) alogc b = blogc a

 x  y , if m  1
(ix) If logm x > logm y   (x) logm a = b  a = mb
 x  y , if 0  m  1

a  mb , if m  1  a  mb , if m 1

(xi) logm a > b   (xii) logm a < b  
 a  mb , if
b
a  m , if 0  m  1 0  m 1

(b) Exponential Function

y = ax where a > 1 or 0 < a < 1 is an exponential function of x.
This function is the inverse of logarithmic function i.e. it can be
obtained by interchanging x and y in y = loga x.
As observed from the graph, if a > 1, then y increases as x increases.
If 0 < a < 1, then y decreases as x increases.

Continuity :
The graph of f (x) = ax is continuous (i.e. no break in the curve)
everywhere.
Domain and Range:
The domain of the function f (x) is x  R and Range is y > 0.

(c) Trigonometric Functions

(i) y = A sin (mx)
(ii) y = A cos (mx)
(iii) y = A tan (mx)
Period :
Period of y = A sin (mx) and
y = A cos (mx) is T = (2)/m
Period of y = A tan (mx) is T = /m

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Continuity :
The graph of y = A sin (mx) and y = A cos (mx) is continuous
(i.e. no break in the curve) every where.
The graph of y = A tan (mx) is discontinuous

(i.e. break in the curve) at x = (2n + 1)
2m
Domain and Range :
The Domain of y = A sin (mx) and y = A cos (mx) is x  R
and Range is y [A, A].
The Domain of y = A tan (mx) is

x  R   2n  1 and Range is y  R.
2m
(iv) y = A cot mx
Period :
It is a periodic function with period = /m
Continuity :
It can be observe that y = A cot mx is discontinuous at
x = n/m where n  I
Domain and Range :
n
The domain of y = A cot mx is x  R – and the range is y R.
m

(v) y = A sec mx
Period :
It is a periodic function with period = 2/m.
Continuity :
It can be observe that y = A sec mx is
discontinuous at x = (2n + 1)/2m, n  I

Domain and Range :

The domain of y = A sec mx

is x  R   2n  1
2m
and the Range is y     ,  A   A,   .

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(vi) y = A cosec mx

Period :
It is a periodic function with period = 2/m.

Continuity :
It can be observe that y = A cosec mx is discontinuous at x
= n/m.

Domain and Range :

The domain of y = A cosec mx is R – {n /m} and the
range is y     ,  A   A,  

(d) Inverse Trigonometric Function :

(To be discussed in the later module of Function).

3.3 Piecewise Defined functions :

(a) Modulus or Absolute Value Function
Modulus function is a numerical value function or we also call it as absolute value function.
5   5  5, 1  1,  2.53  2.53 …..etc.
it can also be understood as the distance defined with respect to origin.
For example, if x = 1 means distance covered is one unit on right hand side or left side of origin.
 x =1
 x=+1

x < 1 means distance covered is less than one unit on right hand side or left hand side of origin
as shown in the following figure

Similarly x > 1, means distance covered is more than one unit on right hands side or left hand
side of origin as shown in the following figure

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Modulus of x
f (x) = x = magnitude of x or the positive value of x.
The expression x can be further split as follows :
x ; x  0
y= x = 
 x ; x  0

Continuity :
The graph of y = x is continuous (i.e. no break in the curve) but has a corner at origin as shown.

Domain & Range :

The Domain of the function f (x) is x  R and Range is y  [0, ).

Results :
(A) If a > 0 then;
(i) | f (x) | = a  f (x) = + a
(ii) | f (x) | < a  – a < f (x) < a
(iii) | f (x) | > a  f (x) < – a or f (x) > a

(B) If a < 0 then ;

(i) | f (x) | = a  no solution
(ii) | f (x) | < a  no solution
(iii) | f (x) | > a  all real values of x in domain of f (x)

(b) Greatest integer function (unit step function) :

y = [ x ] = the greatest integer less than or equal to x.
It can also be simplified as :
y=[x]=n if n  x < n + 1
where n is an integer.

Continuity :
The graph of f (x) is discontinuous (i.e. break in the curve) at integral values of x.
Domain & Range :
The Domain of the function f (x) is x  R and Range is y  I (integer).

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Note : (a) [ x ] = x holds if x is integer. (b) [ x + a] = [ x ] + a if a is integer.

(c) x = [x] + {x}, {x} denotes the fractional part of x. (d) [ x ] =  [ x ], x  I
(e) [ x ] =  [ x ]  1, x  I (f) [x + y]  [x] + [y]

 [ x]   x 
(g)  n  = n , n  N
   

(c) Fractional part function :

Fractional part function is represented as y   x  x   x 

If we express y = {x} in intervals  x  0, 4  , we get :

 x 0  x 1

 x 1 1 x  2

y   x  x   x    x  2 2x3

x3 3 x4

 0 x4
Now plot the graph of above definition as shown in the figure.
We can extend graph for other values of x.

Continuity :
If we observe graph, we can see that graph has breaks at all integer values. Hence y = {x} is discontinuous
 x I.
Periodicity :
From graph, we can see that y = {x} repeats after interval 1. Therefore y = {x}is a periodic function with
period 1.
Domain & Range
Domain of y   x is x  R and range is y  0 , 1

(d) Least Integer function : (To be discussed in the later module of Function).

(e) Signum Function :

1 , x  0

y = sgn (x) = 0 , x  0
 1 , x  0


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This can also be written as

x
 , x0
y = sgn (x) = x

0 , x0

Domain and Range

Domain of y = sgn (x) is x  R and range is y  1, 0, 1
Continuity
The graph of f (x) is continuous for all values of x except at x = 0 where there is a break in the curve.

GRAPHS TO REMEMBER Section - 4

You are supposed to remember graphs of all functions that we studied in earlier sections. There are other
graphs that you should learn and remember. See the following graphs that you have to always remember.

1. y=x 2. y=–x

3. y=|x| 4. y=–|x|

5. |y|=x 6. –|y|=x

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7. y2 = 4ax 8. y2 = – 4ax

9. x2 = 4ay 10. x2 = – 4ay

11. y = ax2 greater is ‘a’ steeper is curve.

1 1 1 1
12. y  n , n is odd integer  1 [i.e. graph of y  , , , . . . . can be
x x x3 x5

considered same (for rough sketching)].

1 1 1 1
13. y  n , n is even integer > 1 [i.e. graph of y  , , , . . . . can
x x2 x4 x6

be considered same (for rough ketching)].

14. y  xn , n is even integer  2 [i.e. graph of y  x 2 , x 4 , x6 , . . . . can be

considered same (for rough sketching)].

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15. y  xn , n is odd integer > 1 [i.e. graph of y  x3 , x5 , x7 , . . . . can be

considered same (for rough sketching)].

TRANSFORMATIONS Section - 5

5.1. Transformations
(i) Transformation 1

(a) y = f (x) 
 y = f (x + a)
To draw y = f (x + a), shift the graph of
y = f (x) through ‘a’ units towards left.

(b)  y = f (x  a)
y = f (x) 
To draw y = f (x – a) shift the graph of
y = f (x) through ‘a’ unit towards right.

Illustrating the concept :

Plot the graph of following :
(i) 4y = 2x (ii) y = 4.2x
(i) 4y =2x  y = 2x – 2 . (ii) y = 4.2x  y = 2x + 2

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(ii) Transformation 2
(a) y = f (x) 
 y + a = f (x)
To draw y + a = f (x), shift graph of
y = f (x) by a units downward.

(b)  y  a = f (x)
y = f (x) 
To draw y  a = f (x), shift the
graph of y = f (x) by a units upward.

Illustrating the concept :

(i) Draw the graph of : y = cos2x.

1  cos 2 x 1 cos 2 x  cos 2 x

y=  y= + y  a  f ( x ) y=
2 2 2 2

Shift 1/2 unit up

Observation:
In f (x) = cos2 x maximum value of f (x) is 1 and minimum value of f (x) is 0.

(ii) Plot the graph of following : (a) y = ex –1 (b) y  log e x  1

(a) y = ex –1  y + 1 = ex  y = ex

y  a  f ( x)

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(b) y = logex –1  y + 1 = logex  y = logex

y  a  f ( x)

(iii) Transformation 3
y = f (x) 
 y = f (x)
To plot y = f ( x), Draw y = f (x) first then take the mirror image of y = f (x) in the yaxis.

Illustrating the concept :

 π π
Sketch the graph of following function : (i) y –e–x = 0 , (ii) y = cosec (x)  x   – 2 , 2 
 

(i) y = e–x 

f ( x)  f ( x)
 y = ex

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(ii) y = cosec (–x) 

f ( x)  f ( x)
 y = cosec (x)

(iv) Transformation 4
 y =  f (x)
y = f (x) 

To plot y =  f (x), Draw y = f (x) first and then take the mirror image of y = f (x) in x-axis.

Illustrating the concept :

Sketch the graph of following functions :
(i) x2y = 1 ; (ii) y = log 1/x
1 1
(i) y   y
x2 f  x   f  x x2

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1
(ii) y  log   log x  y  log x
x f  x   f  x

(v) Transformation 5
y = f (x) 
 y = f (| x |)
To plot y = f (| x |), draw the graph of y = f (x) first, then remove the portion of the graph in left half
and after that take the mirror image of portion of the graph in right half in the Y-axis. Also include the
right portion of the graph of y = f (x).
Illustrating the concept :

Sketch the graph of following functions :

1
(i) y = |x|+ 1 ; (ii) y = x2 – 2 | x | + 3
1 1
(i) y = | x | + 1 
f ( x )  f (| x |) y =
 y = 1
x 1 f ( x )  f ( x  a)
x

Shift 1 unit left Reject left half and take reflection of right half in left half

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(ii) y = x2 – 2 | x | + 3 
f ( x)  f (| x |)  y = x – 2x + 3
2

Reject left half and take reflection of right half in left half

(vi) Transformation 6
y = f (x) 
 y = | f (x) |
To plot y = | f (x) |, draw the curve y = f (x), then take the mirror image of the lower portion of the
curve (the curve below xaxis) in xaxis and then reject the lower part (or flip lower part into upper)
Also include the upper protion of the curve y  f ( x).

Illustrating the concept :

Draw the graph of the following curves :
(i) y = | x2 – 2x – 3| ; (ii) y = | log x | .
(i) y = | x2 – 2x – 3 |   2
f ( x)  | f ( x)| y = x – 2x – 3

Flip lower part into upper

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(ii) y = | log x | 

f ( x)  | f ( x)| y = log x


Flip lower part into upper

(vii) Transformation 7
y = f (x) 
 | y | = f (x)
To plot | y | = f (x), draw y = f (x) first then remove the lower portion of the curve (part of the
curve below x-axis) and then take the mirror image of upper portion (part of the curve above
x-axis) in xaxis. Also include the upper portion of the curve y = f (x).

Illustrating the concept :

Draw the graph of the following curves : (i) | y | = (x – 1) (x – 2) ; (ii) | y | = sin x

(i) | y | = (x – 1) (x – 2) 
| y |  f ( x) y = (x – 1) (x – 2)


Reject the lower part and then take reflection of upper half into lower half

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(ii) | y | = sin

Reject the lower part and then take reflection of upper half into lower half

(viii) Transformation 8
y = f (x) 
 y = [ f (x) ].
To plot y = [ f (x) ] use the following working rule :
(i) Draw y = f (x).
(ii) Draw y = k horizontal lines separated by unit distance till they intersect the graph,
where k  I (integer)
(iii) From the point of intersection of above horizontal lines with the graph of y = f(x) draw
vertical line.
(iv) From each intersection point draw horizontal lines upto nearest right vertical line such
that the horizontal line is always below the graph.
(v) Horizontal step drawn in step-(iv) is graph of y = [f (x)]

Illustrating the concept :

Draw the graph of the following curves : (i) y = [2 sin x] in [0, 2] ; (ii) y = [x/2]

Draw horizontal line at y I From each intersection point draw horizontal lines upto
nearest right vertical line such that the horizontal line
is always below the graph

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(ii) y = [x/2]

Draw horizontal line at y I From each intersection point draw horizontal lines upto
nearest right vertical line such that the horizontal line is
always below the graph

COMPOSITE FUNCTION Section - 6

Let A, B and C be three non-empty sets.

Let f : A  B and g : B  C be two functions then gof : A  C. This function is called Composite
Function.
Note :
(i) fog(x) = f (g (x)) ; (ii) fof (x) = f (f (x)) ;
(iii) gog(x) = g(g(x)) ; (iv) gof (x) = g (f (x)).

Explanation :
(i) To undrestand the concept of composite function consider fog(x) :

In the above diagram, for Ist block, ‘x’ is the independent variable and corresponding g(x) is the dependent
variable. But for IInd block, g(x) i.e. the dependent variable of Ist block is independent variable corresponding
fog(x) is the dependent variable.

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(ii) fof (x) is

(iii) gof (x) is

(iv) gog(x) is

Illustrating the concept :

If f (x) and g(x) are two functions such that f (x) = sin x and g(x) = x2 then
(i) Show that fog  gof ; (ii) Find the values of (fog)  π /2  and (gof) (/3).
(i) As (fog) (x) = f {g (x)} = f (x2) = sin x2 . . . . . . (i)
and (gof) (x) = g {f (x)} = g (sin x) = sin2 x . . . . . . (ii)
From (i) and (ii), (fog) (x)  (gof) (x)

   1
(ii) From (i),  fog  x  sin x 2   fog   2   sin    
   4 2

2
  2    3   3
and From (ii),  gof  x  sin2 x   gof     sin
3 3  2  4

Illustration - 1 The period of the following functions f (x) = sin4 x  cos 4 x is :

 
(A) 2 (B)  (C) (D)
2 4
SOLUTION : (C)
f (x) = sin4 x  cos 4 x 1 3 1
= 1
4
1  cos 4 x    cos 4 x
4 4
2
=  sin2 x  cos 2 x   2sin2 x cos 2 x 2 
Period = 
1 1 4 2
2
= 1  2 sin x cos x   1  sin2 2 x
2 2
[Using sin 2 = 2 sin cos)

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Illustration - 2 The period of the functions f (x) = a sin kx + b cos kx is :

2  
(A) (B) (C) (D) None of these
k k 2k

SOLUTION : (A)
f (x) = a sin kx + b cos kx
 a b 
f  x   a2  b2  sin kx  cos kx 
 a2  b2 a 2  b2 

 a 2  b2 sin  kx    , where   tan1 b

a
2
Period of f  x  
k

Another Approach :
2 2
We know, period of sin kx  and period of coskx 
k k
 2 2  2
Hence period of f  x   a sin kx  b cos kx = LCM of  ,  =
 k k  k

Illustration - 3 sin 2nx 

If f  x   , n  N has as it’s fundamental period. then n.=
1  cos 2 nx 6
(A) 2 (B) 3 (C) 6 (D) 12

SOLUTION : (C)
sin 2nx 2 sin 2nx 
f  x   Hence period of f  x  
1  cos 2nx  3  cos 2nx n
1  
 2  
But period of f (x) = (given)
2  6
T1 : period of sin 2nx  
2n n  n=6
T2 : period of cos 2nx  2  
2n n

Illustration - 4 If f  x   sin    x  is a periodic function with period  where [.] denotes greatest inte-
ger function, then   ;
(A) (4, 5) (B) [4, 5) (C) (2, 3) (D) [2, 3)

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SOLUTION : (B)
2
Period = 

 [ ] = 4  4    5

Illustration - 5 The domain of the function f  x   x  1  x 2 is x 

 1   1   1   1   1 
(A)  1,  , 1 (B) 1, 1 (C)   ,   ,  (D)  , 1
 2  2   2  2   2 

SOLUTION : (D)

For f  x  to be defined, we must have x  1  x2  0 or x  1  x2

1 1 1
 x2  1  x2  x2  x or x
2 2 2

But x  1 x2  x0

1
 x ....(i)
2

Also, 1 x2  0  x2  1  x 2  1  0   x  1  x  1  0  x 1, 1 ....(ii)

 1 
From (i) and (ii) we get : x   , 1
 2 

Illustration - 6
 x  1  x  3
The domain of the function f  x    x  2
is x 

(A) 1, 2 3,  (B) 1, 2 3,  (C) 1, 2  3,  (D) None of these

SOLUTION : (A)

For f  x  to defined

 x  1  x  3
0
 x  2
 x 1, 2 3, 
 Critical points are -1, 2, 3

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Illustration - 7
1 x
The domain of the function f  x   2  x is x  

(A) 1, 1    ,  2  2,  (B) 1, 1   ,  2   2, 

(C) 1, 1  ,  2   2,  (D) None of these

SOLUTION : (C)

1 x Put x  t
f  x  is defined, if 2  x  0 and 2  x  0
 t  1 t  2  0


 1 x  2  x   0 and x  2, 2  t  1 or t  2
2
2  x   x  1 or x  2

  x  1 x  2  0 and x  2, 2  1 x  1 or  x  2 or x  2

Domain of f is 1, 1    ,  2  2, 

Illustration - 8
The domain of the function f  x   log10  3  x  is x  
 x 

 3  3  3
(A)  0, (B) (0, 3) (C)   , (D)  0,
 2   2   2 

SOLUTION : (D)

 3 x   3 x  For f ( x )  loga x is defined for

log10   is defined for log10  0
 x   x   x  0, a  0, a  1

3 x  3  3 x 
  100  1  x   0,  ....(i) log10   is defined for
x  2  x 

3 x
0  0 x3 ....(ii)
x

 3
From (i) and (ii), we get domain of f is x   0,  .
2  

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Illustration - 9 1
The domain of the function f  x   is x  
 x 2   x   6
(A)   ,  2 4,  (B)   ,  2  4, 

(C)   ,  2 4,  (D) None of these

SOLUTION : (A)

f  x  is defined for  x  2

Also  x  3 for  x   4, 5, 6 ...

 x2   x  6  0   x  3  x  2  0
 x4
  x  2 or  x  3
Domain of f is x  ,  2 4, .
But  x  2 for  x   3,  4,  5, ...

Illustration - 10 x
The range of the function y  is y 
1  x2

 1  1 1  1 
(A) 0, 2  (B)  2 , 2  (C)   2 , 0 (D) None of these
    

SOLUTION : (B)
Clearly, y is defined for all real x. 1  4 y 2  0, y  0

 Domain of y is x  ,.  1  2 y  1  2 y   0, y  0

x
We have, y   y  x2 y  x  2y  1  2 y  1  0, y  0
1 x 2
1 1
   y  , y 0
 x2 y  x  y  0 2 2
For x to be real, discriminant of Also, if y = 0, we get x = 0 so that y = 0 is
also in the rnge.
above quadratic equation  0
 1 1
 Range of y    , 
 2 2

Illustration - 11
x2
The range of the function y  is y 
1  x2
(A) [0, 1) (B) (0, 1) (C) [0, 1] (D) None of these

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SOLUTION : (A)
2 y y y 1  y 
y is defined for all real x.  x  1 y  x  
1 y 1  y
 Domaain of y is x    ,   For x to be real,
x 1  y  0 i.e., y  1 and y 1  y   0
we have, y   x2 y  y  x 2
2
1 x
 y  y  1  0 and y  1
Range of y  [0, 1]

Illustration - 12 If f  x  is an odd function with period 2, find the value of f  4.

(A) 0 (B) 1 (C) 2 (D) None of these

SOLUTION : (A)
As f  x  is an odd function f  x  2  f  x  4 ...(iii)

 f  x    f  x  Using (ii) and (iii) we hae :

f  x   f  x  4
Put x  0  f  0    f  0
 f 0  0 ...(i) Put x  0
 f 0  f  0  4
As f  x  is periodic with period 2
But f  0  0 ....[Using (i)]
 f  x   f  x  2 ...(ii)
Replace x by x  2 Hence value of f  4  0

Illustration - 13 2
The domain of the definition of the function f  x   x 2   x 2  , where [x] is greatest
 
integer less than or equal to x is :
(A) R (B) 0,   (C)   , 0  (D) None of these
SOLUTION (D) :
For f (x) to take real values, expression inside
square root should be non-negative.
2
i.e., x 2   x   0

2
Let us draw graph of y = x2 and y   x  and
observe graphically where their difference can be
non-negative. From graaph it is obvious that
difference of x 2 and x 2 is non-negative for all
 
x  0,   i.e. region right of y-axis as well as  domain is x  [0, )  n where n  N
negative integers.

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Illustration - 14   2 
The domain of the function f  x   log10 1  log10  x  5x  16   is x
 
(A) (2, 3) (B) [2, 3] (C) (2, 3] (D) [2, 3)

SOLUTION : (A)
Domain of y  log10 x is x > 0 and x 2  5 x  16  10
 For f (x) to be defined,  x R and x 2  5x  6  0
1  log10  x 2  5 x 16   0  x R and  x  2  x  3  0
 

log10  x 2  5 x  16   1  x  R and x   2 , 3
  
On combining we get :
 x 2  5 x 16  0
Domain of f (x) is x   2, 3 .
Illustration - 15 The domain of the function f  x   log x cos x  x    ,   is x 
         
(A)  2 , 2   1 (B) ,   1 (C)

 , (D) None of these
 
 2
 2  2 2 
SOLUTION : (D)
log g  x  f  x  is defined for

f  x   0 and g  x    0, 1  1,   [Observe part of

the graph that
 cos x > 0 and
is above x-axis]
x   0, 1  1,   . . . . .(i)
From graph, cos x > 0 Combining (i) and (ii),

 x  
  
,  . . . . . .(ii) Domain of f (x) is x   0,   - {1}
 2 2  2

Illustration - 16  2 
The range of the function is : f  x   log e  3x  4 x  5 

 11   11   11 11
(A)   , log e (B)  log e ,  (C)   log e 3 , log e 3  (D) None of these
 3   3   
SOLUTION : (B)
  16  60  
 D  3x 2  4 x  5   , 
Range of 3x 2  4 x  5 is  4a ,    43 
  
Where D is the discriminant and a is coeffi
cient of x2 of quadratic expression  Range of 3x 2  4 x  5  11 ,  
3 
3x 2  4 x  5 From graph of y  log e t , we can observe that
 Range of

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11 
   3x 2  4 x  5    ,  
  3 

 11 
log  3x 2  4 x  5   log e ,  
   3 
 2 
11   11  Hence range of log  3x  4 x  5  is
 t   ,  , log e t  log e ,  
 3   3  [loge 11/3,  )

Illustration - 17
The range of the function is : f  x   3x 2  4 x  5 is :

 11   11   11   11 
(A)  ,  (B)   ,  (C)  ,   (D)  ,  
 3   3   3  3
    

SOLUTION : (C)
 D 
Range of 3x 2  4 x  5 is  4a ,  
 

Where D is the discriminant and a is coefficient of

x2 of quadratic expression 3x 2  4 x  5
 Range of
11 
  16  60      3x 2  4 x  5    , 
2  ,     3 
3x  4 x  5  4  3 
   11 
3x 2  4 x  5   ,  
11   3 
 Range of is 3x 2  4 x  5 is  3 ,  
  Hence range of is 3x 2  4 x  5 is
From graph of, we can observe that  11 
 ,  
 11   3 
11 
 t   , , t   ,  
3   3 

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IN - CHAPTER EXERCISE

1. Draw the following curves :

(i) y  x2  2x  3

(ii) x  y 1

(iii) y  log x

(iv) y  2  x2
(v) y[sin[ x]]
(vi) y  [ x]  x  x[ x]
x
(vii) y  1 e  e x (where [.] is greatest integer function)

2. Construct the graph of the function

1  x where x  1
y  f ( x  1)  f ( x  1) wheree f ( x)  
 0 where x  1

NOW ATTEMPT OBJECTIVE WORKSHEET BEFORE PROCEEDING AHEAD IN THIS EBOOK

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THINGS TO REMEMBER

1. Intervals and Notations

Interval Notation

Open interval a<x <b x  (a, b)

Closed interval a< x< b x  [a, b]

Half-open interval a< x  b x  (a, b]

a < x<b x  [a, b)

Infinite intervals –<x< x  (– , )

Other Notations a<x x  (a, )

a<x x  [a, )
x< b x  (– , b)
x<b x  (– , b]

2. Domain
The set of values of x for which y takes real values (so that the function is well defined) is known as
the Domain for that function.

3. Range
The set of values which y can take is known as the set of Range for that function.

4. Nature of a function
Even function f (x) = f (x) eg : f (x) = x2, cos x
Odd function f (x) =  f (x) eg : f (x) = x3, sin x

5. Properties of Even and Odd function :

(a) Sum : (i) even + even = even
(ii) even + odd = neither even nor odd
(iii) odd + even = neither even nor odd
(iv) odd + odd = odd
(b) Difference : (i) even – even = even
(ii) even – odd = neither even nor odd

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(iii) odd – even = neither even nor odd

(iv) odd – odd = odd
(c) Product : (i) even × even = even
(ii) even × odd = odd
(iii) odd × even = odd
(iv) odd × odd = even
(d) Division : (i) even  even = even
(ii) even  odd = odd
(iii) odd  even = odd
(iv) odd  odd = even
(e) (i) if f (x) + f (– x) = 0  f is odd function
(ii) if f (x) – f (– x) = 0  f is even function
(f) The square of an even or an odd function is always an even function.

6. Periodic function
A function f (x) is said to be periodic function of x, if there exists a positive real number T such that
f (x+ T) = f (x), for all values of x in the domain of f (x).
The smallest value of T is called the fundamental period or period of the function.

7. Standard Results on Periodic Functions

Function Period
1. sin n x, cos n x , If n is even
secn x, cosecn x 2, If n is odd or fraction

2. tan n x , cot n x , n is even or odd

3. | sinx |, | cos x |
| tan x |, | cot x | 
| sec x |, | cosec x |

4. x   x    x 1

5. x , x 2 , x3  5 etc. These functions are not periodic

8. Properties of Periodic Function

(i) If f (x) has period T , then

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(a) cf (x) has period T also

(b) f (x + c) is periodic with period T
(c) f (x)  c is periodic with period T
(d) If constant is added, subtracted, multiplied or divided in periodic function, period remains
same.
(e) Every constant function is always periodic, with no fundamental period.
(f) Inverse of a periodic function does not exist. But in case of trigonometric function since
domain and range is restricted and defined then inverse exists.
(iii) If f (x) is periodic with period T, then
T
k f (cx + d) has period c , hence period is affected by coefficient of x only .
(iv) If f (x) and g(x) are two functions with period T1 and T2 respectively and h  x   a f  x   b g  x 
then h (x) has period = LCM of {T1, T2}

9. Constant function It is a function of the form y  f  x   b , where b is a constant.

Consider y  f  x   x
Identify function As y = x, both x and y take identical values. That is why this function is called
an identify function.
Both x and y are defined for all real values of x. So domain and range of the
identify function is all real values of x i.e., x  R and y  R .

Two functions y  f  x  and y  g  x  are said to be equal if

(i) both functions f (x) and g (x) have same domain and same range,
Equal Functions (ii) As well as they should be equal for all values of x in domain i.e.,
f  x  g  x  x 
domain
Equal functions are also known as identical functions.

10. Classification of function :

(i) Algebraic functions

Function Definition
Monomial function Any function of the form f  x   kx n where k is constant and
n  N is known as monomial function.

Domain Range Continuity Periodic

x  R Depends upon k and n always continuous not periodic

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Function Definition
Polynomial function A function f (x) = a0x + a1xn  1 +… + an , (a0  0)
n

where a0, a1, a2, . . ., an are real constants and n is non-negative

integer, then f (x) is called polynomial function where n is the degree
of polynomial function.
Domain Range Continuity Periodic
xR depends always continuous not periodic
p ( x)
Rational function A function of the form f (x) = , where p(x) and q(x) are
q ( x)
polynomial and q (x)  0, is called rational function.

Domain Range Continuity Periodic

xR  {q (x) = 0} depends discontinuous at not
x q x   0 periodic
Irrational function An algebraic function or rational function containing one or more
radicals (non-integral rational power of x) is called irrational function.
Domain Range Continuity Periodic
xR  {q (x) = 0} depends discontinuous at not
x q x   0 periodic
(ii) Transcendental functions
Function Definition
Logarithmic Logarithmic function is represented as y = loga x where x > 0,
function a  (0, 1) (1, )
If a > 1, y increases as x increases, if 0 < a < 1, y decreases as x
increases.
Domain Range Continuity Periodic
x>0 yR always continuous not periodic

Exponential y = ax where a > 1 or 0 < a < 1

function This function is the inverse of logarithmic function i.e. it can be obtained
by interchanging x and y in y = loga x.
If a > 1, then y increases as x increases, if 0 < a < 1, then y decreases
as x increases.
Domain Range Continuity Period
xR y>0 always continuous not periodic

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Trigonometric Domain Range Continuity Period

functions
y  sin x xR y   1, 1 always continuous periodic with
period 2
y  cos x xR y   1, 1 always continuous periodic with
period 2

y  tan x x R   2n  1  yR discontinuous at periodic with
 2

x   2n  1 period 
2
y  cosecx x  R  n  y  R   1, 1 discontinuous at periodic with
x = n period 2

y = sec x x R   2n  1 y  R   1, 1 discontinuous at periodic with
2

x   2n  1 period 2
2
y  cot x x  R  n  yR discontinuous at periodic with
x = n period 

(iii) Piecewise defined function :

Function Definition
Modulus f (x) = x = magnitude of x or the positive value of x.
The expression x can be further split as follows :
x ; x  0
y= x = 
 x ; x  0
Domain Range Continuity Periodic
xR y0 always continuous not periodic

Function Definition
Greatest Integer y = [ x ] = the greatest integer less than or equal to x.
It can also be simplified as : y = [ x ] = n if n  x < n + 1 where n is
an integer.
Domain Range Continuity Periodic
xR yI discontinuous at not periodic
xI

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Function Definition

Fractional part Fractional part function is represented as y   x  x   x

Domain Range Continuity Periodic

xR y   0, 1 discontinuous at periodic with
xI period 1
Function Definition

1 , x  0

0 , x0
Signum y = sgn (x) = 
1 , x  0


Domain Range Continuity Periodic

xR 1, 0, 1 discontinuous at not periodic
f (x) = 0

11. Transformations
S.No. Transformation How to transform
1. (a) y = f (x) 
 y = f (x + a) Shift the graph of y = f (x) through ‘a’ units towards left.
(b)  y = f (x  a)
y = f (x)  Shift the graph of y = f (x) through ‘a’ unit towards right.

2. (a) y = f (x) 
 y + a = f (x) Shift graph of y = f (x) by ‘a’ unit downward.

(b)  y  a = f (x)
y = f (x)  Shift the graph of y = f (x) by ‘a’ unit upward.

3. y = f (x) 
 y = f (x) Draw y = f (x) first then take the mirror image of y = f (x) in the y
axis.

4.  y =  f (x)
y = f (x)  Draw y = f (x) first and then take the mirror image of y = f (x) in x-
axis.

5. y = f (x) 
 y = f (| x |) Draw the graph of y = f (x) first, then remove the left portion of the
graph after that take the mirror image of the right portion of the
curve in the Y-axis. Also include the right portion of the graph of
y = f (x).

6. y = f (x) 
 y = | f (x) | Draw the curve y = f (x), then take the mirror image of the lower
portion of the curve (the curve below xaxis) in xaxis and reject
the lower part (or flip lower part into upper)

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7. y = f (x) 
 | y | = f (x) Draw y = f (x) first then remove the lower portion of the curve and
then take the mirror image of upper portion of the curve in xaxis.
Also include the upper portion of the curve y = f (x).

8. y = f (x) 
 y = [ f (x) ]. (i) Draw y = f (x).

(ii) Draw y = k horizontal lines separated by unit distance till

they intersect the graph, Where k  I (integer)
(iii) From the point of intersection of above horizontal lines with
the graph of y = f(x) draw vertical line.
(iv) From each intersection point draw horizontal lines upto
nearest right vertical line such that the horizontal line is always
below the graph.
(v) Horizontal step drawn in step-(iv) is graph of y = [f (x)]
12. Composite functions
Let A, B and C be three non-empty sets.
Let f : A  B and g : B  C be two Functions then gof : A  C. This function is called Composite
Function.
Note :
(i) fog(x) = f (g (x)) ; (ii) fof (x) = f (f (x)) ;
(iii) gog(x) = g(g(x)) ; (iv) gof (x) = g (f (x)).

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SOLUTION - IN - CHAPTER EXERCISE

1. (i) y = | | x2 | – 2x – 3 | = | x2 – 2x – 3 | let f (x) = x2 – 2x–3 = (x –3) (x + 1)

(ii) | x | + | y | = 1 . . . . (iii) |x|+y=1 . . . . (ii) x+y=1 . . . .(i)

 y |y|  x |x|

(iii) | y | = | log | x || . . . .(iv) y = |log (| x |)| . . . . (iii)

y  | y |  f(x)  | f (x)|
y = log | x | . . . . (ii) y = logx . . . . (i)
x|x|

(iv) y  2  x2

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(v) y = [sin[x]] . . . . (iii) y = sin[x] . . . . (ii) y = sinx . . . . (i)

 f (x)  (f(x))  x  [x]

(vi) y  [ x]  x  [ x]
y 0 x 0  x 1
1  x 1 1  x2

 2  x 1 2 x  3

(vii) | y | = | 1 + e| x | – e–x| . . . . (iv) y = 1 + e| x | – e–x . . . . (iii)

y|y|  f (x) | f (x)|

y  1  e| x|  e x  x x 
. . . . (ii) y = e| x | – e–x  y  e  e x  0  . . . . (i)
 y  y 1  0 x  0 

2.

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1  | x  1 |  | x  1|  1
f  x  1  
 0  | x  1|  1
 0 x0
 x 0 x  1


2  x 1  x  2 . . . . . (1)
 0 x2

1  | x  1|  | x  1|  1
f  x  1  
 0  | x  1|  1
 0 x2
2  x 2  x   1


 x 1 x  0 . . . . . (2)
 0 x0

Combining graph (1) and (2)

Self Study Course for IITJEE with Online Support Solutions 45