fo/u fopkjr Hkh# tu] ugha vkjEHks dke] foifr ns[k NksM+s rqjar e/;e eu dj ';keA
iq#"k flag ladYi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA
jfpr% ekuo /keZ iz.ksrk
ln~xq# Jh j.kNksM+nklth egkjkt
STUDY PACKAGE
Subject : Mathematics
Topic: FUNCTION
Index
1. Theory
2. Short Revision
3. Exercise (1 To 5)
4. Assertion & Reason
5. Que. from Compt. Exams
ENJOY
MATHEMATICS
WITH
SUHAAG SIR
Student’s Name :______________________
Class
:______________________
Roll No.
:______________________
Head Office:
243-B, III- Floor,
Near Hotel Arch Manor, Zone-I
M.P. NAGAR, Main Road, Bhopal
!:(0755) 32 00 000, 98930 58881
Free Study Package download from website : www.iitjeeiitjee.com,
www.tekoclasses.com
Functions
2 of 41
Definition :
Every function say f : A → B satisfies the following conditions:
(a)
f ⊆ A x B,
(b)
∀ a ∈ A ⇒ (a, f(a)) ∈ f and
(c)
(a, b) ∈ f & (a, c) ∈ f ⇒ b = c
Illustration # 1: (i)
Which of the following correspondences can be called a function ?
(A)
f(x) = x3
;
{–1, 0, 1} → {0, 1, 2, 3}
(B)
f(x) = ± x
;
{0, 1, 4} → {–2, –1, 0, 1, 2}
(C)
f(x) = x
;
{0, 1, 4} → {–2, –1, 0, 1, 2}
98930 58881
Function is a special case of relation, from a non empty set A to a non empty set B, that
associates each member of A to a unique member of B. Symbolically, we write f: A → B. We read it as "f is a
function from A to B".
Set 'A' is called domain of f and set 'B' is called co-domain of f.
For example, let A ≡ {–1, 0, 1} and B ≡ {0, 1, 2}. Then A × B ≡ {(–1, 0), (–1, 1), (–1, 2), (0, 0), (0, 1), (0, 2), (1, 0),
(1, 1), (1, 2)}
Now, " f : A → B defined by f(x) = x 2 " is the function such that
f ≡ {(–1, 1), (0, 0), (1, 1)}
f can also be show diagramatically by following picture.
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
A.
(D)
f(x) = – x
;
{0, 1, 4} → {–2, –1, 0, 1, 2}
Solution:
f(x) in (C) & (D) are functions as definition of function is satisfied. while in case of (A) the given relation is
not a function, as f(–1) ∉ codomain. Hence definition of function is not satisfied.
While in case of (B), the given relation is not a function, as f(1) = ± 1 and f(4) = ± 2 i.e. element 1 as well as 4 in
domain is related with two elements of codomain. Hence definition of function is not satisfied.
(ii)
Which of the following pictorial diagrams represent the function
(A)
(B)
(C)
(D)
Solution:
B & D. In (A) one element of domain has no image, while in (C) one element of domain has two images
in codomain
Assignment: 1.
Let g(x) be a function defined on [−1, 1]. If the area of the equilateral triangle with two of its
vertices at (0,0) & (x,g(x)) is 3 / 4 sq. units, then the function g(x) may be.
2.
(A) g(x)= ± (1 − x 2 )
(B*) g(x) = (1 − x 2 ) (C*) g(x) = − (1 − x 2 )
Represent all possible functions defined from {α, β} to {1, 2}
Answer
(1)
B
(2)
B.
(i)
(ii)
(iii)
(D) g(x) =
(1 + x 2 )
(iv)
Domain, Co-domain & Range of a Function :
Let f: A → B, then the set A is known as the domain of f & the set B is known as co−domain of f. If a member 'a'
of A is associated to the member 'b' of B, then 'b' is called the f-image of 'a' and we write
b = f (a). Further 'a' is called a pre-image of 'b'. The set {f(a): ∀ a ∈A} is called the range of f and is denoted
by f(A). Clearly f(A) ⊂ B.
Sometimes if only definition of f (x) is given (domain and codomain are not mentioned), then domain is set of
those values of ' x' for which f (x) is defined, while codomain is considered to be (– ∞, ∞)
A function whose domain and range both are sets of real numbers is called a real function. Conventionally the
word "FUNCTION” is used only as the meaning of real function.
Illustration # 2 :
Find the domain of following functions :
(i)
Solution :(i)
(ii)
sin–1 (2x – 1)
f(x) =
x2 − 5
f(x) =
2
x 2 − 5 is real iff x – 5 ≥ 0
(ii)
⇒
|x| ≥
5
⇒
x ≤ – 5 or x ≥
5
∴
the domain of f is (–∞, – 5 ] ∪ [ 5 , ∞)
–1 ≤ 2x – 1 ≤ + 1
∴
domain is x ∈ [0, 1]
Algebraic Operations on Functions :
If f & g are real valued functions of x with domain set A and B respectively, then both f & g are defined in A ∩ B.
Now we define f + g, f − g, (f. g) & (f /g) as follows:
f
f( x )
  (x) =
g( x ) domain is {x  x ∈ A ∩ B such that g(x) ≠ 0}.
g
Note : "
For domain of φ(x) = {f(x)}g(x) , conventionally, the conditions are f(x) > 0 and g(x) must be defined.
"
For domain of φ(x) = f(x)Cg(x) or φ(x) = f(x)Pg(x) conditions of domain are f(x) ≥ g(x) and f(x) ∈ N and g(x) ∈
W
Illustration # 3:
Find the domain of following functions :
(iii)
Solution:
f(x) =
(i)
sin x − 16 − x 2
(ii)
f(x) =
3
log(x 3 − x)
4−x
sin x is real iff sin x ≥ 0 ⇔ x∈[2nπ, 2nπ + π], n∈I.
2
(iii)
f(x) = x cos
−1
x
3 of 41
(i)
Illustration # 4:
Solution
x2 + x − 1
{x2 + x + 1 and x 2 + x – 1 have no common factor}
x2 + x − 1
x2 + x + 1
Illustration # 5:
Solution
∴
(iii)
x2 + x + 1
x2 + x + 1
⇒
yx2 + yx – y = x2 + x + 1
x2 + x − 1 2
⇒
(y – 1) x + (y – 1) x – y – 1 = 0
If y = 1, then the above equation reduces to –2 = 0. Which is not true.
Further if y ≠ 1, then (y – 1) x2 + (y – 1) x – y – 1 = 0 is a quadratic and has real roots if
(y – 1)2 – 4 (y – 1) (–y – 1) ≥ 0
i.e.
if y ≤ –3/5 or y ≥ 1 but y ≠ 1
Thus the range is (–∞, –3/5] ∪ (1, ∞)
Graphical Method : Values covered on y-axis by the graph of function is range
y=
(ii)
f(x) =
Find the range of f(x) =
f(x) =
Find the range of f(x) =
x2 − 4
x−2
x2 − 4
x−2
= x + 2; x ≠ 2
graph of f(x) would be
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
98930 58881
2
16 − x 2 is real iff 16 − x ≥ 0 ⇔ − 4 ≤ x ≤ 4.
Thus the domain of the given function is {x : x∈[2nπ, 2nπ + π], n∈I }∩[−4, 4] = [−4, −π] ∪ [0, π].
(ii)
Domain of 4 − x 2 is [−2, 2] but 4 − x 2 = 0 for x = ± 2
⇒
x ∈ (–2, 2)
log(x3 − x) is defined for x3 − x > 0 i.e. x(x − 1)(x + 1) > 0.
∴
domain of log(x3 − x) is (−1, 0 ) ∪ (1, ∞).
Hence the domain of the given function is {(−1, 0 ) ∪ (1, ∞)}∩ (−2, 2) =
(−1, 0 ) ∪ (1, 2).
(iii)
x > 0 and –1 ≤ x ≤ 1
∴
domain is (0, 1]
Assignment :
3.
Find the domain of following functions.
1
–1 2 x − 1
(i)
f(x) = log( 2 − x ) + x + 1
(ii)
f(x) = 1− x – sin
3
Ans. (i)
[–1, 1) ∪ (1, 2)
(ii)
[–1, 1]
Methods of determining range :
(i)
Representing x in terms of y
Definition of the function is usually represented as y (i.e. f(x) which is dependent variable) in terms of an expression
of x (which is independent variable). To find range rewrite given definition so as to represent x in terms of an
expression of y and thus obtain range (possible values of y).
If y = f(x) ⇔
x = g(y), then domain of g(y) represents possible values of y, i.e. range of f(x).
Thus the range of f(x) is R – {4}
Using Monotonocity/Maxima-Minima
(a)
Continuous function: If y = f(x) is continuous in its domain then range of f(x) is y ∈ [min f(x), max. f(x)]
(b)
Sectionally continuous function: In case of sectionally continuous functions, range will be union of
[min f(x), max. f(x)] over all those intervals where f(x) is continuous, as shown by following example.
Let graph of function y = f(x) is
Then range of above sectionally continuous function is [y2, y3] ∪ (y4, y5 ] ∪ (y6, y 7]
Note : " In case of monotonic functions minimum and maximum values lie at end points of interval.
Illustration # 6 :
Find the range of following functions :
(i)
y = !n (2x – x2)
(ii)
y = sec–1 (x2 + 3x + 1)
Solution :
(i)
Step – 1
Using maxima-minima, we have
2x – x2 ∈ (–∞, 1]
{i.e. domain (0, 1]}
Step – 2
For log to be defined accepted values are 2x – x2 ∈ (0, 1]
Now, using monotonocity
!n (2x – x 2) ∈ (–∞, 0]
∴
range is (– ∞, 0]
Ans.
y = sec–1 (x2 + 3x + 1)
Let
t = x2 + 3x + 1 for x ∈ R
 5

 5 
then
t ∈ − , ∞ 
but y = sec–1 (t)
⇒
t ∈ − , − 1 ∪ [1, ∞)
4
4




4 of 41
(ii)
Answer
C.
(i)
1
domain R; range R
(iii)
y=
(iv)
y = cot–1 (2x – x2)
(v)
3
 2
y = !n sin –1  x + x +  Answer
4

x2 − x
Answer
Answer
Classification of Functions :
(ii)
3 − 5 3 + 5 
domain R ; range  2 , 2 


domain R – [0, 1] ; range (0, ∞)
π 
domain R ; range  , π 
4 
− 2 − 5 − 2 + 5 
π
 π
,
domain x ∈ 
 ; range !n 6 , !n 2 


4
4


TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Assignment:
4.
Find domain and range of following functions.
x 2 − 2x + 5
(i)
y = x3
(ii)
y= 2
x + 2x + 5
98930 58881

 π
−1  5  
from graph range is y ∈ 0,  ∪ sec  − 4 , π

 
 2

Functions can be classified as :
(i)
One − One Function (Injective Mapping) and Many − One Function:
One − One Function :
A function f : A → B is said to be a one-one function or injective mapping if different elements of A have
different f images in B.
Thus for x1, x 2 ∈ A & f(x1), f(x2) ∈ B, f(x1) = f(x2) ⇔ x1 = x2 or x1 ≠ x2 ⇔ f(x1) ≠ f(x 2).
Diagrammatically an injective mapping can be shown as
OR
Many − One function :
A function f : A → B is said to be a many one function if two or more elements of A have the same f
image in B.
Thus f : A → B is many one iff there exists atleast two elements x 1 , x 2 ∈ A, such that
f(x1) = f(x2) but x1 ≠ x2.
Diagrammatically a many one mapping can be shown as
OR
Note : "
If a function is one−one, it cannot be many−one and vice versa.
Methods of determining whether function is ONE-ONE or MANY-ONE :
(a)
If x1, x2 ∈ A & f(x1), f(x 2) ∈ B, f(x 1) = f(x 2) ⇔ x1 = x 2 or x1 ≠ x 2 ⇔ f(x1) ≠ f(x2), then function is ONE-ONE
otherwise MANY-ONE. (b)
If there exists a straight line parallel to x-axis, which cuts the graph of
the function atleast at two points, then the function is MANY-ONE, otherwise ONE-ONE. (c) If either f′(x)
≥ 0, ∀ x ∈ complete domain or f′(x) ≤ 0 ∀ x ∈ complete domain, where equality can hold at discrete
point(s) only, then function is ONE-ONE, otherwise MANY-ONE.
(ii)
Onto function (Surjective mapping) and Into function :
Onto function :
If the function f : A → B is such that each element in B (co−domain) must have atleast one pre−image in
A, then we say that f is a function of A 'onto' B. Thus f : A → B is surjective iff ∀ b ∈ B, there exists some
a ∈ A such that f (a) = b.
Diagrammatically surjective mapping can be shown as
OR
Method of determining whether function is ONTO or INTO :
Find the range of given function. If range ≡ co−domain, then f(x) is onto, otherwise into
Into function :
If f : A → B is such that there exists atleast one element in co−domain which is not the image of any
element in domain, then f(x) is into.
5 of 41
Diagrammatically into function can be shown as
OR
(a)
one−one onto (injective & surjective)
(b)
one−one into (injective but not surjective)
(c)
many−one onto (surjective but not injective)
(d)
many−one into (neither surjective nor injective)
Note : "
98930 58881
Note : "
If a function is onto, it cannot be into and vice versa.
Thus a function can be one of these four types:
− ve ; 0 ≤ x < 1
f′(x) = 
+ ve ; 1 < x < 3
∴
f(x) is not monotonic. Hence it is not injective.
For f(x) to be surjective, A should be equal to its range. By graph range is [2, 6]
∴
A ≡ [2, 6]
∴
Assignment:
5.
For each of the following functions find whether it is one-one or many-one and also into or onto
(i)
f(x) = 2 tan x; (π/2, 3π/2) → R
Answer
(ii)
(iii)
D.
one-one onto
1
; (–∞, 0) → R
1+ x2
Answer
one-one into
f(x) = x2 + !n x
f(x) =
Answer
one-one onto
Various Types of Functions :
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
If f is both injective & surjective, then it is called a bijective mapping. The bijective functions are also
named as invertible, non singular or biuniform functions.
"
If a set A contains 'n' distinct elements then the number of different functions defined from
A → A is n n and out of which n! are one one.
Illustration # 7
(i)
Find whether f(x) = x + cos x is one-one.
Solution
The domain of f(x) is R.
f′ (x) = 1 − sin x.
∴
f′ (x) ≥ 0 ∀ x ∈ complete domain and equality holds at discrete points only
∴
f(x) is strictly increasing on R. Hence 3f(x) is2one-one.
(ii)
Identify whether the function f(x) = –x + 3x – 2x + 4 ; R → R is ONTO or INTO
Solution
As codomain ≡ range, therefore given function is ONTO
(iii)
f(x) = x2 – 2x + 3; [0, 3] → A. Find whether f(x) is injective or not. Also find the set A, if f(x) is surjective.
Solution
f′(x) = 2(x – 1); 0 ≤ x ≤ 3
Polynomial Function : If a function f is defined by f (x) = a 0 xn + a1 xn−1 + a2 xn−2 +... + a n−1 x + an where
n is a non negative integer and a0 , a1, a2,........., a n are real numbers and a0 ≠ 0, then f is called a
polynomial function of degree n.
Note : "
There are two polynomial functions, satisfying the relation; f(x).f(1/x) = f(x) + f(1/x), which are
f(x) = 1 ± xn
(ii)
Algebraic Function : y is an algebraic function of x, if it is a function that satisfies an algebraic equation
of the form, P0 (x) yn + P1 (x) yn−1 +....... + Pn−1 (x) y + Pn (x) = 0 where n is a positive integer and P0 (x), P1
(x)....... are polynomials in x. e.g. y = x is an algebraic function, since it satisfies the equation y² − x² = 0.
Note : "
All polynomial functions are algebraic but not the converse.
"
A function that is not algebraic is called Transcendental Function.
g( x )
(iii)
Fractional / Rational Function : A rational function is a function of the form, y = f (x) =
, where g (x)
h( x )
& h (x) are polynomials and h (x) ≡/ 0.
(iv)
Exponential Function :
A function f(x) = ax = ex In a (a > 0, a ≠ 1, x ∈ R) is called an exponential function. Graph of exponential
function can be as follows :
Case - Ι
Case - ΙΙ
For a > 1
For 0 < a < 1
(i)
(vi)
Case- ΙΙ
For 0 < a < 1
Absolute Value Function / Modulus Function :
 x if
The symbol of modulus function is f (x) = x and is defined as: y = x= 
− x if
(vi)
Signum Function :
A function f (x) = sgn (x) is defined as follows :
 1 for x > 0

f (x) = sgn (x) =  0 for x = 0
− 1 for x < 0

| x |

; x≠0
It is also written as sgn x =  x
 0 ; x = 0
| f ( x) |
; f ( x) ≠ 0

Note : sgn f(x) =  f ( x )
 0 ;
f (x) = 0
(vii)
x≥0
.
x<0
98930 58881
Case- Ι
For a > 1
6 of 41
Logarithmic Function : f(x) = logax is called logarithmic function where a > 0 and a ≠ 1 and x > 0. Its
graph can be as follows
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
(v)
Greatest Integer Function or Step Up Function :
The function y = f (x) = [x] is called the greatest integer function where [x] equals to the greatest integer
less than or equal to x. For example :
for − 1 ≤ x < 0 ; [x] = − 1 ;
for 0 ≤ x < 1 ; [x] = 0
for
1 ≤ x < 2 ; [x] = 1 ;
for 2 ≤ x < 3 ; [x] = 2
and so on.
Alternate Definition :
The greatest integer occur on real number line while moving L.H.S. of x (starting from x) is [x]
(a)
(c)
(viii)
Properties of greatest integer function :
x − 1 < [x] ≤ x
(b)
[x ± m] = [x] ± m iff m is an integer.
 0 ; if x is an int eger
[x] + [y] ≤ [x + y] ≤ [x] + [y] + 1 (d)
[x] + [− x] = 
 − 1 otherwise
Fractional Part Function:
It is defined as, y = {x} = x − [x].
e.g. the fractional part of the number 2.1 is 2.1 − 2 = 0.1 and the fractional part of − 3.7 is 0.3. The period
of this function is 1 and graph of this function is as shown.
7 of 41
Identity function :
The function f : A → A defined by, f(x) = x ∀ x ∈ A is called the identity function
on A and is denoted by ΙA. It is easy to observe that identity function is a bijection.
(x)
Constant function : A function f : A → B is said to be a constant function, if every element of A has the
same f image in B. Thus f : A → B; f(x) = c, ∀ x ∈ A, c ∈ B is a constant function.
Illustration # 8 (i)
Let {x} & [x] denote the fractional and integral part of a real number x respectively. Solve
4{x} = x + [x]
Solution
As x = [x] + {x}
2 [ x]
∴
Given equation ⇒
4{x} = [x] + {x} + [x]
⇒
{x} =
3
As [x] is always an integer and {x} ∈ [0, 1), possible values are
[x]
{x}
x = [x] + {x}
0
0
0
2
5
1
3
3
5
∴
There are two solution of given equation x = 0 and x =
3
(ii)
Draw graph of f(x) = sgn (!n x)
Solution
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
98930 58881
(ix)
Assignment: 6.
If f : R → R satisfying the conditions f(0) = 1, f(1) = 2 and f(x + 2) = 2f (x) + f(x + 1), then find f (6).
Answer
64
7.
Draw the graph of following functions where [.] denotes greatest integer function
(i)
y = [2 x] + 1
(ii)
y = x [x], 1 ≤ x ≤ 3
(iii) y = sgn (x2 – x)
Answer (i)
E.
Odd & Even Functions : (i)
(ii)
(iii)
If f (−x) = f (x) for all x in the domain of ‘f’ then f is said to be an even
function. If f (x) − f (−x) = 0 ⇒ f (x) is even. e.g. f (x) = cos x; g (x) = x² + 3.
(ii)
If f (−x) = −f (x) for all x in the domain of ‘f’ then f is said to be an odd function.
If f (x) + f (−x) = 0 ⇒ f (x) is odd.
e.g. f (x) = sin x; g (x) = x3 + x.
Note : "
A function may neither be odd nor even. (e.g. f(x) = ex , cos–1x)
"
If an odd function is defined at x = 0, then f(0) = 0
Properties of Even/Odd Function
(a)
Every even function is symmetric about the y−axis & every odd function is symmetric about the origin.
For example graph of y = x2 is symmetric about y-axis, while graph of y = x3 is symmetric about origin
(b)
All functions (whose domain is symmetrical about origin) can be expressed as the sum of an even & an
odd function, as follows
f(x) =
(c) The only function which is defined on the entire number line and is even & odd at the same time is f(x) = 0.
If f and g both are even or both are odd then the function f.g will be even but if any one of them is odd then
f.g will be odd. (e)
If f(x) is even then f′(x) is odd but converse need not be true.
 x + x 2 + 1
 is an odd function.
Illustration- 9: Show that log 


2
2




Then f(–x) = log  − x + ( − x ) + 1 
Solution
Let f(x) = log  x + x + 1  .




 x 2 + 1 − x  x 2 + 1 + x 



1
2





= log
= log
– log  x + x + 1  = –f(x)
2


2
+
+
x
1
x
x +1+ x
Hence f(x) is an odd function.
Illustration - 10
Show that ax +a–x is an even function.
Solution
Let f(x) = ax + a–x
Then f(–x) = a–x + a–(–x) = a–x +ax = f(x).
Hence f(x) is an even function
Illustration - 11
Show that cos–1 x is neither odd nor even.
Solution
Let f(x) = cos–1x. Then f(–x) = cos–1 (–x) = π – cos–1 x which is neither equal to f(x) nor equal to f(–x).
Hence cos–1 x is neither odd nor even
Assignment: 8.
Determine whether following functions are even or odd?
e x + e −x
(i)
Answer
Odd
e x − e−x
 2

(ii)
log  x + 1 − x 
Answer
Odd


2


(iii)
x log  x + x + 1 
Answer
Even


TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
98930 58881
8 of 41
(d)
(iv)
sin–1 2x 1− x 2
Answer
Odd
Even extension / Odd extension :
Let the defincition of the function f(x) is given only for x ≥ 0. Even extension of this function implies to define the
function for x < 0 assuming it to be even. In order to get even extension replace x by –x in the given defincition
Similarly, odd extension implies to define the function for x < 0 assuming it to be odd. In order to get odd
extension, multiply the definition of even extension by –1
Illustration - 12
What is even and odd extension of f(x) = x3 – 6x2 + 5x – 11 ; x > 0
Solution
Even extension
;x<0
f(x) = –x3 – 6x2 + 5x – 11
Odd extension
f(x) = x3 + 6x2 + 5x + 11
;x<0
F.
Periodic Function : A function f(x) is called periodic with a period T if there exists a real number T >
0 such that for each x in the domain of f the numbers x – T and x + T are also in the domain of f and f(x) = f(x +
T) for all x in the domain of 'f'. Domain of a periodic function is always unbounded. Graph of a periodic function
with period T is repeated after every interval of 'T'.
e.g. The function sin x & cos x both are periodic over 2π & tan x is periodic over π.
The least positive period is called the principal or fundamental period of f or simply the period of f.
Note : "
f (T) = f (0) = f (−T), where ‘T’ is the period.
"
Inverse of a periodic function does not exist. " Every constant function is always periodic, with no
fundamental period.
Properties of Periodic Function
1
and f( x ) also have a period T.
(a)
If f(x) has a period T, then
f( x )
T
(b)
If f(x) has a period T then f (ax + b) has a period | a | .
f (x)
is L.C.M.
(c)
If f (x) has a period T1 & g (x) also has a period T2 then period of f(x) ± g(x) or f(x) . g(x) or
g( x )
of T1 & T2 provided their L.C.M. exists. However that L.C.M. (if exists) need not to be fundamental period.
f (x)
If L.C.M. does not exists f(x) ± g(x) or f(x) . g(x) or
is aperiodic.
g( x )
e.g. |sinx| has the period π, | cosx | also has the period π
π
∴
|sinx| + |cosx| also has a period π. But the fundamental period of |sinx| + |cosx| is .
2
Illustration - 13
Find period of following functions
x
x
(i)
f(x) = sin
+ cos
(ii)
f(x) = {x} + sin x
3
2
x
2x
3x
(iii)
f(x) = cos x . cos 3x
(iv)
f(x) = sin
– cos
– tan
3
3
2
x
x
x
x
Solution
(i)
Period of sin
is 4π while period of cos is 6π . Hence period of sin
+ cos
is 12 π
3
3
2
2
{L.C.M. of 4 & 6 is 12}
(ii)
Period of
Period of {x} = 1
∴
it is aperiodic
sin
x
=
2p
but L.C.M. of 2π & 1 is not possible
f(x) = cos x . cos 3x
2π 

 = 2π
period of f(x) is L.C.M. of  2π,
3 

9 of 41
(iii)
2π
, where n ∈ N. Hence crossn
checking for n = 1, 2, 3, ....we find π to be fundamental period f(π + x) = (– cos x) (– cos 3x) = f(x)
2π
2π
π
(iv)
Period of f(x) is L.C.M. of
,
,
3 / 2 1/ 3 3 / 2
4π
2π
= L.C.M. of
, 6π ,
= 12π
3
3
 a p !  L.C.M.(a,p,! )
L.C.M.of , ,  =
NOTE :
(b, q,m)
 b q m  H.C.F.
Assignment: 9.
Find the period of following function.
(i)
f(x) = sin x + | sin x |
Answer
2π
x
(ii)
f(x) = 3 cos x – sin
Answer
6π
3
2x
3x
(iii)
sin
– cos
Answer
70 π
5
7
2
4
Answer
π
(iv)
f(x) = sin x + cos x
G.
Composite Function :
(ii)
f(x) = x , g(x) = x2 − 1.
Domain of f is [0, ∞), range of f is [0, ∞).
Domain of g is R, range of g is [−1, ∞).
Since range of f is a subset of the domain of g,
∴
domain of gof is [0, ∞) and g{f(x)}= g(√x) = x − 1. Range of gof is [−1, ∞)
Further since range of g is not a subset of the domain of f
i.e. [−1, ∞) ⊄ [0, ∞)
∴
fog is not defined on whole of the domain of g.
Domain of fog is {x∈R, the domain of g : g(x)∈ [0, ∞), the domain of f}.
Thus the domain of fog is D = {x∈R: 0 ≤ g(x) < ∞}
i.e. D = { x∈R: 0 ≤ x 2 − 1}= { x∈R: x ≤ −1 or x ≥ 1 }= (−∞, −1] ∪ [1, ∞)
fog (x) = f{g(x)} = f(x2−1) = x 2 − 1 Its range is [0, ∞).
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
98930 58881
but 2π may or may not be fundamental periodic, but fundamental period =
Let f: X→Y1 and g: Y2→ Z be two functions and the set D = {x∈ X: f(x)∈ Y2}. If D ≡/ φ, then the function h defined on
D by h(x) = g{f(x)} is called composite function of g and f and is denoted by gof. It is also called function of a function.
Domain of gof is D which is a subset of X (the domain of f ). Range of gof is a subset of the range of g. If D =
Note : "
X, then f(x) ⊂ Y2.
Properties of Composite Functions :
(a)
In general gof ≠ fog (i.e. not commutative)
(b)
The composite of functions are associative i.e. if three functions f, g, h are such that
fo (goh) & (fog) oh are defined, then fo (goh) = (fog) oh.
(c)
If f and g both are one-one, then gof and fog would also be one-one.
(d)
If f and g both are onto, then gof or fog may or may not be onto.
(e)
The composite of two bijections is a bijection iff f & g are two bijections such that gof is defined, then gof
is also a bijection only when co-domain of f is equal to the domain of g.
(f)
If g is a function such that gof is defined on the domain of f and f is periodic with T, then gof is also
periodic with T as one of its periods. Further if
#
g is one-one, then T is the period of gof
#
g is also periodic with T′ as the period and the range of f is a sub-set of [0, T′ ], then T is the
period of gof
Illustration # 14
Describe fog and gof wherever is possible for the following functions
(i)
f(x) = x + 3 , g(x) = 1 + x2
(ii)
f(x) = x , g(x) = x2 − 1.
Solution
(i)
Domain of f is [−3, ∞), range of f is [0, ∞).
Domain of g is R, range of g is [1, ∞).
Since range of f is a subset of domain of g,
∴
domain of gof is [−3, ∞)
{equal to the domain of f }
gof (x) = g{f(x)} = g ( x + 3 ) = 1 + (x+3) = x + 4. Range of gof is [1, ∞).
Further since range of g is a subset of domain of f,
∴
domain of fog is R
{equal to the domain of g}
2
2
fog (x) = f{g(x)}= f(1+ x ) = x + 4 Range of fog is [2, ∞).
 π π
Let f(x) = ex ; R + → R and g(x) = sin–1 x; [–1, 1] → − ,  . Find domain and range of fog (x)
 2 2
Solution
 π π
Domain of f(x) : (0, ∞)
Range of g(x) : − , 
 2 2
(iii)
 π
values in range of g(x) which are accepted by f(x) are  0, 
 2
π
π
⇒
0 < g(x) ≤
0 < sin–1x ≤
0<x≤1
2
2
10 of 41
Hence domain of fog(x) is x ∈ (0, 1]
Domain :
(0, 1]
Range :
(1, eπ/2]
Example of composite function of non-uniformly defined functions :
Illustration # 15
If
f(x)
= | |x – 3| – 2 |
0≤x≤4
g(x)
= 4 – |2 – x|
–1≤x≤3
then find fog(x) and draw rough sketch of fog(x).
Solution
f(x) = | | x – 3| – 2|
0 ≤ x ≤ 4
| x − 1 | 0 ≤ x < 3

=
| x − 5 | 3 ≤ x ≤ 4
=
=
1 − (2 + x )

 2 + x −1
5 − (2 + x )

 1− 6 + x
 6 − x −1

 5 − 6 + x
0 ≤ 2 + x < 1 and − 1 ≤ x < 2
1 ≤ 2 + x < 3 and − 1 ≤ x < 2
3 ≤ 2 + x ≤ 4 and − 1 ≤ x < 2
0 ≤ 6 − x < 1 and
1 ≤ 6 − x ≤ 3 and
2≤x≤3
2≤x≤3
3 ≤ 6 − x ≤ 4 and
2≤x≤3
−2 ≤ x <1
− 1 − x

−1≤ x < 1
 1+ x
 3 − x
1≤ x ≤ 2

 x − 5 − 6 ≤ − x < −5
 5 − x − 5 ≤ − x < −3

 x − 1 − 3 ≤ − x ≤ −2
 − 1 − x − 2 ≤ x < −1

−1≤ x < 1
 1+ x
 3 − x
1≤ x ≤ 2

=
−
<x≤6
x
5
5

5−x
3<x≤5

 x − 1
2≤x≤3
1 + x − 1 ≤ x < 1

3 − x 1 ≤ x < 2
=
x −1 2 ≤ x ≤ 3

Alternate method for finding fog
2 + x − 1 ≤ x < 2
g(x) = 
6 − x 2 ≤ x ≤ 3
graph of g(x) is
and − 1 ≤ x < 2
and − 1 ≤ x < x
and − 1 ≤ x < 2
and
and
2≤x≤3
2≤x≤3
and
2≤x≤3
and − 1 ≤ x < 2
and − 1 ≤ x < 2
and − 1 ≤ x < 2
and 2 ≤ x ≤ 3
and
2≤x≤3
and
2≤x≤3
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
1 − x 0 ≤ x < 1

x −1 1≤ x < 3
=
5 − x 3 ≤ x ≤ 4

g(x) = 4 – |2 – x|
−1 ≤ x ≤ 3
4 − ( 2 − x ) − 1 ≤ x < 2

=
4 − ( x − 2) 2 ≤ x ≤ 3
2 + x − 1 ≤ x < 2

=
6 − x 2 ≤ x ≤ 3
1 − g( x ) 0 ≤ g( x ) < 1

g( x ) − 1 1 ≤ g( x ) < 3
∴
fog (x) = 
5 − g( x ) 3 ≤ g( x ) ≤ 4

98930 58881
Therefore
 π π
domain : − , 
 4 4
range : [0, 1]
11 of 41
1 − g( x ) for no value
 2 + x −1 −1≤ x < 1
x +1 −1≤ x < 1



g
(
x
)
1
1
x
1
5
(
2
x
)
1
x
2
3 − x 1≤ x < 2
−
−
≤
<
−
+
≤
<
= 
= 
= 
5 − g( x )


1≤ x ≤ 3

5 − (6 − x ) 2 ≤ x ≤ 3
x −1 2 ≤ x ≤ 3
Assignment: 10.
Define fog(x) and gof(x). Also their Domain & Range.
(i) f(x) = [x], g(x) = sin x
(ii) f(x) = tan x, x ∈ (–π/2, π/2); g(x) = 1− x 2
Answer
(i)
gof = sin [x]
domain : R
range { sin a : a ∈ Ι}
fog = [ sin x]
domain : R
range : {–1, 0, 1}
Answer
(ii)
gof = 1 − tan 2 x
98930 58881
∴
1 − g( x ) 0 ≤ g( x ) < 1

g( x ) − 1 1 ≤ g( x ) < 3
fog(x) = 
5 − g( x ) 3 ≤ g( x ) ≤ 4

fog = tan 1− x 2
domain : [–1, 1]
range [0, tan 1]
Let f(x) = ex : R+ → R and g(x) = x2 – x : R → R. Find domain and range of fog (x) & gof (x)
Answer
fog (x)
gof f(x)
Domain : (–∞, 0) ∪ (1, ∞)
Domain : (0, ∞)
 1 
Range : [1, ∞)
Range : − , ∞ 
 4 
H.
Inverse of a Function :
Let f : A → B be a function. Then f is invertible iff there is a function g : B
→ A such that go f is an identity function on A and fog is an identity function on B. Then g is called inverse of
f and is denoted by f−1.
For a function to be invertible it must be bijective
Note : "
The inverse of a bijection is unique.
"
Inverse of an even function is not defined.
Properties of Inverse Function :
(a)
The graphs of f & g are the mirror images of each other in the line y = x. For example f(x) = a x and g(x)
= loga x are inverse of each other, and their graphs are mirror images of each other on the line y = x as
shown below.
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
11.
Normally
points of intersection of f and f–1 lie on the straight line y = x. However it must be noted that f(x)
–1
and f (x) may intersect otherwise also.
(c)
In general fog(x) and gof(x) are not equal but if they are equal then in majority of cases either f and g are
inverse of each other or atleast one of f and g is an identity function.
(d)
If f & g are two bijections f : A → B, g : B → C then the inverse of gof exists and
(gof)−1 = f−1 o g−1.
1
(e)
If f(x) and g are inverse function of each other then f′(g(x)) = g′( x )
2x + 3
Illustration # 16
(i)
Determine whether f(x) =
; R → R, is invertible or not? If so find it.
4
2x + 3
Solution:
As given function is one-one and onto, therefore it is invertible. y =
4
4y − 3
4x − 3
–1
⇒
x=
∴
f (x) =
2
2
–1 
2 
(ii)
Is the function f(x) = sin  2 x 1 − x  invertible?


Solution:
Domain of f is [–1, 1] and f is continuous
(b)
−1
1
 2
<x<
if

2
2
2
 1− x
f ′( x ) = 1 − 2x 2 1 − x 2 =  − 2
−1
1

or x >
if x <
2
 1 − x
2
2
 −1 1 
 and is decreasing in each of the intervals
,
f(x) is increasing in 
2
 2
(
2 1 − 2x 2
∴
)

 −1 
−1 
 − 1,
 and 
, 1
2

 2 
f(x) is not one-one, so is not invertible.
Let f(x) = x2 + 2x; x ≥ –1. Draw graph of f–1(x) also find the number of solutions of the equation,
f(x) = f–1(x)
12 of 41
∴
(iii)
f(x) = f–1(x) is equilavent to solving y = f(x) and y = x
⇒
x2 + 2x = x
⇒
x(x + 1) = 0
⇒
x = 0, –1
Hence two solution for f(x) = f–1(x)
2
(iv)
If y = f(x) = x – 3x + 1, x ≥ 2. Find the value of g′(1) where g is inverse of f
Solution
y=1
⇒
x2 – 3x + 1 = 1
But
x≥2
∴
x=3
Now
g(f(x)) = x
Differentiating both sides w.r.t. x
⇒
⇒
g′(f(x)). f′(x) = 1
1
f ′(3)
Alternate Method
y = x2 – 3x + 1
x2 – 3x +1 – y = 0
⇒
⇒
g′(f(3)) =
x=
=
x (x – 3) = 0
1
f ′( x )
1
g′ (1) = =
6−3
⇒
x = 0, 3
g′(f(x)) =
(As f′(x) = 2x – 3)
=
1
3
3 ± 9 − 4(1 − y )
2
3 ± 5 + 4y
x≥2
x=
⇒
2
3 + 5 + 4y
g(x) =
2
3 + 5 + 4y
2
1
g′(x) = 0 +
x
g′(1) =
1
x 5 + 4 x–1
5+4
12.
Determine f (x), if given function is invertible
2
f : (–∞, –1) → (–∞, –2) defined f(x) = –(x + 1) – 2
π
 π 7π 

(ii)
f:  ,
 → [–1, 1] defined by f(x) = sin  x + 3 
6 6 


2π
–1
Answer
(i) – 1 + − x − 2
(ii)
– sin x
3
1
=
9
=
1
3
Assignment:
(i)
I.
Equal or Identical Function :
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
98930 58881
Solution
Two functions f & g are said to be identical (or equal) iff :
(i)
The domain of f ≡ the domain of g.
(ii)
The range of f ≡ the range of g and
x
1
(iii)
f(x) = g(x), for every x belonging to their common domain. e.g. f(x) = & g(x) = 2 are identical functions.
x
x
2
x
But f(x) = x and g(x) =
are not identical functions.
x
Illustration # 17
Examine whether following pair of functions are identical or not
x2
& g(x) = x Answer
No, as domain of f(x) is R – {0} while domain of g(x) is R
x
2
2
2
(ii)
f(x) = sin x + cos x & g(x) = sec x – tan2x
π


Answer
No, as domain are not same. Domain of f(x) is R while that of g(x) is R – (2n + 1) ; n ∈ I
2


Assignment: 13.
Examine whether following pair of functions are identical or not
 x
x≠0

(i)
f(x) = sgn (x) & g(x) =  | x |
 0
x=0
(i)
f(x) =
J.
f(x) = sin–1x + cos –1x & g(x) =
 1
 1
If f(x) is a polynomial function satisfying f(x) . f   = f(x) + f   ∀ x ∈ R – {0} and
x
x
f(2) = 9, then find f (3)
Solution
f(x) = 1 ± xn
As f(2) = 9
∴
f(x) = 1 + x3
Hence f(3) = 1 + 33 = 28
 1
 1
Assignment: 14.
If f(x) is a polynomial function satisfying f(x) . f   = f(x) + f   ∀ x ∈ R – {0} and f(3) = –8,
x
 
x
then find f(4)
Answer
– 15
13 of 41
π
Answer
(i)
Yes
(ii)
No
2
General : If x, y are independent variables, then:
(i)
f (xy) = f (x) + f (y) ⇒ f (x) = k ln x or f (x) = 0. (ii)
f (xy) = f (x). f (y) ⇒ f (x) = xn , n ∈ R
(iii)
f (x + y) = f (x). f (y) ⇒ f (x) = akx.
(iv)
f (x + y) = f (x) + f (y) ⇒ f(x) = kx, where k is a constant.
 1
 1
(v)
f(x) . f   = f(x) + f  
⇒
f(x) = 1 ± xn where n ∈ N
x
 
x
(ii)
If f(x + y) = f(x) . f(y) for all real x, y and f(0) ≠ 0 then prove that the function, g(x) =
f (x)
1 + f 2 (x)
is an even function
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
15.
98930 58881
Illustration # 18
:
GENERAL DEFINITION :
If to every value (Considered as real unless other−wise stated) of a variable x, which belongs to some collection
(Set) E, there corresponds one and only one finite value of the quantity y, then y is said to be a function (Single
valued) of x or a dependent variable defined on the set E ; x is the argument or independent variable .
If to every value of x belonging to some set E there corresponds one or several values of the variable y, then y is
called a multiple valued function of x defined on E.Conventionally the word "FUNCTION” is used only as the
meaning of a single valued function, if not otherwise stated.
x
Pictorially : 

→
input
f (x ) = y
output
 → , y is called the image of x & x is the pre-image of y under f.
Every function from A → B satisfies the following conditions .
(i)
f⊂ AxB
(ii)
∀ a ∈ A ⇒ (a, f(a)) ∈ f
(iii)
(a, b) ∈ f & (a, c) ∈ f ⇒ b = c
and
98930 58881
1.
14 of 41
THINGS TO REMEMBER
Short Revision
DOMAIN, CO−
−DOMAIN & RANGE OF A FUNCTION :
Let f : A → B, then the set A is known as the domain of f & the set B is known as co-domain of f . The set of
all
f images of elements of A is known as the range of
f . Thus :
Domain of f = {a  a ∈ A, (a, f(a)) ∈ f}
Range of f = {f(a)  a ∈ A, f(a) ∈ B}
It should be noted that range is a subset of co−domain . If only the rule of function is given then the domain of the function
is the set of those real numbers, where function is defined. For a continuous function, the interval from minimum to
maximum value of a function gives the range.
3.
(i)
IMPORTANT TYPES OF FUNCTIONS :
POLYNOMIAL FUNCTION :
If a function f is defined by f (x) = a0 xn + a1 xn−1 + a2 xn−2 + ... + an−1 x + an where n is a non negative integer and a0,
a1, a2, ..., an are real numbers and a0 ≠ 0, then f is called a polynomial function of degree n .
NOTE : (a)
A polynomial of degree one with no constant term is called an odd linear
function . i.e. f(x) = ax , a ≠ 0
(b)
There are two polynomial functions , satisfying the relation ;
f(x).f(1/x) = f(x) + f(1/x). They are :
&
(ii) f(x) = 1 − xn , where n is a positive integer .
(i) f(x) = xn + 1
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
2.
(ii)
ALGEBRAIC FUNCTION :
y is an algebraic function of x, if it is a function that satisfies an algebraic equation of the form
P 0 (x) y n + P 1 (x) y n−1 + ....... + P n−1 (x) y + P n (x) = 0 Where n is a positive integer and
P0 (x), P1 (x) ........... are Polynomials in x.
e.g. y = x is an algebraic function, since it satisfies the equation y² − x² = 0.
Note that all polynomial functions are Algebraic but not the converse. A function that is not algebraic is called
TRANSCEDENTAL FUNCTION .
(iii)
FRACTIONAL RATIONAL FUNCTION :
A rational function is a function of the form. y = f (x) =
(IV)
g(x)
h (x )
, where
g (x) & h (x) are polynomials & h (x) ≠ 0.
EXPONENTIAL FUNCTION :
A function f(x) = ax = ex ln a (a > 0 , a ≠ 1, x ∈ R) is called an exponential function. The inverse of the exponential
function is called the logarithmic function . i.e. g(x) = loga x .
Note that f(x) & g(x) are inverse of each other & their graphs are as shown .
+∞
(1, 0)
y
f(x) = ax , 0 < a < 1
→
log a
)=
x
(
g
%
x
y
=
x
)45º
(1, 0)
%
g(x) = loga x
98930 58881
ABSOLUTE VALUE FUNCTION :
A function y = f (x) = x is called the absolute value function or Modulus function. It is defined as : y = x=
 x if x ≥ 0
− x if x < 0

(vi)
SIGNUM FUNCTION :
A function y= f (x) = Sgn (x) is defined as follows :
1
y = f (x) =  0

y
y = 1 if x > 0
for x > 0
for x = 0
> x
O
 − 1 for x < 0
y = Sgn x
y = −1 if x < 0
It is also written as Sgn x = |x|/ x ;
x ≠ 0 ; f (0) = 0
GREATEST INTEGER OR STEP UP FUNCTION :
The function y = f (x) = [x] is called the greatest integer function where [x] denotes the greatest integer less than
or equal to x . Note that for :
−1 ≤ x < 0
;
[x] = − 1
0≤x< 1
;
[x] = 0
1≤x< 2
;
[x] = 1
2≤x < 3
;
[x] = 2
and so on .
Properties of greatest integer function :
y$
(a)
(b)
(c)
(d)
(viii)
4.
[x] ≤ x < [x] + 1 and
x − 1 < [x] ≤ x , 0 ≤ x − [x] < 1
[x + m] = [x] + m if m is an integer .
[x] + [y] ≤ [x + y] ≤ [x] + [y] + 1
[x] + [− x] = 0 if x is an integer
= − 1 otherwise .
3
•
2
1
−3
3
−2
•
•
•
º
−1
−1
•
º
−2
º
º
1
y$
•
−1
1− − −
º
º
−−
•
•
1
DOMAINS AND RANGES OF COMMON FUNCTION :
Domain
(i.e. values taken by x)
Range
(i.e. values taken by f (x) )
R = (set of real numbers)
R,
if n is odd
+
R ∪ {0} , if n is even
Algebraic Functions
(i)
xn , (n ∈ N)
º
%
x
2
−3
FRACTIONAL PART FUNCTION :
It is defined as :
g (x) = {x} = x − [x] .
e.g. the fractional part of the no. 2.1 is
2.1− 2 = 0.1 and the fractional part of − 3.7 is 0.3. The period
of this function is 1 and graph of this function is as shown .
Function
(y = f (x) )
A.
graph of y = [x]
graph of y = {x}
º
•
2
º
−−−−
−−
(vii)
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
(v)
=
$
(0, 1)
)45º
x
15 of 41
+∞
$
1
a>
x ,
(0, 1)
=a
)←
f(x 
•
%x
R – {0} , if n is odd
R+ ,
(iii)
(iv)
B.
1
x1 / n
, (n ∈ N)
R – {0} , if n is odd
R – {0} , if n is odd
R+ ,
R+ ,
if n is even
sin x
cos x
R
R
(iii)
tan x
R – (2k + 1)
if n is even
[–1, + 1]
[–1, + 1]
π
, k ∈I
2
R
π
, k ∈I
2
(v)
cosec x
R – kπ , k ∈ I
(vi)
cot x
R – kπ , k ∈ I
Inverse Circular Functions (Refer after Inverse is taught )
sec x
R – (2k + 1)
(– ∞ , – 1 ] ∪ [ 1 , ∞ )
(– ∞ , – 1 ] ∪ [ 1 , ∞ )
R
 π π
− 2 , 2 
[ 0, π]
(i)
sin–1 x
[–1, + 1]
(ii)
cos–1 x
[–1, + 1]
(iii)
tan–1 x
R
 π π
− , 
 2 2
(iv)
cosec –1x
(– ∞ , – 1 ] ∪ [ 1 , ∞ )
 π π
− 2 , 2  – { 0 }


(v)
sec–1 x
(– ∞ , – 1 ] ∪ [ 1 , ∞ )
(vi)
cot –1 x
R
Domain
(i.e. values taken by x)
π 
[ 0, π] –  
2 
( 0, π)
Range
(i.e. values taken by f (x) )
Exponential Functions
(i)
(ii)
(iii)
(iv)
E.
R,
if n is odd
+
R ∪ {0} , if n is even
(i)
(ii)
Function
(y = f (x) )
D.
R,
if n is odd
+
R ∪ {0} , if n is even
Trigonometric Functions
(iv)
C.
x1 / n , (n ∈ N)
if n is even
ex
e1/x
ax , a > 0
a1/x , a > 0
R
R–{0}
R
R –{0}
R+
R+ – { 1 }
R+
R+ – { 1 }
Logarithmic Functions
(i)
logax , (a > 0 ) (a ≠ 1)
R+
R
(ii)
1
logxa = log x
a
R+ – { 1 }
R–{0}
(a > 0 ) (a ≠ 1)
16 of 41
R – {0}
98930 58881
1
, (n ∈ N)
xn
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
(ii)
H.
I.
[x]
R
I
(ii)
1
[x]
R – [0, 1 )
1

 , n ∈ I − {0} 
n

Fractional Part Functions
(i)
{x}
R
[0, 1)
(ii)
1
{x}
R–I
(1, ∞)
Modulus Functions
(i)
|x|
R
R+ ∪ { 0 }
(ii)
1
|x|
R–{0}
R+
R
{–1, 0 , 1}
R
{c}
Signum Function
|x|
,x ≠0
x
=0,x=0
sgn (x) =
J.
Constant Function
say f (x) = c
5.
(i)
(ii)
(iii)
6.
17 of 41
(i)
98930 58881
G.
Integral Part Functions Functions
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
F.
EQUAL OR IDENTICAL FUNCTION :
Two functions f & g are said to be equal if :
The domain of f = the domain of g.
The range of f = the range of g
and
f(x) = g(x) , for every x belonging to their common domain. eg.
x
1
f(x) =
& g(x) = 2 are identical functions .
x
x
CLASSIFICATION OF FUNCTIONS :
One − One Function (Injective mapping) :
A function f : A → B is said to be a one−one function or injective mapping if different elements of A have
different f images in B . Thus for x1, x2 ∈ A & f(x1) ,
f(x2) ∈ B , f(x1) = f(x2) ⇔ x1 = x2 or x1 ≠ x2 ⇔ f(x1) ≠ f(x2) .
Diagramatically an injective mapping can be shown as
OR
Note : (i)
Any function which is entirely increasing or decreasing in whole domain, then
f(x) is one−one .
(ii)
If any line parallel to x−axis cuts the graph of the function atmost at one point,
then the function is one−one .
Diagramatically a many one mapping can be shown as
18 of 41
Many–one function :
A function f : A → B is said to be a many one function if two or more elements of A have the same f image
in B . Thus f : A → B is many one if for ; x1, x2 ∈ A , f(x1) = f(x2) but x1 ≠ x2 .
OR
(ii)
Any continuous function which has atleast one local maximum or local minimum, then f(x) is many−one .
In other words, if a line parallel to x−axis cuts the graph of the function atleast at two points, then
f is many−one .
If a function is one−one, it cannot be many−one and vice versa .
98930 58881
Note : (i)
Diagramatically surjective mapping can be shown as
OR
Note that : if range = co−domain, then f(x) is onto.
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Onto function (Surjective mapping) :
If the function f : A → B is such that each element in B (co−domain) is the f image of atleast one element in A, then
we say that f is a function of A 'onto' B . Thus f : A → B is surjective iff ∀ b ∈ B, ∃ some a ∈ A such that f (a)
=b.
Into function :
If f : A → B is such that there exists atleast one element in co−domain which is not the image of any element in
domain, then f(x) is into .
Diagramatically into function can be shown as
OR
Note that : If a function is onto, it cannot be into and vice versa . A polynomial of degree even will always be into.
Thus a function can be one of these four types :
(a)
one−one onto (injective & surjective)
(b)
one−one into (injective but not surjective)
(c)
many−one onto (surjective but not injective)
(d)
many−one into (neither surjective nor injective)
Note : (i)
(ii)
If f is both injective & surjective, then it is called a Bijective mapping.
The bijective functions are also named as invertible, non singular or biuniform functions.
If a set A contains n distinct elements then the number of different functions defined from A → A is nn
& out of it n ! are one one.
19 of 41
Identity function :
The function f : A → A defined by f(x) = x ∀ x ∈ A is called the identity of A and is denoted by IA.
It is easy to observe that identity function is a bijection .
Constant function :
A function f : A → B is said to be a constant function if every element of A has the same f image in B . Thus f : A
→ B ; f(x) = c , ∀ x ∈ A , c ∈ B is a constant function. Note that the range of a constant function is a singleton
and a constant function may be one-one or many-one, onto or into .
(iii)
8.
f
f (x )
  (x) =
 g
g (x )
domain is {x  x ∈ A ∩ B s . t g(x) ≠ 0} .
98930 58881
ALGEBRAIC OPERATIONS ON FUNCTIONS :
If f & g are real valued functions of x with domain set A, B respectively, then both f & g are defined in
A ∩ B. Now we define f + g , f − g , (f . g) & (f/g) as follows :
(i)
(f ± g) (x) = f(x) ± g(x)
(ii)
(f . g) (x) = f(x) . g(x)
COMPOSITE OF UNIFORMLY & NON-UNIFORMLY DEFINED FUNCTIONS :
Let f : A → B & g : B → C be two functions . Then the function gof : A → C defined by
(gof) (x) = g (f(x)) ∀ x ∈ A is called the composite of the two functions f & g .
f (x )
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
7.
x→
Diagramatically 
→ g (f(x)) .
 →
Thus the image of every x ∈ A under the function gof is the g−image of the f−image of x .
Note that gof is defined only if ∀ x ∈ A, f(x) is an element of the domain of g so that we can take its g-image.
Hence for the product gof of two functions f & g, the range of f must be a subset of the domain of g.
PROPERTIES OF COMPOSITE FUNCTIONS :
(i)
The composite of functions is not commutative i.e. gof ≠ fog .
(ii)
The composite of functions is associative i.e. if f, g, h are three functions such that fo (goh) & (fog) oh are
defined, then fo (goh) = (fog) oh .
(iii)
The composite of two bijections is a bijection i.e. if f & g are two bijections such that gof is defined, then
gof is also a bijection.
9.
10.
11.
12.
HOMOGENEOUS FUNCTIONS :
A function is said to be homogeneous with respect to any set of variables when each of its terms is of
the same degree with respect to those variables .
For example 5 x2 + 3 y2 − xy is homogeneous in x & y . Symbolically if ,
f (tx , ty) = tn . f (x , y) then f (x , y) is homogeneous function of degree n .
BOUNDED FUNCTION :
A function is said to be bounded if f(x) ≤ M , where M is a finite quantity .
IMPLICIT & EXPLICIT FUNCTION :
A function defined by an equation not solved for the dependent variable is called an
IMPLICIT FUNCTION . For eg. the equation x3 + y3 = 1 defines y as an implicit function. If y has been expressed
in terms of x alone then it is called an EXPLICIT FUNCTION.
INVERSE OF A FUNCTION :
Let f : A → B be a one−one & onto function, then their exists a unique function
g : B → A such that f(x) = y ⇔ g(y) = x, ∀ x ∈ A & y ∈ B . Then g is said to be inverse of f . Thus g = f−
1
: B → A = {(f(x), x)  (x, f(x)) ∈ f} .
PROPERTIES OF INVERSE FUNCTION :
(i)
The inverse of a bijection is unique .
(ii)
If f : A → B is a bijection & g : B → A is the inverse of f, then fog = IB and
gof = IA , where IA & IB are identity functions on the sets A & B respectively.
Note that the graphs of f & g are the mirror images of each other in the
line y =
2
x . As shown in the figure given below a point (x ',y ' ) corresponding to y = x (x >0) changes to (y ',x ' )
corresponding to y = + x , the changed form of x = y .
20 of 41
ODD & EVEN FUNCTIONS :
If f (−x) = f (x) for all x in the domain of ‘f’ then f is said to be an even function.
e.g. f (x) = cos x ; g (x) = x² + 3 .
If f (−x) = −f (x) for all x in the domain of ‘f’ then f is said to be an odd function.
e.g. f (x) = sin x ; g (x) = x3 + x .
NOTE : (a)
(b)
(c)
(d)
(e)
f (x) − f (−x) = 0 => f (x) is even & f (x) + f (−x) = 0 => f (x) is odd .
A function may neither be odd nor even .
Inverse of an even function is not defined .
Every even function is symmetric about the y−axis & every odd function is symmetric about the origin .
Every function can be expressed as the sum of an even & an odd function.
e.g. f ( x) =
(f)
(g)
98930 58881
13.
The inverse of a bijection is also a bijection .
If f & g are two bijections f : A → B , g : B → C then the inverse of gof exists and
(gof)−1 = f−1 o g−1 .
f ( x ) + f ( − x ) f ( x) − f ( − x )
+
2
2
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
(iii)
(iv)
The only function which is defined on the entire number line & is even and odd at the same time is f(x) = 0.
If f and g both are even or both are odd then the function f.g will be even but if any one of them is odd
then f.g will be odd .
14.
PERIODIC FUNCTION :
A function f(x) is called periodic if there exists a positive number T (T > 0) called the period of the function such
that f (x + T) = f(x), for all values of x within the domain of x.
e.g. The function sin x & cos x both are periodic over 2π & tan x is periodic over π .
NOTE : (a)
f (T) = f (0) = f (−T) , where ‘T’ is the period .
(b)
Inverse of a periodic function does not exist .
(c)
Every constant function is always periodic, with no fundamental period .
(d)
If f (x) has a period T & g (x) also has a period T then it does not mean that f (x) + g (x) must have
a period T . e.g. f (x) = sinx + cosx.
15.
1
and
f (x)
(e)
If f(x) has a period p, then
(f)
if f(x) has a period T then f(ax + b) has a period T/a (a > 0) .
f (x) also has a period p .
GENERAL :
If x, y are independent variables, then :
(i)
f(xy) = f(x) + f(y) ⇒ f(x) = k ln x or f(x) = 0 .
(ii)
f(xy) = f(x) . f(y) ⇒ f(x) = xn , n ∈ R
kx
(iii)
f(x + y) = f(x) . f(y) ⇒ f(x) = a .
(iv)f(x + y) = f(x) + f(y) ⇒ f(x) = kx, where k is a constant .
EXER
CISE–1
EXERCISE–1
Q.1
Find the domains of definitions of the following functions :
(Read the symbols [*] and {*} as greatest integers and fractional part functions respectively.)
(i) f (x) = cos2x + 16 − x 2
(iii) f (x) = ln  x 2 − 5x − 24 − x − 2 


(ii) f (x) = log7 log5 log3 log2 (2x3 + 5x2 − 14x)
(iv) f (x) =
1 − 5x
7 −x −7
4x 2 − 1
+ ln x(x 2 − 1)
(viii) f (x) =
log 1
2
1
(ix) f (x) = x 2 − x +
x
x −1
2
(x) f (x) = ( x 2 − 3x − 10) . ln 2 ( x − 3)
9 − x2
cos x −
(xi) f(x) = logx (cos 2πx)
(xiii) f(x) =
(
log1 / 3 log 4
(xii) f (x) =
( [x]
2
−5
))
(xiv) f(x) =
1
2
6 + 35x − 6 x 2
1
1
+ log(2{x}− 5) (x 2 − 3x + 10) +
[x]
1− x
,
(xv) f(x) = logx sin x


1

(xvi) f(x) = log2  − log1/ 2  1 +
x°
sin 100


( )
(xvii) f (x) =
(xviii) f (x) =

 +


1
+ log1 – {x}(x2 – 3x + 10) +
[x ]
(5x − 6 − x ) [{ln{x}}]
2
log10 (log10 x) − log10 (4 − log10 x) − log10 3
1
2−| x|
1
+
sec(sin x)

+ (7 x − 5 − 2x ) +  ln

2
7

 − x  
2

(xix) If f(x) = x 2 − 5 x + 4 & g(x) = x + 3 , then find the domain of
Q.2
5
(
)
2 (sin x − cos x) + 3
x
(iv) f (x) = 1+ | x |
Q.5
(ii) y =
2x
1+ x2
(iii) f(x) =
x 2 − 3x + 2
x2 + x − 6
(v) y = 2 − x + 1 + x
x +4 −3
x −5
Draw graphs of the following function , where [ ] denotes the greatest integer function.
(i) f(x) = x + [x]
(ii) y = (x)[x] where x = [x] + (x) & x > 0 & x ≤ 3
(iii) y = sgn [x] (iv) sgn (x −x)
Classify the following functions f(x) definzed in R → R as injective, surjective, both or none .
(vi) f (x) = log(cosec x - 1) (2 − [sin x] − [sin x]2)
Q.4
f
(x) .
g
Find the domain & range of the following functions .
( Read the symbols [*] and {*} as greatest integers and fractional part functions respectively.)
(i) y = log
Q.3
−1
(vii) f (x) =
(a) f(x) =
x 2 + 4x + 30
x 2 − 8x + 18
Let f(x) =
1
. Let f2(x) denote f [f (x)] and f3(x) denote f [f {f(x)}]. Find f3n(x) where n is a natural number. Also
1− x
(b) f(x) = x3 − 6 x2 + 11x − 6
(c) f(x) = (x2 + x + 5) (x2 + x − 3)
state the domain of this composite function.
Q.6
Q.7
98930 58881
1
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
(vii) f (x) =
21 of 41
 2 log10 x + 1 

(vi) f (x) = log100 x 
−x


(v) y = log10 sin (x − 3) + 16 − x 2
π
π


5
If f(x) = sin²x + sin²  x +  + cos x cos x +  and g   = 1 , then find (gof) (x).
3
3


4
The function f(x) is defined on the interval [0,1]. Find the domain of definition of the functions.
(a) f (sin x)
(b) f (2x+3)
Q.8(i) Find whether the following functions are even or odd or none
(b) f(x) =
(
)
x ax +1
(c) f(x) = sin x + cos x
a −1
x
(1 + 2 )
(f) f(x) =
2
x
(d) f(x) = x sin x − x
2
(g) f(x)=
(e) f(x)= sin x − cos x
3
x
x
+ +1
e −1 2
22 of 41
(a) f(x) = log  x + 1 + x 2 
2x
(h) f(x) = [(x+1)²]1/3 + [(x −1)²]1/3
x
(ii) If f is an even function defined on the interval (−5, 5), then find the 4 real values of x satisfying the equation f (x) =
Q.10
Show if f(x) = n a − x n , x > 0 n ≥ 2 , n ∈ N , then (fof) (x) = x . Find also the inverse of f(x).
Q.11
(a)
Represent the function f(x) = 3x as the sum of an even & an odd function.
(b)
For what values of p ∈ z , the function f(x) = n x p , n ∈ N is even.
A function f defined for all real numbers is defined as follows for x ≥ 0 : f (x) = [1x,,x0>≤1x≤1
How is f defined for x ≤ 0 if : (a) f is even
Q.13

If f (x) = max  x ,
(b) f is odd?
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Q.12
98930 58881
Q.9
 x +1 
f
 ..
 x+2
Write explicitly, functions of y defined by the following equations and also find the domains of definition of the given
implicit functions :
(a) 10x + 10y = 10
(b) x + y= 2y
1 
 for x > 0 where max (a, b) denotes the greater of the two real numbers a and b. Define the
x 
function g(x) = f(x) . f  1  and plot its graph.
 x
Q.14
The function f (x) has the property that for each real number x in its domain, 1/x is also in its domain and f (x) +
Q.15
1
f   = x. Find the largest set of real numbers that can be in the domain of f (x)?
x
Compute the inverse of the functions:
x
(a) f(x) = ln  x + x 2 + 1
1

x −1
(b) f(x) = 2
3
(c) y =
10 x − 10 − x
10 x + 10 − x

Q.16
A function f :  , ∞ →  , ∞ defined as, f(x) = x2 − x + 1. Then solve the equation f (x) = f −1 (x).


2
4
Q.17
Function f & g are defined by f(x) = sin x, x∈R ; g(x) = tan x , x∈R −  K + 1  π
where K ∈ I . Find
Q.18

(i) periods of fog & gof.
Find the period for each of the following functions :
(a) f(x)= sin4x + cos4x (b) f(x) = cosx
(d) f(x)= cos
2
(ii) range of the function fog & gof .
(c) f(x)= sinx+cosx
3
2
x − sin x .
5
7
Q.19
Prove that the functions ;
(c) f(x) = x + sin x
(a) f(x) = cos x
(d) f(x) = cos x2
(b) f(x) = sin x
are not periodic .
Q.20
Find out for what integral values of n the number 3π is a period of the function :
f(x) = cos nx . sin (5/n) x.
EXER
CISE–2
EXERCISE–2
Let f be a one−one function with domain {x,y,z} and range {1,2,3}. It is given that exactly one of the following
statements is true and the remaining two are false .
f(x) = 1 ; f(y) ≠ 1
; f(z) ≠ 2 . Determine f−1(1)
Q.2
Solve the following problems from (a) to (e) on functional equation.
23 of 41
Q.1
The function f (x) defined on the real numbers has the property that f ( f ( x ) )· (1 + f ( x ) ) = – f (x) for all x in the
domain of f. If the number 3 is in the domain and range of f, compute the value of f (3).
(b)
Suppose f is a real function satisfying f (x + f (x)) = 4 f (x) and f (1) = 4. Find the value of f (21).
(c)
Let 'f' be a function defined from R+ → R+ . If [ f (xy)]2 = x ( f ( y) )2 for all positive numbers x and y and f (2) =
(d)
Let f (x) be a function with two properties
(i)
for any two real number x and y, f (x + y) = x + f (y) and
Find the value of f (100).
(e)
(ii)
f (0) = 2.
Let f be a function such that f (3) = 1 and f (3x) = x + f (3x – 3) for all x. Then find the value of f (300).
Q.3(a) A function f is defined for all positive integers and satisfies f(1) = 2005 and f(1)+ f(2)+ ... + f(n) = n2f(n)
for all n > 1. Find the value of f(2004).
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
6, find the value of f (50).
98930 58881
(a)
(b) If a, b are positive real numbers such that a – b = 2, then find the smallest value of the constant L for which
x 2 + ax − x 2 + bx < L for all x > 0.
(c) Let f (x) = x2 + kx ; k is a real number. The set of values of k for which the equation f (x) = 0 and f ( f ( x) ) = 0
have same real solution set.
(d) If f (2x + 1) = 4x2 + 14x, then find the sum of the roots of the equation f (x) = 0.
ax + b
5
for real a, b and c with a ≠ 0. If the vertical asymptote of y = f (x) is x = – and the vertical
Q.4
Let f (x) =
4x + c
4
3
asymptote of y = f –1 (x) is x = , find the value(s) that b can take on.
4
Q.5
A function f : R → R satisfies the condition, x2 f (x) + f (1 – x) = 2x – x4 . Find f (x) and its domain and range.
Q.6
Suppose p(x) is a polynomial with integer coefficients. The remainder when p(x) is divided by x – 1 is 1 and the
remainder when p(x) is divided by x – 4 is 10. If r (x) is the remainder when p(x) is divided by (x – 1)(x – 4), find
the value of r (2006).
− | ln{ x }|
Q.7
e

Prove that the function defined as , f (x) = 
 {x}
− {x}
1
| ln{ x }|
where ever it exists
otherwise , then
f (x) is odd as well as even. ( where {x} denotes the fractional part function )
Q.8
Q.9
 1

 
1 
πx
π
In a function 2 f(x) + xf   − 2f  2 sin  π  x +    = 4 cos2
+ x cos
2
x
 x
4 
 

Prove that
(i) f(2) + f(1/2) = 1
and
(ii) f(2) + f(1) = 0
A function f , defined for all x , y ∈ R is such that f (1) = 2 ; f (2) = 8
& f (x + y) − k xy = f (x) + 2 y2 , where k is some constant . Find f (x) & show that :
 1 
 = k for x + y ≠ 0.
 x + y
f (x + y) f 
Q.10
Let ‘f’ be a real valued function defined for all real numbers x such that for some positive constant ‘a’ the equation
1
2
f (x + a ) = + f (x) − (f (x))
2
holds for all x . Prove that the function f is periodic .
Q.12
f (x) = −1 + x − 2 , 0 ≤ x ≤ 4
g (x) = 2 − x , − 1 ≤ x ≤ 3
Then find fog (x) & gof (x) . Draw rough sketch of the graphs of fog (x) & gof (x) .
Find the domain of definition of the implicit function defined by the implicit equation ,
Q.13
3y + 2x = 24 x − 1 .
Let {x} & [x] denote the fractional and integral part of a real number x respectively. Solve 4{x}= x + [x]
If
24 of 41
Q.11
2
4
98930 58881
 1 
 2 
 3 
 2005 
9x
+f 
 +f 
 + ....+ f 

Q.14 Let f (x) = x
then find the value of the sum f 
 2006 
 2006 
 2006 
 2006 
9 +3
Q.15 Let f (x) = (x + 1)(x + 2)(x + 3)(x + 4) + 5 where x ∈ [–6, 6]. If the range of the function is
[a, b] where a, b ∈ N then find the value of (a + b).
Q.16 Find a formula for a function g (x) satisfying the following conditions
(a)
domain of g is (– ∞, ∞)
(b)
range of g is [–2, 8]
(c)
g has a period π and
(d)
g (2) = 3
Q.18
3 4
The set of real values of 'x' satisfying the equality   +   = 5 (where [ ] denotes the greatest integer function)
x x
b
 b
belongs to the interval  a ,  where a, b, c ∈ N and is in its lowest form. Find the value of a + b + c + abc.
c
 c
Find the set of real x for which the function f(x) =
[
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Q.17
1
is not defined, where [x] denotes the
x − 1 + 12 − x − 11
] [
]
greatest integer function.
A is a point on the circumference of a circle. Chords AB and AC divide the area of the circle into three equal parts
. If the angle BAC is the root of the equation, f (x) = 0 then find f (x) .
Q.20 If for all real values of u & v, 2 f(u) cos v = f(u + v) + f(u − v), prove that, for all real values of x
(i) f(x) + f(− x) = 2a cos x
(ii) f(π − x) + f(− x) = 0
(iii) f(π − x) + f(x) = − 2b sin x . Deduce that f(x) = a cos x − b sin x, a, b are arbitrary constants.
Q.19
EXER
CISE–3
EXERCISE–3
Q.1
If the functions f , g , h are defined from the set of real numbers R to R such that ;
 0, if x ≤ 0
f (x)= x − 1, g (x) = x + 1 , h (x) = 
; then find the composite function ho(fog) & determine
 x , if x ≥ 0
2
2
whether the function (fog) is invertible & the function h is the identity function.
(
)
[REE '97, 6]
2
Q.2(a) If g (f(x)) = sin x & f (g(x)) = sin x , then :
(A) f(x) = sin2 x , g(x) = x
(B) f(x) = sin x , g(x) = x
(C) f(x) = x2 , g(x) = sin x
(D) f & g cannot be determined
(b) If f(x) = 3x − 5, then f−1(x)
(A) is given by
1
3x − 5
(B) is given by
x +5
3
(C) does not exist because f is not one−one (D) does not exist because f is not onto [JEE'98, 2 + 2]
Q.3
If the functions f & g are defined from the set of real numbers R to R such that f(x) = e x,
g(x) = 3x − 2, then find functions fog & gof. Also find the domains of functions (fog)−1 & (gof)−1. [ REE '98, 6 ]
Q.4
If the function f : [1, ∞) → [1, ∞) is defined by f(x) = 2x (x − 1), then f−1(x) is :
 1
(A)  
 2
Q.5
x (x − 1)
(B)
(
1
1 + 1 + 4 log2 x
2
)
(C)
(
1
1 − 1 + 4 log2 x
2
)
[ JEE '99, 2 ]
(D) not defined
The domain of definition of the function, y (x) given by the equation, 2x + 2y = 2 is :
(A) 0 < x ≤ 1
(B) 0 ≤ x ≤ 1
(C) − ∞ < x ≤ 0
(D) − ∞ < x < 1
− 1 , x < 0

Q.7(a) Let g (x) = 1 + x − [ x ] & f (x) =  0 , x = 0 . Then for all x , f (g (x)) is equal to
1 , x>0

(B) 1
(C) f (x)
(D) g (x)
1
(b) If f : [1 , ∞) → [2 , ∞) is given by , f (x) = x + , then f −1 (x) equals
x
2
x
x+ x −4
x − x2 − 4
(A)
(B)
(C)
(D) 1 −
1 + x2
2
2
log (x + 3)
(c) The domain of definition of f (x) = 2 2
is :
x + 3x + 2
x2 − 4
(d)
(A) R \ {− 1, − 2}
(B) (− 2, ∞)
(C) R\{− 1, − 2, − 3} (D) (− 3, ∞) \ {− 1, − 2}
Let E = {1, 2, 3, 4 } & F = {1, 2}. Then the number of onto functions from E to F is
(A) 14
(B) 16
(C) 12
(D) 8
(e)
Let f (x) =
αx
, x ≠ − 1 . Then for what value of α is f (f (x)) = x ?
x +1
98930 58881
(A) x
25 of 41
Given x = {1, 2, 3, 4}, find all one−one, onto mappings, f : X → X such that,
f (1) = 1 , f (2) ≠ 2 and f (4) ≠ 4 .
[ REE 2000, 3 out of 100 ]
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Q.6
(B) − 2
(C) 1
(D) − 1.
(A) 2
2
Q.8(a) Suppose f(x) = (x + 1) for x > –1. If g(x) is the function whose graph is the reflection of the graph of f(x) with
respect to the line y = x, then g(x) equals
1
, x > –1 (C) x + 1 , x > –1
(D) x – 1, x > 0
(x + 1) 2
(b) Let function f : R → R be defined by f (x) = 2x + sinx for x ∈ R. Then f is
(A) one to one and onto
(B) one to one but NOT onto
(C) onto but NOT one to one 2
(D) neither one to one nor onto
x +x+2
Q.9(a) Range of the function f (x) = 2
is
x + x +1
 7
 7
(A) [1, 2]
(B) [1, ∞ )
(C) 2 , 
(D) 1, 
 3
 3
x
(b) Let f (x) =
defined from (0, ∞ ) → [ 0, ∞ ) then by f (x) is
1+ x
(A) one- one but not onto
(B) one- one and onto
(C) Many one but not onto
(D) Many one and onto
[JEE 2003 (Scr),3+3]
Q.10 Let f (x) = sin x + cos x, g (x) = x2 – 1. Thus g ( f (x) ) is invertible for x ∈
(A) – x – 1, x > 0
(B)
 π 
 π 
 π π
(A) − , 0
(B) − , π (C) − , 
(D)
 2 
 2 
 4 4
Q.11(a) If the functions f (x) and g (x) are defined on R → R such that
0,

f (x) = 
 x,
x ∈ rational
0,

, g (x) = 
 x,
x ∈ irrational
(A) one-one and onto (B) neither one-one nor onto
 π
0, 2  [JEE 2004 (Screening)]


x ∈ irrational
x ∈ rational
then (f – g)(x) is
(C) one-one but not onto (D) onto but not one-one
(b) X and Y are two sets and f : X → Y. If {f (c) = y; c ⊂X, y ⊂ Y} and {f –1(d) = x; d ⊂ Y, x ⊂ X}, then the true
statement is
(
(C) f (f
)
( b) ) = b , b⊂ y
(A) f f −1 ( b) = b
−1
(B) f −1 (f (a ) ) = a
(D) f −1 (f (a ) ) = a , a ⊂ x
[JEE 2005 (Scr.)]
26 of 41
Exercise-4
Part : (A) Only one correct option
(A) (1, 4)
3.
(B) (– 2, 4)
( x +3 )x +
(B) (0, 3)
cot−1
6.
7.
8.
9.
10.
11.
12.
(D) [2, ∞)
(C) (2, 4)
2
 π
(B) 0, 
 2
Range of f(x) = log
(A) [0, 1]
5.
cos−1
x +3 x +1 is defined on the set S, where S is equal to:
(C) {0, − 3}
(D) [− 3, 0]
1
 2 1
 2
The range of the function f (x) = sin−1  x +  + cos−1  x −  , where [ ] is the greatest integer function, is:
2
2




The function f(x) =
(A) {0, 3}
π 
(A)  , π 
2 
4.
is
x 2 + 2x + 8
5
(C) { π }
 π
(D)  0, 
 2
 3
(C) 0, 
 2
(D) none of these
{ 2 (sinx – cosx) + 3} is
(B) [0, 2]
Range of f(x) = 4x + 2x + 1 is
(A) (0, ∞)
(B) (1, ∞)
(C) (2, ∞)
(D) (3, ∞)
If x and y satisfy the equation y = 2 [x] + 3 and y = 3 [x – 2] simultaneously, the [x + y] is
(A) 21
(B) 9
(C) 30
(D) 12
The function f : [2, ∞) → Y defined by f(x) = x2 − 4x + 5 is both one−one & onto if
(A) Y = R
(B) Y = [1, ∞)
(C) Y = [4, ∞)
(D) Y = [5, ∞)
Let S be the set of all triangles and R + be the set of positive real numbers. Then the function,
f : S → R+, f (∆) = area of the ∆, where ∆ ∈ S is :
(A) injective but not surjective
(B) surjective but not injective
(C) injective as well as surjective
(D) neither injective nor surjective
Let f(x) be a function whose domain is [– 5, 7]. Let g(x) = |2x + 5|. Then domain of (fog) (x) is
(A) [– 4, 1]
(B) [– 5, 1]
(C) [– 6, 1]
(D) none of these
The inverse of the function y =
ex − e−x
is
e x + e −x
1
1+ x
1
2+x
1
1− x
(A)
log
(B)
log
(C)
log
(D) 2 log (1 + x)
2
1− x
2
2−x
2
1+ x
The fundamental period of the function,
f(x) = x + a − [x + b] + sin πx + cos 2πx + sin 3πx + cos 4πx +...... + sin (2n − 1) πx
+ cos 2 nπx for every a, b ∈ R is: (where [ ] denotes the greatest integer function)
(A) 2
(B) 4
(C) 1
(D) 0
The period of e cos
(A) 1
4
πx + x − [ x ] + cos πx
(B) 2
is ______(where [ ] denotes the greatest integer function)
(C) 3
(D) 4
(
)
1
13.
If y = f(x) satisfies the condition f x + x1 = x2 + 2 (x ≠ 0) then f(x) =
x
(A) − x 2 + 2
(B) − x 2 − 2
(C) x2 + 2
(D) x2 − 2
14.
Given the function f(x) =
15.
16.
a x + a −x
(a > 0). If f(x + y) + f(x − y) = k f(x). f(y) then k has the value equal to:
2
(A) 1
(B) 2
(C) 4
(D) 1/2
A function f : R → R satisfies the condition, x2 f(x) + f(1 − x) = 2x − x4. Then f(x) is:
(A) – x2 – 1
(B) – x2 + 1
(C) x2 − 1
(D) – x4 + 1
The domain of the function, f (x) =
(A) (− 1, 0)
17.
18.
q 

 2p 

)
− 1 cos
(2 x + 1) tan 3 x is:
 π
(B) (− 1, 0) −  − 
 6
π
 π
(C) (− 1, 0] −  − , − 
2
 6
 π 
(D)  − , 0 
 6 


 q 
q 
  (C) R −  (− ∞ , − 1) ∩ −   (D) none of these
 2p 
 2p 


(B) R −  ( − ∞ , − 1] ∪ −

If [ 2 cos x ] + [ sin x ] = − 3, then the range of the function, f (x) = sin x + 3 cos x in [0, 2 π] is:
(where [. ] denotes greatest integer function)
(A) [− 2, − 1)
20.
(x
1
−1
If f (x) = 2 [x] + cos x, then f: R → R is: (where [. ] denotes greatest integer function)
(A) one−one and onto
(B) one−one and into (C) many−one and into
(D) many−one and onto
If q2 − 4 p r = 0, p > 0, then the domain of the function, f (x) = log (p x3 + (p + q) x2 + (q + r) x + r) is:
(A) R −  −
19.
98930 58881
2.
− log0.3 ( x − 1)
The domain of the function f(x) =
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
1.
(B) (− 2, − 1]
(C) (− 2, − 1)




The domain of the function f (x) = log1/2  − log2  1 +

(A) 0 < x < 1
(B) 0 < x ≤ 1


1 
 − 1 is:

x

(D) [–2, – 3 )
4
(C) x ≥ 1
(D) null set
The range of the functions f (x) = log
(A) (− ∞, 1)
22.
(B) (− ∞, 2)
(2 − log (16sin x + 1)) is
2
2
2
(C) (− ∞, 1]
(D) (− ∞, 2]
 1 + x3 
 + sin (sin x) + log
The domain of the function, f (x) = sin−1 
(x 2 + 1),
(3{x} + 1)
 2 x3/ 2 


27 of 41
21.
23.
where {x} represents fractional part function is:
(A) x ∈ {1}
(B) x ∈ R − {1, − 1}
(C) x > 3, x ≠ I
(D) none of these
The minimum value of f(x) = a tan2 x + b cot2 x equals the maximum value of g(x) = a sin2x + b cos2x where
a > b > 0, when
(A) 4a = b
(B) 3a = b
(C) a = 3b
(D) a = 4b
24.
Let f (2, 4) → (1, 3) be a function defined by f (x) = x −   (where [. ] denotes the greatest integer function), then
2
x
 
(x) is equal to :
x
(C) x + 1
(D) x − 1
(B) x +  
2
The image of the interval R when the mapping f: R → R given by f(x) = cot–1 (x2 – 4x + 3) is
 π 3π 
 3π 
π 
(A)  , 
(B)  , π 
(C) (0, π)
(D)  0, 
4 4 
 4
4 
(A) 2x
25.
26.
If the graph of the function f (x) =
ax − 1
x n (a x + 1)
is symmetric about y-axis, then n is equal to:
(A) 2
98930 58881
−1
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
f
(B) 2 / 3
(C) 1 / 4
(D) – 1 / 3
 π
: R+ →  0, 
27.
If f(x) = cot–1x
 2
and g(x) = 2x – x2
: R → R. Then the range of the function f(g(x)) wherever define is
 π
 π
π π
 π
(A)  0, 
(B)  0, 
(C)  , 
(D)  
2
4
4
2






4
28.
Let f: (e2, ∞) → R be defined by f(x) =!n (!n(!n x)), then
(A) f is one one but not onto
(B) f is on to but not one - one (C) f is one-one and onto (D) f is neither one-one nor onto
29.
Let f: (e, ∞) → R be defined by f(x) =!n (!n(!n x)), then
(A) f is one one but not onto
(B) f is on to but not one - one
(C) f is one-one and onto
(D) f is neither one-one nor onto
30.
Let f(x) = sin x and g(x) = |!n x| if composite functions fog(x) and gof (x) are defined and have ranges
R1 & R2 respectively then.
(A) R1 = {u: – 1 < u < 1}
R2 = {v: 0 < v < ∞}
R2 = {v: –1< v < 1}
(B) R1 = {u: – ∞ < u < 0}
(C) R1 = {u: 0 < u < ∞}
R2 = {v:– 1 < v < 1; v ≠ 0}
(D) R1 = {u: –1 < u < 1}
R2 = {v:0 < v < ∞}
−( x 2 −3 x + 2 )
is
31.
Function f : (– ∞, 1) → (0, e5] defined by f(x) = e
(A) many one and onto (B) many one and into (C) one one and onto
(D) one one and into
32.
The number of solutions of the equation [sin–1 x] = x – [x], where [ . ] denotes the greatest integer function is
(A) 0
(B) 1
(C) 2
(D) infinitely many
x
x
33.
The function f(x) = x
+
+ 1 is
e −1
2
(A) an odd function
(B) an even function
(C) neither an odd nor an even function
(D) a periodic function
Part : (B) May have more than one options correct
(sin −1 !og2 x),
34.
For the function f(x) = !n
35.
π

(B) Range is  − ∞ , !n 
(C) Domain is (1, 2]
2

A function ' f ' from the set of natural numbers to integers defined by,
1



(A) Domain is  , 2
2
 n − 1 , when n is odd

f (n) =  2n
is:
, when n is even
 − 2
(A) one-one
(B) many-one
(C) onto
(D) Range is R
(D) into
36.
Domain of f(x) = sin −1 [2 − 4x2] where [x] denotes greatest integer function is:



3 3
3
3
3
3


 − {0} (B) − 3 , 3  − {0}

,
,
(A)  −
(C) − 2 , 2 
(D)  −


2 
2 
 2
 2
 2 2 
37.
If F (x) =
(A)
(C)
sin π [x]
, then F (x) is:
{x}
periodic with fundamental period 1
range is singleton
(B)
even



identical to sgn  sgn
{x } 
− 1, where {x} denotes fractional part function and [ . ] denotes greatest
{x } 
28 of 41
(D)
integer function and sgn (x) is a signum function.
38.
D ≡ [− 1, 1] is the domain of the following functions, state which of them are injective.
(A) f(x) = x2
(B) g(x) = x3
(C) h(x) = sin 2x
(D) k(x) = sin (πx/2)
Exercise-5
2.
Find the domain of the function f(x) =
3.
Find the inverse of the following functions. f(x) = !n (x +
4.
 π π
Let f : − ,  → B defined by f (x) = 2 cos2x +
 3 6
f –1 (x).
 3x − 1
–1
1− 2x + 3 sin  2 
1+ x 2 )
3 sin2x + 1. Find the B such that f
–1
5.
Find for what values of x, the following functions would be identical.
 x −1
f (x) = log (x - 1) - log (x - 2) and g (x) = log  x − 2  .


6.
If f(x) =
7.
 1
 1
Let f(x) be a polynomial function satisfying the relation f(x). f   = f(x) + f   ∀ x ∈ R – {0} and
x
x
f(3) = –26. Determine f′(1).
8.
Find the domain of definitions of the following functions.
9.
4x
4x + 2
, then show that f(x) + f(1 – x) = 1
3 − 2 x − 21 − x
(i)
f (x) =
(iii)
f (x) = !og10 (1 – !og10(x2 – 5x + 16))
x−2
+
x+2
1− x
1+ x
(ii)
f (x) =
(ii)
 4 − x2

f (x) = sin !og  1 − x

f (x) = sin2 x + cos4x
Find the range of the following functions.
x 2 − 2x + 4
(i)
f (x) =
(iii)
f (x)= x4
x 2 + 2x + 4
− 2 x2 + 5
(iv)




98930 58881
x+2
Find the domain of the function f(x) =
exists. Also find
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
1
+
log10 (1 − x )
1.
10.
Solve the following equation for x (where [x] & {x} denotes integral and fractional part of x)
2x + 3 [x] – 4 {–x} = 4
11.
Draw the graph of following functions where [.] denotes greatest integer function and { .} denotes fractional part
function.
(i) y = {sin x }
(ii) y = [x] + { x }
12.
13.
2
Draw the graph of the function f(x) = x – 4 | x | + 3
a has exactly four distinct real roots.
Examine whether the following functions are even or odd or none.
x | x |,
x ≤ −1


(1 +2 x )7
[1 + x ] + [1 − x ], − 1 < x < 1
(i)
f (x) =
(ii)
f (x) = 
 − x | x |,
2x
x ≥1

(iii)
14.
and also find the set of values of ‘a’ for which the equation f(x) =
f (x) =
2x (sinx + tanx )
, where [ ] denotes greatest integer function.
x + 2π
2  π  − 3


Find the period of the following functions.
sin2 x
cos 2 x
− 1 + cot x − 1 + tanx
(i)
f (x) = 1
(ii)
f (x) = tan
π
[ x ] , where [.] denotes greatest integer function.
2
f (x) =
 |sinx | sinx 


+
 cos x | cos x | 
1
2
(iv)
f (x) =
sin x + sin 3 x
cos x + cos 3 x
29 of 41
(iii)
ANSWER KEY FUNCTIONS
EXER
CISE–1
EXERCISE–1
 5π − 3π 
 π π
 3π 5 π 
1

,
Q 1. (i) −
∪ − ,  ∪  ,  (ii)  − 4 , −  ∪ (2, ∞) (iii) (– ∞ , – 3]
2
4 
4 
 4
 4 4
 4
(iv) (– ∞, – 1) ∪ [0, ∞)
1 
 1   1
,

∪
(v) (3 − 2π < x < 3 − π) U (3 < x ≤ 4) (vi)  0,
 100   100 10 
(vii) (−1 < x < −1/2) U (x > 1)
(x) { 4 } ∪ [ 5, ∞ )
1 − 5 
1 + 5

, 0 ∪ 
, ∞ (ix) (−3, −1] U {0} U [ 1,3 )


 2
 2
(viii) 
(xi) (0 , 1/4) U (3/4 , 1) U {x : x ∈ N, x ≥ 2}
(xiii) [– 3,– 2) ∪ [ 3,4)
(xiv) φ
(xv) 2Kπ < x < (2K + 1)π but x ≠ 1 where K is non−negative integer
 1 π   5π 
(xii)  − ,  ∪  , 6 
 6 3  3

 5
(xviii) (1, 2) ∪  2, 
 2
(xvi) {x 1000 ≤ x < 10000} (xvii) (–2, –1) U (–1, 0) U (1, 2)
(xix) (− ∞ , −3) ∪ (−3 , 1] ∪ [4 , ∞)
Q 2.
(i) D : x ε R
R : [0 , 2]
(ii) D = R ; range [ –1 , 1 ]
(iii)
D : {xx ∈ R ; x ≠ −3 ; x ≠ 2} R : {f(x)f(x) ∈R , f(x) ≠ 1/5 ; f(x) ≠ 1}
(iv)
D : R ; R : (–1, 1)
(vi)
D : x ∈ (2nπ, (2n + 1)π) − 2 nπ + π6 , 2 nπ + π2 , 2 nπ + 56π , n ∈ I and
(v) D : −1 ≤ x ≤ 2 R :
{
[
R : loga 2 ; a ∈ (0, ∞) − {1} ⇒ Range is (–∞, ∞) – {0}
3, 6
}
]
(vii)
 1   1 1
D : [– 4, ∞) – {5}; R :  0,  ∪  , 
 6   6 3
Q.4
(a) neither surjective nor injective (b) surjective but not injective (c) neither injective nor surjective
Q.5
Q.6
f3n(x) = x ; Domain = R − {0 , 1}
1
Q.7 (a) 2Kπ ≤ x ≤ 2Kπ + π where K ∈ I (b) [−3/2 , −1]
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
98930 58881
1 + x 2
x ≤1
15.
If f(x) = 
and g(x) = 1 – x ; – 2 < x < 1 then define the function fog(x).
 x + 1 1 < x ≤ 2
1
16.
Find the set of real x for which the function, f (x) =
is not defined, where [x] denotes the
[| x − 1 |] + [| 12 − x | ] −11
greatest integer not greater than x.
 4 − 2cosx 
cos −1 (sin (x + π ))
3
 & the function
17.
Given the functions f(x) = e
, g(x) = cosec −1 
3


h(x) = f(x) defined only for those values of x, which are common to the domains of the functions f(x) and g(x).
Calculate the range of the function h(x).
18.
Let ‘f’ be a real valued function defined for all real numbers x such that for some positive constant ‘a’ the
1
equation f ( x + a) = + f ( x ) − (f ( x ))2 holds for all x. Prove that the function f is periodic.
2
19.
If
f (x) = −1 +  x − 2, 0 ≤ x ≤ 4
g (x) = 2 − x, − 1 ≤ x ≤ 3
Then find fog (x), gof (x), fof(x) & gog(x). Draw rough sketch of the graphs of fog (x) & gof (x).
20.
Find the integral solutions to the equation [x] [y] = x + y. Show that all the non-integral solutions lie on exactly
two lines. Determine these lines. Here [ .] denotes greatest integer function.
__________________________________________________________________________________________________
(b) even, (c) neither odd nor even, (d) odd,
(g) even,
(h) even;
(ii)
(e) neither odd nor even, (f) even,
−1 + 5 −1 − 5 −3 + 5 −3 − 5
,
,
,
2
2
2
2
(a) y = log (10 − 10x) , − ∞ < x < 1
(b) y = x/3 when − ∞ < x < 0 & y = x when 0 ≤ x < + ∞
Q.10 f−1(x) = (a − xn)1/n
Q.12 (a) f(x) = 1 for x < −1 & −x for −1 ≤ x ≤ 0; (b) f(x) = −1 for x < −1 and x for −1 ≤ x ≤ 0
30 of 41
Q.8 (i) (a) odd,
Q.9
2
(b)
{–1, 1}
1
log2 x
1+ x
; (c) log
2
log2 x − 1
1− x
Q.16 x = 1
Q.17 (i) period of fog is π , period of gof is 2π ; (ii) range of fog is [−1 , 1] , range of gof is [−tan1, tan1]
Q.18 (a) π/2 (b) π (c) π/2 (d) 70 π
Q.20 ± 1, ± 3, ± 5, ± 15
EXER
CISE–2
EXERCISE–2
Q 1. f−1(1) = y
Q.2
(a) – 3/4, (b) 64, (c) 30, (d) 102, (e) 5050
1
, (b) 1, (c) [0, 4), (d) – 5
1002
Q.3
(a)
Q 4.
b can be any real number except
Q.6
6016
15
Q5. f (x) = 1 – x2, D = x∈ R ; range =(– ∞, 1]
4
Q 9. f (x) = 2 x2
x +1 , 0 ≤ x < 1
Q 11. fog (x) =
− (1 + x) ,
x−1
−1 ≤ x ≤ 0
, 0<x≤2
3− x , 1 ≤ x ≤ 2
; gof (x) =
x −1 , 2 < x ≤ 3
;
5−x , 3 < x ≤ 4
x
,
0≤x≤1
fof (x) = 4 − x , 3 ≤ x ≤ 4 ;

Q 12.  −

Q.14
−x
1002.5
Q.15
,
0<x≤2
4−x ,
2<x≤3
gog (x) = x
 3 −1
3 + 1 1 − 3
 ∪ 
,
,
2
2 
2

, −1 ≤ x ≤ 0
3 + 1

2 
5049
Q.13
Q.16
g (x) = 3 + 5 sin(nπ + 2x – 4), n ∈ I
Q 18. (0 , 1) ∪ {1, 2, ....., 12} ∪ (12, 13)
Q.17 20
x = 0 or 5/3
Q 19. f (x) = sin x + x −
EXER
CISE–3
EXERCISE–3
π
3
Q.1 (hofog)(x) = h(x2) = x2 for x ∈ R , Hence h is not an identity function , fog is not invertible
Q.2 (a) A, (b) B
Q.3 (fog) (x) = e3x − 2 ; (gof) (x) = 3 ex − 2 ;
Domain of (fog)–1 = range of fog = (0, ∞); Domain of (gof)–1 = range of gof = (− 2, ∞)
Q.4 B
Q.5 D
Q.6 {(1, 1), (2, 3), (3, 4), (4, 2)} ; {(1, 1), (2, 4), (3, 2), (4, 3)} and {(1, 1), (2, 4), (3, 3), (4, 2)}
Q.7 (a) B, (b) A, (c) D, (d) A, (e) D
Q.8
(a) D ; (b) A
Q.9 (a) D , (b) A
Q.10 C
Q.11 (a) A ; (b) D
EXER
CISE–4
EXERCISE–4
1. D
2. C
3. C
4. B
5. B
6. C
7. B
8. B
9. C
10. A
11. A
12. B
13. D
14. B
98930 58881
x
−x
Q.15 (a) e − e ;
Q.14
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Q.13
1
if 0< x ≤1
 x2
g( x ) = 
 2
 x if x >1
17. C 18. B
31. D 32. B
37. ABCD
1. [–2, 0) ∪ (0, 1)
23. D
24. C
25. D
26. D
27. C
28. A
EXER
CISE–5
EXERCISE–5
 1 1
− 3 , 2 


1
2
 −1 x − 2  π 
 sin 
 − 
 2  6

7. – 3 8. (i) [0, 1]
(ii) φ
(iii) (2, 3)
9. (i)
1 
 3 , 3


(ii) [− 1, 1] (iii) [4, ∞)
3 
(iv)  , 1
4 
3 
 
2
11. (i)
(ii)
a ∈ (1, 3) ∪ {0}
12.
13. (i) neither even nor odd (ii) even (iii) odd
14. (i) π (ii) 2 (iii) 2 π (iv) π
2 − 2x + x 2
15. f(g(x)) = 
 2 − x
0 ≤ x ≤1
−1≤ x < 0
 π


π
16. (0, 1) U {1, 2,......., 12} U (12, 13) 17. e 6 , e 


18. Period 2 a
− (1 + x ) , − 1 ≤ x ≤ 0
19. fog (x) = 
;
 x −1 , 0 < x ≤ 2
x +1

3 − x
gof(x) = 
x −1
5 − x
, 0 ≤ x <1
, 1≤ x ≤ 2
, 2<x≤3
, 3<x≤4
, 0≤x≤2
 x
fof (x) = 
;
4 − x , 2 < x ≤ 2
 − x , −1≤ x ≤ 0

, 0<x≤2
gog(x) =  x
4 − x , 2 < x ≤ 3

20. Integral solution (0, 0); (2, 2). x + y = 6, x + y = 0
***Best of LUCK***
10.
98930 58881
e x − e −x
2
4. B = [0, 4] ; f –1 (x) =
5. (2, ∞)
22. D
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
3. f–1 =
2.
19. D 20. D 21. D
33. B 34. BC
38. BD
31 of 41
15. B 16. D
29. C 30. D
35. AC 36. B