Function



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 Function

Definition of Set:
A set is a collection of distinct objects, considered
as an object in its own right.
Ex.
A = {1, 2, 3, 4}; B = {a, e, i, o, u}.

Note:
1. A set is generally represented by a capital Know the facts
letter.
2. The elements of set are written within the
There is no set like {1, 2, 1}.
braces.
3. The numbers 2, 4 and 6 are distinct objects
when considered separately, but when they
are considered collectively, they form a single
set of size three, written {2, 4, 6}.
4. No element in the set is repeated.
5. Set is a collection in which order of elements
is not important.

Ex. {1, 2, 3} ≡ {3, 2, 1} .

Roster form
Representation of a set that lists all the elements in
the set, separated by commas, within braces.
Ex. {–3, –2, –1, 0, 1, 2}

Set Builder form

Mathematical notation for describing a set by
enumerating its elements or stating the properties
Function

that its members must satisfy.

Ex. {x | – 4 ≤ x < 3, x ∈ R}
1.
 Note:
1. Here ‘|’ means ‘‘such that’’ and after ‘‘|’’, the
property or definition is written.
2. We can also use ‘:’ instead of ‘‘|’’.

Ex.
{x|x is a vowel} → Set Builder form
{a, e, i, o, u} → Roster form.
Both sets are same.

Ordered Pair
An ordered pair (a, b) is a pair of objects. The
order in which the objects appear in the pair is
significant.

Note:
1. The ordered pair (a, b) is different from the ordered
pair (b, a) unless a = b.
2. It can be understood if the pair is taken as a point.
3. Here, (1, 2) and (2, 1) represent different points, so
cannot be considered equal.

Know the facts

Point to Remember!!!
Cartesian product
It is product of 2 sets. (
n A ×B
 =
) n (A)

× n (B)

no of elements no of elements no of elements
A
 × B = {( a,b ) | a  A and b  B} in set A ×B in set A in set B

new set 

element of new set
Using this 2C1 × 3C1 = 2 × 3 = 6

Ex. A = {a, b} and B = {1, 2, 3}

A × B = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}
B × A = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}
Function

2.
 Note:
It can be seen that all elements of
A × B and B × A are not equal.
∴ A×B ≠ B×A

A brief introduction to ‘‘Relation’’

Any subset of A × B is a ‘‘Relation’’, from A → B Point to Remember!!!
(pronounced A to B).
If possible, n(A) = p; n(B) = q, then
Ex.
number of possible relations in
If n (A× B) = 45 , then number of possible relations
A → B is 2pq.
in A → B is
45
C0 + 45C1 + ... + 45C45 = 245

Function
Definition:
A relation from a set A to a set B is called a function if
(i) Each element of set ‘A’ is associated with some
element in set ‘B’.
(ii) Each element of set ‘A’ has unique image in set ‘B’.
Ex.
∴ f ≡ {(1, a), (2, b), (3, c)}
So, it can be said that f ⊂ A × B.

Image and Pre-image

If an element (a ∈ A) is associated with an element
(1 ∈ B), then ‘1’ is called, the

“f image of a” or “image of a under f”

or
“the value of the function f at a”
or
“argument of a under the function f”.
Function

3.
 Ex.
f = {(a, 1), (b, 2), (c, 3), (d, 4)}
A = {a, b, c, d}, B = {1, 2, 3, 4, 5}
f: A → B

Ex.

A = {a, b, c, d}, B = {1, 2, 3, 4, 5}

f: A → B

(i) is a function.
Every element in A has a unique image in B.

(ii) is a function.
Every element in A has a unique image in B.

(iii) is not a function.

(‘d’ has no image in B)

(iv) is not a function.

(‘d’ does not have a unique image in B)
Function

4.
Domain, Codomain and Range
A = {a, b, c, d}, B = {1, 2, 3, 4, 5}
Domain ® {a, b, c, d}
Codomain ® {1, 2, 3, 4, 5}
Range ® {1, 2, 3, 4}
If

Range can be said to be the collection of

functional outputs.

Ex.
Domain → {a, b, c, d}
Codomain → {1, 2, 3, 4}
Range → {1, 2, 3}

Note: Range ⊆ Codomain

Note:
When only ‘‘rule of function’’ is given
(i) It is called ‘‘Real valued function’’.
(ii) Domain ≡ Set of real ‘x’ for which y is real
(Input values).
(iii) Range ≡ Set of all real y values obtained
after putting real x in domain
(All output values).

Find domain of
Q. (i) y = x

Here, it can be seen that all the values

A. in (- ∞ , ∞ ) can be used as input as we
move from left to right of graph.
At the same time, y achieves all values
in (- ∞ , ∞ ) as we move from bottom to
top of graph.
∴ Domain ≡ x ∈ R
Range ≡ y ∈ R
Function

5.
 Q. (ii) y = 2x – 1

A. Domain ≡ x ∈ R
M-1: For range, it can be seen from graph
∴ Range ≡ y ∈ R
M-2: y = 2x – 1
Since x ∈ (–∞, ∞)
y = 2(–∞, ∞) – 1
= (–∞, ∞) –1
= (–∞, ∞).
∴ y ∈ R.

Q. (iii) y = 3x + 4

A. Similarly,
Domain ≡ x ∈ R
Range ≡ y ∈ R
So, In general, for y = ax + b
(linear, a ≠ 0) ,
Domain ≡ x ∈ R
Range ≡ y ∈ R.

Solving Trick (for domain):

Try to think of values of x for which real
value of y does not exist.

Q. Find the domain of

1
(i) y =
x

A. Domain: x ∈ (-∞, 0) ∪ (0, ∞)

(at x = 0, y does not exit)
For range, draw the graph and check
especially for endpoints of intervals.
f(-∞) ≈ 0¯
f(0¯) = –∞
f(0+) = +∞
f(∞) ≈ 0+
Function

y≠0

6.
 ∴ Range: y ∈ R – {0}
1
Alternate Method: x = ⇒y≠0
y
So, graph of xy = 1 or xy = c2 is given:

Q. (ii) y =
1
2x − 1

A. Domain: 2x − 1 ≠ 0
1
or x  R −  
2
For range,

 1 1 1  1
y x −  = ⇒ YX =  Y = y; X = x − 
 2 2 2  2
So, using shifting of origin, it can be seen that
function is similar to xy = c2
∴ Range: y ∈ R – {0}

1
Q. (iii) y =
3x + 4

A. 3x + 4 ≠ 0 ⇒ x ≠
−4
3
Point to Remember!!!
 −4 
∴ Domain: x  R −  
3  1
Range of f ( x ) = , (a ≠ 0)
ax + b
Similarly, range: y ∈ R – {0}
will always be R– {0}.

Q. (iv) y = x

A. Since square root of a negative value is not real,

in case of y = f ( x ) , to find domain, make f(x)
≥ 0.
Using concept, Domain of y = x is x ∈ [0, ∞).
From graph
Range: y ∈ [0, ∞)
Function

7.
 Point to Remember!!!

y2 = x is curve not function.

For x ≥ 0
⇒y=± x
 o, for same value of x, there is two
S

values of y. So, it cannot be a function.

Point to Remember!!!

If any vertical line cuts a curve at at

least two different points, then the
curve cannot be a function.

Q. (v) y = 2x − 1

A. 2x − 1 ≥ 0 ⇒ x ≥
1
2
1 
Domain: x   , ∞ 
 2 
1 
Trick for range: 2  , ∞ − 1
2 
= 1, ∞ ) − 1 = 0, ∞ ) = 0, ∞ )
So, range ≡ [0, ∞)

Q. (vi) y = 3x + 4

A.  −4 
Domain: x   , ∞ 
3 
Range: y  0, ∞ )
Function

8.
 Point to Remember!!!

ax + b , a ≠ 0 has its range

[0, ∞)

1
Q. (vii) y =
x

A. Here, it is similar to
1
.
ax + b
So ax + b > 0
∴ Domain: x ∈ (0, ∞)
 1 1 
Range: y ∈ (0, ∞)  → → ( 0, ∞ ) 
 (0, ∞ ) (0, ∞ ) 

1
Q. (viii) y =
3x − 4

A. 4 
Similarly, Domain: x   , ∞ 
3  1
Range: y ∈ (0, ∞) or R+ 3x–4

Algebraic operations on functions

(i) Let f and g be functions with domain D1 and D2
then the function f + g, is defined as
(f + g) (x) = f(x) + g(x); Domain: D1 ∩ D2
In this case, both functions f(x) and g(x) must be
real simultaneously. Only then, the overall function
will be real.
(ii) Let f and g be function with domain D1 and D2,
then the function f– g is defined as
(f – g) (x) = f(x) – g(x); Domain: D1 ∩ D2
Function

9.
 Again, both functions should be real at the same time. So, the domain
is set of all the values of x common to both of their domain.
f
(iii) Let f and g be functions with domain D1 and D2, then the function   ( x )
g
f (x) f f (x)
is defined as i.e.,   ( x ) =
g (x) g g (x)
Domain:
D1 ∩ D2 : {x | g ( x ) ≠ 0} or

D1 ∩ D2 − {x | g ( x ) = 0}

In this case, Denominator, i.e., g(x) must not be zero.

Ex.
f(x) = x; g(x) = x2-1
Df = R; Dg = R
f x
  (x) = 2
g
  x −1
x2 – 1 ≠ 0 ⇒ x ≠ 1, –1.
f
∴ Domain of   ( x ) is x ∈ R – {1, –1}.
g

(iv) Let f and g be functions with domain D1 and D2, then the function fg is
defined as
( fg ) ( x ) = f ( x ) .g ( x ) ; Domain: D1 ∩ D2

Q. f(x) = x3 + 2x2 and g(x) = 3x2 – 1. Find domain of f ± g,fg and f/g.

A. Df = R; Dg = R
Df ∩ Dg = R
(i) f ± g : Df ∩ Dg ⇒ Domain = R ∩ R = R
(ii) fg : Df ∩ Dg ⇒ Domain = R ∩ R = R
(iii) f/g : Df ∩ Dg ⇒ Denominator ≠ 0
1
⇒x≠±
3
 1 
∴ Domain: R − ± 
 3
Function

10.
Continuous Functions
If graph of a function can be drawn without taking up the pen, then function
is continuous.

Ex.
(i) Graph of sin x is continuous ∀ x ∈ R

π
(ii) Graph of tan x, is discontinuous at ( 2n + 1) ,n∈I
2

sinx π
tanx = ⇒ cosx ≠ 0 ⇒ x ≠ ( 2n + 1) , n ∈ I
cosx 2
Function

11.
 Domain, Range and Graph of Trigonometric Functions
(i) y = sin x
Domain : x ∈ R; Range ∈ [–1, 1]

Point to Remember!!!

y
sinθ =
r

π
Maximum value of y = sin x is 1 at x = .
2
3π
Minimum value of y = sin x is –1 at x = .
2
Domain: x ∈ R
Range: y ∈ [–1, 1]

(ii) y = cos x

Point to Remember!!!

x
cosθ =
r

Domain : x ∈ R ; Range : [–1, 1]

Function

12.
(iii) y= tan x

π
Domain: x ∈ R − ( 2n + 1) ; n∈I
2

Range: R

(iv) y = cot x
cosx
cotx =
sinx
⇒ sin x ≠ 0

⇒ x ≠ nπ, n ∈ I

∴ Domain: x ∈ R – nπ, n ∈ I

Range: R

(v) y = cosec x
Graph can be constructed by observing the
graph of y = sin x.

1
Since cosecx = , sin x ≠ 0
sinx
∴ x ≠ nπ
Domain: R – nπ
From graph it can be seen that
range ∈ (–∞, –1] ∪ [1, ∞)
Function

13.
 (vi) y = sec x

Similarly, like cosec x, graph of sec x can be drawn by the help from
graph of cosx.
1 π
Since secx = , cosx ≠ 0 ⇒ x ≠ ( 2n + 1) , n ∈ I
cosx 2
π
∴ Domain: x ∈ R − ( 2n + 1) ,nI
2
Range: (–∞, –1] ∪ [1, ∞)

Q. Find range of y :
(i) y = sin (2x)

A. x ∈ (- ∞ , ∞ )

⇒ 2x ∈ (- ∞ , ∞ )
As different values of x are used in sin(2x), input moves from (- ∞ , ∞ ).
So, sin(x) ∈ [-1, 1]
Function

14.
Q. (ii) y = sin (x2)

A. x ∈ (- ∞ , ∞ )

⇒ x2 ∈ [0, ∞ )

So, range is [-1, 1]

Q. (iii) y = sin ( x )

A. x ∈ [0, ∞ )
Input for sin x is [0, ∞ )
∴ Range is [–1, 1]

x x
Q. (iv) y = cos4
2
− sin4
2

A.  x x x x
y =  cos2 − sin2   cos2 + sin2 
 2 2  2 2 
 x x 
=  cos2 − sin2 
 2 2
 x
= cos  2·  = cos x.
 2
Range of cos x is [–1, 1]
So, y ∈ [–1, 1]

Q. (v) y = (sin x + 2)2 + 1

A. sin x ∈ [–1, 1]
sin x + 2 ∈ [1, 3]
(sin x + 2)2 ∈ [1,9]
y = (sin x + 2)2 +1 ∈ [2, 10]
Function

15.
 Q. (vi) y = 4 tan x cos x

A. There is loss of domain because of presence of tanx.

Due to presence of tanx, cos x ≠ 0
sinx
y = 4 tanx cosx = 4 .cosx ; (cosx ≠ 0)
cosx
y = 4 sinx (cosx ≠ 0 so sinx ≠ -1 and 1)
∴ y ∈ (–4, 4) (⸪ sin x ∈ (-1, 1)) Point to Remember!!!

a0 is called leading coefficient.

Definition of Polynomial Function
an is called constant term.
If a function f is defined by f(x) = a0xn + a1xn-1 + f(0) = an.
a2xn-2 + … + an-1x + 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 Points to Remember!!!
degree n.
Ex. f(x) = 2 is zero degree polynomial (i) A
 polynomial function is always
continuous.
Ex. (ii) 
A polynomial of degree one is
f(x) = ax2 + bx + c … Quadratic Polynomial called a linear function.
f(x) = ax3 + bx2 + cx + d … Cubic Polynomial (iii) A polynomial of degree one with
Similarly, polynomial of degree 4 is called no constant term is called an
‘Biquadratic’. odd linear function. y = mx is an
odd linear function.
(iv) A polynomial of degree odd has
its range ‘R’ but a polynomial of
degree even has a range which
is always a subset of R.
In case of polynomial, the term
with highest degree is always
dominant when compared to
other terms, whenever x=± ∞
(other terms are negligible
compared to it at x = ± ∞ ).
So, in case of odd degree
polynomial, its range is R.
In case of even degree polynomial,
the range is only a subset of R.
Function

16.
Ex. y = x3 + 2x2 – 7x + 3
as x → ∞ ; y → ∞ Point to Remember!!!
as x → −∞ ; y → - ∞
 D 
Range of parabola (a > 0):  − , ∞
Since, polynomial is continuous, the range is  4a 
(- ∞ , ∞ )
y = ax2 + bx + c
 D
If a < 0, than range is  −∞, −
 4a 

Ex. y = 2x100 + 76
x → + ∞ ;y → + ∞
x → - ∞ ;y → + ∞
Here, since leading coefficient is positive, y
can not assume the value of ‘– ∞ ‘. Hence, its
graph will be:
This graph has a minima. so, range is subset
of R.

Definition of Algebraic functions

A function f is called an algebraic function if it can be constructed using
algebraic operations such as addition, subtraction, multiplication or division
or taking radical sign (Starting with polynomial).
Here radian sign refers to operations like:
−2/3
x1/3 ; ( 2x − 1) ; x + 2 etc.
Function

17.
 Ex. f ( x ) = x2 + 1
Domain : x2 + 1 ≥ 0
Since x2 ∈ [0, ∞ ), x2 + 1 ∈ [1, ∞ ) So, x2 + 1 ≥ 0 is true for all real x.
∴ Domain: (– ∞ , ∞ )
Range: [1, ∞ )

Q. f(x) = x2 + ax + 4
(a) Find ‘a’ if range is [2, ∞).
(b) Find ‘a’ if domain is all real.

A. (a) Since, x2 + ax + 4 ∈ [2, ∞ ).

x2 + ax + 4 ∈ [4, ∞ ).
∴ Minimum value of x2 + ax + 4 = 4
−D
⇒ =4
4a
16 − a2
=4 ⇒ a=0
4
(b) Domain is all real.
it means x2 + ax + 4 ≥ 0 ∀ x ∈ R.
∴ D ≤ 0
a2 – 16 ≤ 0
a ∈ [–4, 4]

Definition of Fractional / Rational Function

g (x)
A fractional function is of the form y = f(x) = ,
h (x)

where g(x) and h(x) are polynomial and h(x) ≠ 0.

The domain of f(x) is set of real x such that h(x) ≠ 0.

Ex.
2x4 − x2 + 1
f(x)= ; D = {x|x ≠ ± 2}
x2 − 4
Here, denominator ≠ 0
x2 – 4 ≠ 0
x ≠ 2, -2
Function

18.
Definition of Exponential Function
A function f(x) = ax (a > 0, a ≠ 1, x ∈ R) is
called an exponential function.

Ex.
Point to Remember!!!
y = 2x
Graph of ax
x -∞ -2 -1 0 1 2 ∞
base a > 1
1 1 Domain: x ∈ (- ∞ , ∞ )
y 0+ 1 2 4 ∞
4 2

As x increases, y increases.
As x decreases, y decreases.
1 1 1
as x → - ∞ , y = × × ×…=O+
2 2 2
So, if ax > ay ⇒ x > y

Ex.
x
1 Point to Remember!!!
y=  
2
Graph of ax :
X -∞ -2 -1 0 2 ∞
base 0 < a < 1
Y ∞ 4 2 1 1/4 0+

As x increase, y decreases
So, if ax > ay ⇒ x < y
Domain: (–∞, ∞)
Range: (0, ∞ )
Function

19.
 Ex.

x -2 -1 0 1 2

2x 1/4 1/2 1 2 4

3x 1/9 1/3 1 3 9

10x 1/100 1/10 1 10 100

Plotting these points on graph we can see

that:
(i) At x > 0, if a > b, then ax > bx (a, b> 1)
(ii) At x < 0, if a > b,
then ax < bx (a, b > 1)

Ex.

x -2 -1 0 1 2

2–x 4 2 1 1/2 1/4

3–x 9 3 1 1/3 1/9

10–x 100 10 1 1/10 1/100

Plotting these points on graph, it can be seen

that:
(i) At x > 0, if a > b, then ax > bx (a, b ∈ (0, 1))
(ii) At x < 0, if a > b, then ax < bx (a, b ∈ (0, 1))

Ex.
(i) y = 2 x

Domain: x ≥ 0
x ≥ 0 ⇒ 2 x
≥ 1
Range: [1, ∞ )

2
(ii) y = 2x +1

Domain: x ∈ R
x2 +1 ∈ [1, ∞ )
∴ Range: [2, ∞ )
Function

20.
Definition of Logarithmic function
Point to Remember!!!
y = logax; x > 0, a > 0, a ≠ 1
y = log a x ⇒ a y = x Graph of y = logax

+ ve base a > 1
∴ a > 0 and x > 0 Domain: (0, ∞ )
If a = 1 and x = 1, then log11 has more than Range: (- ∞ , + ∞ )
one value.
So, it will not be a function.
If a = 1 and x ≠ 1, then logax will have no
solution.
∴ a ≠ 1.

Ex.
y = log2x

x 0+ 1/4 1/2 1 2 4 ∞

y –∞ -2 -1 0 1 2 ∞

For base (a) > 1, the graph is increasing.

Ex.
y = lnx, it is also written as y = logex
dy 1
= > 0 ∀ x ∈ R+
dx x

d2 y −1
= < 0 ∀ x ∈ R+
dx2 x2
So, graph is always increasing as well as
concave down.

Point to Remember!!!

Graph of y = logax
base 0 < a < 1
Domain: x ∈ ( 0, ∞ )
Range: (- ∞ , ∞ )
Function

21.
 Q.1 Find the domain and range of y = lnx

A. x > 0 (Rule of logax)

ln x ≥ 0 (in f ( x ) , f(x) ≥ 0)

ln x ≥ ln1
Since base is greater than 1, graph is increasing.
∴ x ≥ 1
∴ Domain: [1, ∞ )
Since ln x ≥ 0, lnx ≥ 0
∴ Range: [0, ∞ )

Q.2 Find domain and range:

(i) y = log 1/3 ( x − 1)

A. Domain conditions:
(i) x-1 > 0 ⇒ x ∈ ( 1, ∞ )
(ii) log 1/3 ( x − 1) ≥ 0
⇒ log 1/3 ( x − 1) ≥ log 1/3 1
⇒ x–1≤1
(from graph of logax, 0 < a < 1)
⇒ x≤2
∴ from (i) and (ii), domain: (1,2]
Range: x ∈ (1,2] ⇒ (x -1) ∈ (0, 1]
⇒ log 1/3 ( x − 1) ∈ [0, ∞ )
∴ Range: [0, ∞ )

Q. (ii) y = log 3 ( cos ( sinx ) )

A. Domain:
log3(cos(sinx)) ≥ 0
⇒ log3(cos(sinx)) ≥ 0 = log31
⇒ cos(sinx) ≥ 1
Since cos θ ∈ [–1, 1], there is only one case
where above situation is possible, i.e, cos
Function

(sinx) = 1

22.
 ⇒ sinx = 2n π , n ∈ I.
sinx= 0, ±2 π , ±4 π , ± 6 π , …
But sin x ∈ [-1, 1]. So, sinx = 0 is only
solution.
∴ x = n π , n ∈ I.
Range:
sinx = 0 ⇒ cos(sinx) = 1
⇒ log3cos(sinx) = log 1 = 0
∴ Range: {0}
This function can also be called many-one
function.

Q. (iii) y =
1
lnx

A. Domain:
Condition are:
(i) x > 0
(ii) lnx ≠ 0 ⇒ x ≠ 1
∴ Domain: x ∈ (0, 1) ∪ (1, ∞ ).
Range:
lnx ∈ (–∞, 0) ∪ (0, ∞).
1
∴ ∈ (- ∞ , 0) ∪ (0, ∞ )
ln x
1
(Similar to function y = )
x

Q.
1
(iv) y = e x

A. Domain: x ≠ 0
⇒ x ∈ (– ∞ , 0) ∪ (0, ∞ )
Range:
1
∈ ( −∞, 0 ) ∪ ( 0, ∞ )
x
⇒ e1/x ∈ ( 0, 1) ∪ ( 1, ∞ )
∴ Range: (0, 1) ∪ (1, ∞).
Function

or (0, ∞ ) – {1}

23.
 Q. (v) y = log10 (log10(1+x3))

A. Domain:
log 10 ( 1 + x3 ) > 0

⇒ 1 + x3 > 1
⇒ x3 > 0 ⇒ x > 0
∴ Domain: x ∈ (0, ∞ )
Range:

(
log10 log 10 ( 1 + x3 )

)
( 0, ∞)

⇒ y ∈ ( −∞, ∞ )
(∴ log(0+) = –∞ and log(∞) = ∞)

1− x  2x 
Q.3 If f(x) = log e   , x < 1 , then f 
1+ x
2 
1+ x 
is equal to:

(A) 2f(x) (B) 2f(x2) (C) (f(x))2 (D) –2f(x)

 2x 
1−
A. 
f
2x 
2 

= log e  1 + x2 

1+ x   1 + 2x 2 
 1+x 
 ( 1 − x )2 
= loge  
 ( 1 + x )2 
 
1− x
= 2 loge  
1+ x
= 2f(x)

Definition of Absolute value function/ Modulus function

 x if x ≥ 0
y= x =
−x if x < 0
Domain: R; Range: [0, ∞)

Ex. |-2| = 2
|+2| = 2
|0| = 0
Function

24.
Ex. Know the facts
y – 2 = |x – 3|
Shifting of origin:
+ ( x − 3 ) ; x ≥ 3 y – b = f(x – α)
y–2 =  Take (0, 0) point of graph to ( α, β ).
 − ( x − 3 ) ; x < 3
In this case, draw the graph y = f(x)
x − 1 x ≥ 3 where X = x – α and Y = y - β
⇒ y=
5 − x x < 3 At x = α , X = 0 and at y = β , Y = 0
So, at ( α , β ), X = 0 and Y = 0.
So, draw the graph Y = f(X) at ( α, β )
Now, if graph of both functions,

y = |x| and y-2=|x-3| is compared, then it as if ( α, β ) was origin.
appears as if graph is shifted from origin
to (3, 2).

Ex.
y + 2 = |x-1|
b = –2
α=1
Sol. y-(-2) = |x-1|
∴ α = 1 and b = –2
So, shift the graph of y = |x| to (1, –2).
Function

25.
 Ex.
 ( x − 1) , x ≥ 1
y = |x–1|= 
− ( x − 1) , x < 1
( α =1; β =0)
So, graph shifted to (1, 0)

Ex. Draw the graph of

(i) y = 1 + |x|
Point to Remember!!!
(ii) y = 2 + |x|
(iii) y = – 1 + |x| Graph of f(x- α ) can be drawn by
Upshift shifting the graph of f(x) to left or
(i) y = 1 + |x| right.
y-1 = |x|; α = 0 and β = 1 So, there is no difference in the
range of f(x) and f(x- α ).
So, graph shifted to (0, 1).

(ii) Similarly, in case of y = 2 + |x|

y – 2 = |x|
graph shifted to (0, 2)

(iii) y = -1+|x|
y-(-1) = |x|
graph shifted to (0, -1)

Q. Find domain, range and graph of y =|x – 1| + 1

A. y – 1 = |x – 1|; α = 1 and β = 1
Graph shifted to (1, 1)
Domain: R
Range: [1, ∞ )
Function

26.
 1
Q. Which of the following graph represents y =
x
?

(A) (B)

(C) (D)

A. Domain: |x| ≠ 0
∴ x≠0
∴ Domain: R – {0}
Range:
x = 0+ y = ∞.
x = 0– y= ∞
x = –∞ y = 0+
x=+∞ y = 0+
|x| ∈ (0, ∞), (|x| ≠ 0)
1
∴ ∈ ( 0, ∞ )
x
∴ Range: (0, ∞)
So, from domain, graphs (A) and (D) are eliminated.
from Range, graph (B) can be eliminated. Graph (C) is the correct answer.
Function

27.
 Definition of Signum Function
A function y = f(x) = sgn(x) is defined as follows:
 1 for x > 0

y = f ( x ) =  0 for x = 0
−1 for x < 0

Domain: R
Range: {–1, 0, 1}

Note:
Hollow circle at (0, 1) and (0, –1) indicates that
these points are excluded.
Know the facts
Solid circle at (0, 0) show that this point is
included. x
 , x≠0
sgn(x) =  x
 0, x = 0


Q. Find domain, range of:

(i) y = sgn(x2 –1)

A.  1 ; x2 − 1 > 0

y =  0 ; x2 − 1 = 0
−1 ; x2 − 1 < 0

 1 ; x ∈ ( −∞, −1) ∪ ( 1, ∞ )

y = 0 ; x = 1, −1
−1 ; x ∈ ( −1, 1)

Domain: x ∈ R
Range: {-1, 0, 1}

Q. (ii) y = log10 sgn(x)

A. sgn(x) > 0 (it is inside logarithm).

∴ sgnx can only be ‘1’.
∴ x>0
∴ Domain: x ∈ (0, ∞ )
Since sgn(x) = 1, log 10 ( sgnx ) = 0
Function

∴ Range: y = {0}

28.
 Q. y = sgn (ln(x2 – x + 2))

 −D 
A. Range for quadratic: 
 4a
, ∞

7 
∴ x2 – x + 2 ∈  , ∞ 
4 
7
∴ minimum value of ln (x2–x+2) is ln
4
7
ln > ln1 which is 0.
4
7
ln is positive
4
So, sgn (ln(x2 – x + 2)) will always give the value ‘1’.
∴ Range: y = {1}
Domain: x ∈ R.

Know the facts

Any real number can be broken

into two parts: integral parts and
fractional parts.
Integral parts is an integer and
fractional parts belongs to [0, 1).
Ex.

(i) 3.2 = 3 + 0.2 (ii) –2.2 = -3 + (0.8)

3 = 3+ 0 –7.2 = –8 + (0.8)
0.9 = 0 + (0.9) – π = –4 + (4 – )
π = 3 + ( π –3)
e = 2 + (e -2)
Function

29.
 Definition of Greatest integer 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.
R = I + f, here I is [R].

Ex. [7.6] = 7
7.6 is not an integer. The integers less than
7.6 are {7, 6, 5, 4, …}. Among these integers,
greatest integer is 7. So, [7.6] = 7.

Ex.

-1 ≤ x < 0 ⇒ [x] = - 1
0 ≤ x<1 ⇒ [x] = 0
1 ≤ x<2 ⇒ [x] = 1
2 ≤ x<3 ⇒ [x] = 2

and so on.
For f(x) = [x], domain is R and range is I.
Here, [0.1] = 0
[0.3] = 0
[0.7] = 0
or y = [x] = 0 ∀ x ∈ [0, 1)

Ex.
1
For f ( x ) = , find the domain and range.
[ ]
x
(where [.] denotes greatest integer function)
Sol. Domain of [x] is R.
1
In case of y = , f(x) ≠ 0.
f (x)
So, in above example, [x] ≠ 0.
⇒ x ∉ [0,1)
∴ Domain: x ∈ R – [0, 1)
1 1 1 1 1
Range: y = …, , , , , , … and so on.
−2 −1 1 2 3
1 
∴ Range: =  : I is an integer, I ≠ 0 
Function

I 

30.
Property:
(a) [x] ≤ x < [x] + 1
(b) x – 1 < [x] ≤ x

Proof:
(a) 0 ≤ x – [x] < 1
⇒ [x] ≤ x < [x] + 1
(b) x–[x] < 1
⇒ x – 1 < [x]
x – [x] ≥ 0
⇒ x ≥ [x]

Property:
(c) [x + m] = [x] + m, where m is an integer.

From the number line, it can be seen that greatest integer on the left
side of x + m is I + m, i.e., [x] + m.

Ex.
[10.7] = 10
[10.7 + 2] = [12.7] = 12 = 10 + 2 = [10.7] + 2.

Property:
 0 if x is an integer
(d) [x] + [ −x ] = 
−1 otherwise

Proof:
Case-I: x = integer.
[x] = x, [–x] = -x
∴ L.H.S = x – x = 0 = R.H.S.
Case-II: x = I + f; f ∈ (0, 1), I is an integer.
LHS = [I + f] + [–I – f]
= I + [(–I – 1) + (1 – f)], 1 – f ∈ (0, 1)
= I + (–I – 1)
= –1 = RHS
Ex. [2.3] + [–2.3] = 2 + (–3) = –1
Function

31.
 Q.1 Let [x] represents the greatest integer less than or equal to x. If all the values of
 1  1
x, such that the product x −  x +  is prime, belongs to the set [x1, x2) ∪
 2  2
[x3, x4), find the value of x21 + x22 + x32 + x42 .

1 1
A. Let x -
2
= R. Then x + = R + 1.
2
 1  1  1
∴ x −  x +  = [R] [R +1] = I (I+1); where [R] = x − =I
 2  2   2 
I (I + 1) is prime.
It is only possible in 2 cases.
Case I: I (I + 1) = 1·2; I =1
Case II: I (I + 1) = (–2)·(–1); I = – 2
1 1
∴ 1 ≤ x– < 2 or –2 ≤ x – <–1
2 2
 3 1  3 5 
⇒ x ∈ − , −  ∪  , 
 2 2  2 2 
3 1 3 5
∴ x 1 = − ; x2 = − ; x3 = ; x4 =
2 2 2 2
44
So, x21 + x22 + x32 + x42 = = 11.
4

Let [t] denotes the greatest integer ≤ t. Then the equation in x,

Q.2
[x]2 + 2[x + 2] - 7 = 0 has:
(A) exactly four integral solutions. (B) infinitely many solutions.
(C) no integral solution. (D) exactly two solutions.

A. Let [x] = I.
∴ I2 + 2 (I + 2) –7 = 0 (  [x + I] = [x] + I)
I2 + 2I –3 = 0
I = 1 or –3
∴ x ∈ [1, 2) ∪ [–3, -2)
So, equation has infinite solutions.
So, (B) is the correct option.
Function

32.
Definition of Fractional part function
It is defined as:
Know the facts
g(x) = {x} = x – [x]
or x = [ x ] + {x} y = x – [x] = {x} The period of this function is 1. For
 
I f
f(x) = {x}, domain is R and range is
x − 0 ; x ∈ [0, 1)  [0, 1).
 
 x − 1 ; x ∈ [ 1, 2 ) 

y = x − 2 ; x ∈ [2, 3) 
x + 1 ; x ∈ [ −1, 0 ) 
 
   
The graph will have parallel lines and it will be discontinuous at all
integers.

Ex.
Know the facts
(i) The fractional part of the number 2.1 is
2.1 – 2 = 0.1
Graph of y={x} is periodic with
(ii) The fractional part of –3.7 is 0.3.
length of period 1.
Property:
(i) {x + n} = {x} ; n∈ I
Proof:
{x + n} = (x +n) – [x + n]
= x + n – ([x] + n)
= x – [x] = {x}
Ex. {–1.5} = {–0.5} = {0.5} = 0.5
Function

33.
 0 , x∈ I
(ii) {x} + {−x} = 
1 , x ∉I
Proof:
Case-I: x is an integer
{x} + {-x} = 0 + 0 = 0.
Case-II: x is not an integer
Let x = I + f ; f ≠ 0, f∈ (0, 1)
{x} + {-x} = {I + f} + {-I + (-f)}
= f + {(-I-1) + (1-f)}
=f+1–f
= 1 Hence proved.

Ex.
{2.8} + {–2.8}
= 0.8 + 0.2 = 1

(iii) {[x]} = 0 Know the facts

Proof: {[x]} = {Integer} = 0
Greatest integer function of any
{∴ [x] is always an integer}
number in this interval is always
zero.
(iv) [{x}] = 0
Proof: {x} ∈ [0, 1)
[[0, 1)] = 0

Find domain and range of:

Q.1 1
(i) f ( x ) = , ({.} denotes fractional part function)
{x}
A. {x} ≠ 0 (  {x} is in denominator)
So, x can not be an integer.
∴ Domain: R – {x | x ∈ I}
or
R – I, where I is an integer.
For range, the denominator has the values in interval (0, 1)
1 1
= ∞; − = 1+
O +
1
∴ Range: (1, ∞ )
Function

34.
Q. (ii) f(x) = log10 {x}, ({.} denotes fractional part function)

A. {x} > 0 as {x} is inside logarithm.

∴ {x} ≠ 0
⇒ x ≠ I, where I is an integer.
∴ Domain: x ∈ R – I.
{x} ∈ (0, 1)
So, log10{x} ∈ (log 10 0+ , log 10 1− )

So, Range is (–∞, 0)

Q. (iii) y = [sin{x}]
(where [.] denotes greatest integer function and {.} denotes
fractional part function)

A. Domain: x ∈ R
{x} ∈ [0, 1)
∴ sin {x} ∈ [sin 0, sin1) = [0, sin1)
1 C ≈ 57 ° . So, 0 < sin1 < 1.

 o, it can be seen that sin{x} has the minimum value 0 and it is

S
always less than 1.
Since, [x] = 0 ∀ x ∈ [0,1), [sin {x}] = 0 ∀ x ∈ R.
So, Range is {0}, a single element.

Q.2 Find domain, range and graph of: y = [x] − 1 + 4 − [x]

(where [.] denotes greatest integer function)

A. [x]–1 ≥ 0 ⇒ [x] ≥ 1
4–[x] ≥ 0 ⇒ [x] ≤ 4
So, 1 ≤ [x] ≤ 4
So, domain: x ∈ [1, 5)
(  [x] = 4 for x ∈ [4 ,5))
For range, firstly [x] = 1, 2, 3 or 4
Putting these values in y,
we get 3 , 1+ 2 , 1+ 2 and 3
respectively.
∴ Range contains only two values, 3 and 1+ 2.
y ∈ { 3,1 + 2 }.
Function

35.
 Q.3 Find domain of y = − 2 [ x ] − x − {x}

(where [.] denotes greatest integer function and {.} denotes fractional part function)

A. |f(x)| ≥ 0 ⇒ – |f(x)| ≤ 0

∴ If we have − f ( x ) , it will give real values only

if |f(x)| = 0
So, in above question, 2[x] -x – {x} = 0
⇒ Range is {0}.
Let x = I + f, [x] = I and {x} = f, f∈ [0,1)
∴ 2I – (I + f) – f = 0
I
⇒ I = 2f ⇒ f =
2
0 1
Since f∈[0, 1), f can only be or .
2 2
∴ If I =0, f = 0 ⇒ I +f = x = 0
1 3
If I = 1, f = ⇒ I +f = x =
2 2
 3
So, x = 0,  (Domain)
 2

Q.4 Find the sum of the given series.

3 3 1  3 2   3 99   3 234 
 8  +  8 + 100  +  8 + 100  + ...  8 + 100  + ...  8 + 100 
(where [.] denotes greatest integer function)

A. Sum = [0.375] + [0.385] + [0.395] + … +[2.715]

3 63
From observation, + = 1.005
8 100
3 163
+ = 2.005
8 100
∴ sum = [0.375] + … + [0.995] + [ 1.005] + … + [ 1.995] + [2.005] +  + [2.715]

  72 terms
63 terms 100 terms

= 0 × 63 + 1 × 100 + 2 × 72
= 0 + 100 + 144 = 244
Function

36.
Q.5 Find the range:
  x    x    x 
f ( x ) = [ 1 + sinx ] + 2 + sin    + 3 + sin  3   + ... + n + sin  n   ;
  2       
x ∈ [0, π] (where [.] denotes greatest integer function)

A.   x 
(1 + 2 + 3 + 4 +…n) + [sinx] + sin    +
  x 
sin  3   + ... +
  x 
sin  n  
  2       
x  π
Since x ∈ [0, π ], ∈ 0, 
3  3
x  3
So, sin   ∈ 0, 
3  2 
 x
∴ sin  = 0 only.
 3 
 x  x  x
Similarly, value of sin  , sin  , … and sin  will always be 0.
 4  5  n
n + 1  x
So, f(x) = n   + [sinx ] + sin 
 2   2 
 x n + 1
At x = π , [sinx]=0 and sin  = 1 ⇒ f(x) = n  +1
 2  2 
π  x n + 1
At x = , [sinx]=1 and sin  = 0 ⇒ f(x) = n  +1
2  2  2 
 x n + 1
At all other values in (0, π ), [sin x] = 0 and sin  = 0 ⇒ f(x) = n  
 2  2 
n2 + n + 2 n2 + n
So, f(x) can assume only two values, and .
2 2
n2 + n n2 + n + 2 
∴ Range =  , 
 2 2 

Point to Remember!!!

1 2 n − 1
[x] + x + +

x + n  + ... +

x + n  = [nx ]
 n 
We can also use it to solve the previous question.
Function

37.
 Definition of Constant function
Constant function is a function whose (output)
value is the same for every input value.
Ex.
For example, the function given is a constant
function because the value is 4 regardless of the
input value (see diagram)
In this type of function, domain is (-∞, ∞), while
range contains only a single value. In above
example, range is {4}

Definition of Identity function

An identity function, also called an identity relation
or identity map or identity transformation, is a
function that always returns the same value that
was used as its argument.

Ex. f(x) = x
Domain = R
Range is (- ∞, ∞ )
This is an increasing function.
it is also represented by Ix.

Q.1 Find domain:

(i) ( (
log 5 log 3 log 2 ( 2x3 + 5x2 − 14x ) ))
A. ( (
log 5 log 3 log 2 ( 2x3 + 5x2 − 14x ) )) ≥ 0 = log51

⇒ log3(log2(2x3 + 5x2 – 14x)) ≥ 1 = log33

⇒ log2(2x3 + 5x2 – 14x) ≥ 3 = log28
⇒ 2x3 + 5x2 – 14x - 8 ≥ 0
Now, we can guess one root through hit
and trial method. Here, it is x = 2.
∴ (x – 2) (2x2 + 9x + 4) ≥ 0
(x – 2) (x +4) (2x +1) ≥ 0
 −1 
⇒ x ∈  −4,  ∪ [2, ∞ )
 2
Function

38.
Q.
4 − 3x x ( x + 8)
(ii) (0.625) − ( 1.6 )

A.
4 − 3x x ( x + 8)
5 8
  ≥ 
8 5
3x − 4 x2 + 8x
8 8
  ≥ 
5 5
⇒ 3x – 4 ≥ x2 + 8x (  ax, a > 1 is an increasing
function)
⇒ x2 + 5x + 4 ≤ 0
⇒ (x + 1) (x + 4) ≤ 0
x ∈ [-4, -1] (domain)

Q. (
(iii) f ( x ) = log x − 1 − x )
A. x− 1−x ≥0 and 1–x ≥ 0
⇒ x ≤ 1 and x ≥ 1−x
Case I: x ≤ 0
LHS = Negative, RHS = Positive
∴ Not possible
Case II: x > 0
Square both sides
x2 ≥ 1 – x
x2 + x – 1 ≥ 0
−1 ± 5
Roots are
2
  −1 − 5     −1 + 5  
x −    x −   ≥0
  2    2  
 −1 − 5   −1 + 5 
⇒ x ∈  −∞, ∪ , ∞
 2   2 

But x > 0 and also x ≤ 1

 −1 + 5 
⇒ x∈  , 1
 2 
Function

39.
 Q. (iv) f ( x ) = sinx + 16 − x2

A. sin x ≥ 0 and 16 – x2 ≥ 0
x2 ≤ 16
x ∈ [–4, 4]
(from 2nd condition)
by graph, y = sinx
Since sinx ≥ 0,
x ∈ [-4, - π ] ∪ [0, π ]

x−2
Q. (v) f ( x ) =
x−3

A. x−2
≥ 0 and x –2 ≥ 0
x−3
Case I: x − 2 =0
x = 2, x – 3 = – 1
0
⇒ =0≥0 So, no problem.
−1
x = {2}
Case II: x−2 > 0
⇒ x>2
in this case, x – 3 must be positive.
So, x>3
∴ x ∈ {2} ∪ (3, ∞ )

Q. (vi) f(x) = (x 2
− 3x − 10 ) ln2 ( x − 3 )

A. (x - 5) (x + 2) ln2 (x-3) ≥ 0 and x – 3 > 0

ln (x-3) ≥ 0 wherever defined
2

Case I: ln2 (x-3) = 0 ⇒ x = 4

f(4) = 0 = 0 so, true.
Case II: x ≠ 4, x > 3
(x – 5) (x + 2) ≥ 0
x ∈ (- ∞ , - 2] ∪ [5, ∞ )
But since x > 3, ∴ x ∈ [5, ∞ )
Case I ∪ Case II
Function

⇒ x ∈ {4} ∪ [5, ∞ )

40.
 1 1 1
Q. (vii) f ( x ) =
[ ]
x
+ log 1−{x} ( x2 − 3x + 10 ) +
2− x
+
sec ( sinx )
, where [·] denotes

greatest integer function and {·} denotes fractional part function.

1
A. Domain of
[x]
→ x ∈ R − [0, 1)

Domain of 2nd function → 1 – {x} > 0

1 – {x} ≠ 1
x2 – 3x + 10 > 0
∴ x ∈R −I
Domain of 3rd function: 2 – |x| > 0
⇒ x ∈ ( −2, 2) Domain of 4th function: sec (sinx) > 0
⇒ cos ( sinx ) > 0
sinx ∈ [-1, 1] and cos θ > 0 for θ ∈ [-1,1]
So, x ∈ R
Taking intersection of all cases, x∈ (-2,-1) ∪ (-1,0) ∪ (1, 2)

Q. (viii) f(x) = 7x − 1

A. 7x – 1 ≥ 0
⇒ 7x ≥ 1
⇒ 7x ≥ 7°
⇒ x ≥ 0 (  increasing function as base > 1)
Domain: x ∈ [0, ∞)

Q. (ix) f(x) = 7x + 1 − 1

A. 7x + 1 ≥ 1
x+1≥0
x≥-1
Domain: x ∈ [-1, ∞)

Q. (x) f(x) :
1 − 5x
7−x − 7

A. 1 − 5x
≥0
7−x − 7

⇒ 7x
(5 x
− 1)
≥0
Function

7 x+1
−1

41.
 7x is always positive.
5x – 1 ≥ 0 ∀x≥0
7x + 1 – 1 ≥ 0 ∀ x + 1 ≥ 0
∴ Sign scheme of ‘5x-1’ and ‘x’ is same. Same is true for ‘7x+1 -1’ and ‘x+1’
∴ We can write the inequality as
x
≥ 0 ⇒ x ∈ (-∞, - 1) ∪ [0, ∞)
x+1

Cr as a function
n

n!
n
Cr is the number of combinations function, defined as nCr = r! n − r ! ,
( )
where n and r are non-negative integer. (n ≥ r)

Q.2 Find range of the following functions:

x x
(i) f(x) = cos4 –sin4
5 5

A. Simplifying, we get
 x x x x
f(x) =  cos2 − sin2   cos2 + sin2 
 5 5  5 5 
2x
= cos · 1
5
∴ Range ∈ [-1, 1]

Q. (ii) f(x) = sin x

A. Range: [-1, 1]

Q. (iii) f(x) = 3 – 2x

A. y = 3 – 2x
2x ∈ (0, ∞)
3-2x ∈ (3 - ∞, 3 – 0)
∴ y ∈ (–∞, 3)
Function

42.
Q. (iv) f(x) = sin (log2x)

A. log2x ∈ (-∞, ∞)
∴ sin(log2x) ∈ [–1, 1]

Q. (v) f(x) = cos 2x – sin 2x

A. Range of a cos θ + b sin θ is  − a2 + b2 , a2 + b2 

 
∴ Range of cos 2x – sin 2x is  − 12 + ( −1) , 12 + ( −1) 
2 2

 
⇒ Range:  − 2, 2 

Q. (vi) y = 2|cosx| – |sin x|

A. y is continuous function. |cosx| is maximum at the same time when |sin x| is

minimum.
Also, |cosx| is minimum when |sinx| is maximum.
∴ In such a case, ymax = 2 × (1) – (0) = 2
ymin = 2 × 0 – 1 = -1
∴ Range: [-1, 2] (⸪ y is a continuous function)

tan ( π [ x − π])
Q. (vii) f ( x ) =
x2 − 3x + 4
where [·] denotes greatest integer function.

A. [x – p] will always be integer.

∴ Numerator is always 0.
∴ Range is {0}.

Q.  π
(viii) f ( x ) = cot 2  x − 
4


A.  π
cot  x −  ∈ (-∞, ∞)
 4
 π
∴ cot2  x −  ∈ [0, ∞)
 4
Function

43.
 x2 − x + 1
Q. (ix) f ( x ) =
x2 + x + 1
=y

A. ⇒ x2 – x + 1 = x2y + xy + y
⇒ x2 (1-y) – x (1+y) + 1 -y = 0
If y is a value in range, then this equation contains solution.
So, D ≥ 0, y ≠ 1 (for y = 1, check separately)
∴ (y + 1)2 ≥ 4(1 – y)2
⇒ (3 – y) (3y – 1) ≥ 0
1
⇒ ( y – 3)  y – ≤0
 3
1 
⇒ y ∈  , 3
3 

Now, for y = 1, x2 – x + 1 = x2 + x + 1 ⇒ x = 0, so value of x exist

1 
Hence y ∈  , 3
3 

x − [x ]
Q. (x) y =
1 + x − [x ]
, where [·] denotes greatest integer function.

1 + x − [x ] 1
A. y= −
1 + x − [x ] 1 + x − [x ]
1
⇒ y = 1-
1 + {x}
{x} ∈ [0, 1) ⇒ 1 + {x} ∈ [1, 2)
1 1   1
⇒ ∈  , 1 ⇒ y ∈ 0, 
1 + {x}  2   2

 π
(xi) f ( x ) = tan  {x} ×  , where {·} denotes fractional part function.
Q.  4

π  π
{x} × ∈ 0, 
A. 4  4 
∴ Range: [0, 1)

π 
Q. (xii) f(x)= tan  sgn ( x2 − 1) 
4 

sgn(x2-1) ∈ {–1, 0, 1}
A.  π π
∴ Range includes tan  −  , tan 0, tan  
 4 4
Function

∴ Range is {-1, 0, 1}.

44.
Q. (xiii) f(x)=|x2 – x – 6|

A.  −D 
x2 − x − 6 ∈  , ∞
 4a 
 −25 
∴ x2 − x − 6 ∈  , ∞
 4 
⇒ |x2 – x - 6| ∈ [0, ∞ )

Q. (xiv) y = |sinx| + |cosx|

A. Squaring both sides.

y2 = sin2x + cos2x + 2 |sinx| |cosx| = 1 + |sin 2x|
∴ y2 ∈ [1, 2] (⸪ |sin 2x| ∈ [0,1])
∴ y ∈ [1, 2 ] (⸪ y is positive)

1
Q. (xv) y =
x +x+1
2

A. x2 + x + 1 ∈ 
 −D 
, ∞
 4a 
3 
x2 + x + 1 ∈  , ∞ 
4 
1 1 1 
∴ ∈ ,
x + x + 1  ∞ 3 / 4 
2

 4
⇒ y ∈  0, 
 3

1
Q. (xvi) y =
sin x + cos4 x
4

A. y=
1
=
1
sin x + cos x ( sin x + cos x ) − 2sin2xcos2x
4 4 2 2 2

1 1
= =
1 1
1 − ( 4sin2 xcos2 x ) 1 − ( sin2x )
2

2 2
1 2  1 1 
1 − ( sin2x ) ∈ 1 − × 1, 1 − × 0
2  2 2 
1 1
( sin2x ) ∈  , 1
2
⇒ 1−
2 2 
Function

∴ y ∈ [1, 2]

45.
 Q. (xvii) f(x) = [sinx], where [·] denotes greatest integer function.

A. sin x ∈ [-1, 1]
⇒ sin x ∈ [-1, 0) ∪ [0, 1) ∪ {1}
⇒ [sin x] ∈ {-1, 0, 1}

Q. (xviii) f(x) = 2 – [sinx] – [sin x]2 = y, where [·] denotes greatest integer function.

A. case I: [sin x] = –1
y = 2- (-1) – (-1)2 = 2 + 1 – 1 = 2
Case II: [sinx] = 0, y = 2 – 0 – 0 = 2
Case III: [sinx] = 1
y=2–1–1=0
∴ y ∈ {0, 2}

Remainder theorem
The polynomial remainder theorem states that
the remainder of the division of a polynomial
f(x) by a linear polynomial x – α is equal to
f(α).
In particular, x – α is a divisor of f(x) if and
only if f(α)=0, a property known as the factor
theorem.
D = (d × q) + R
where, D = dividend, d = divisor, q = quotient, Point to Remember!!!
R = remainder.
(i) If f(x) = g(x), where f(x) and g(x) are
Ex. x3 + 4x2 – 7x + 6 when divided by x – 1 polynomials, then coefficients
Remainder → 13 + 4 · 12 – 7 · 1 + 6 = 4 of all different powers of x are
equal on both sides.
Ex. x3 + x = (x - 1) ( Q ( x ) ) + 2
 (ii) If polynomial is divided by
 Remainder
Quotient quadratic, the remainder is a
linear.
Ex. x2 + 2x – 3 = Ax2 + Bx + c If divided by a cubic, remainder
⇒ A = 1, B =2, C = – 3 is a quadratic.
If divided by a biquadratic,
remainder is a cubic and so on.
Function

46.
 Q. ax4 + bx3–x2 + 2x+3 when divided by x2 + x – 2 gives remainder 4x + 3. Find
a and b.

A. ax4 + bx3 – x2 + 2x + 3 = (x-1) (x + 2) Q(x) + 4x + 3

put x = 1 both side
a + b -1 + 2 + 3 = 0 + 7
⇒ a + b = 3 …(i)
put x = –2
16a - 8b – 4 – 4 + 3 = 0 – 5
⇒ 2a – b = 0 …(ii)
from (i) and (ii), a = 1, b = 2.

Definition of Equal or Identical functions

Two functions f and g are said to be identical if
1. Domain of f = domain of g.i.e., Df = Dg.
2. Range of f = Range of g.
3. f(x) = g(x) ∀ x ∈ Df or x ∈ Dg

Ex.
f(x) = x and g(x) = x2 are not identical function as Df = Dg but Rf = R,
Rg = [0, ∞)
Overall, it can be said that graph of f(x) and g(x) must be same everywhere
for two functions to be identical.

Ex.
f(x) = sinx; g(x)= cos x
Df = Dg = R
Rf = Rg = [-1, 1]
But sinx ≠ cos x everywhere.
∴ graph is not same. So, f(x) and g(x) are not identical.

Q. Check if function are identical or not

(i) y = ln x2; y = 2 ln x

A. Domain of ln x2 : x2 > 0
⇒ x≠0
Domain of 2lnx → x > 0
∴ Domain not same. So not identical.
Function

47.
 1
Q. (ii) y = cosec x ; y =
sinx

1
A. cosec x =
sinx
Graph is same
∴ both functions are identical.

1
Q. (iii) f(x) = tanx; g(x) =
cotx
sinx
A. f(x)=
cosx
, cosx ≠ 0

1
g(x) =
 cosx 
 
 sinx 
sinx ≠ 0 because denominator will not be defined.
also, cos x ≠ 0
as denominator will become 0.
∴ f(x) and g(x) are not identical because domain are not same.

1
Q. (iv) f(x) = secx; g(x) =
cosx

A. Identical

Q. (v) f(x) = cot2(x). cos2x ; g(x) = cot2x – cos2x

A. Both f(x) and g(x) contain cotx, so sinx ≠ 0.

⇒ Domain is same.
g(x) = cot2x – cos2x = cot2x (1-sin2x) = cot2x cos2x = f(x)
∴ graph is same.
∴ Identical

Q. (vi) f(x) = sgn(x2 + 1); g(x) = sin2x + cos2x

A. x2 + 1 > 0 ∀ x ∈ R
∴ f(x) = 1 ∀ x ∈ R
g(x) = 1 ∀ x ∈ R
∴ Both are identical.
Function

48.
Q. (vii) f(x) = tan2x. sin2x; g(x) = tan2x – sin2x

A. Identical

Q. (viii) f(x) = sec2x – tan2x ; g(x)=1

A. f(x) = 1 ∀ x ∈ R. - ( 2n + 1)
π
,n∈I
2
g(x) = 1 ∀ x ∈ R
So, domain is not same.
∴ Not identical

1
Q. (ix) f(x) = logxe ; g(x) =
log e x

A. Df : (0, 1) ∪ (1, ∞)
Dg : (0, 1) ∪ (1, ∞)
Domain is same.
Also, f(x) = g(x)
∴ Identical

1
Q. (x) f(x) = logex ; g(x) =
log xe

A. Df : x > 0
Dg : x > 0, x ≠ 1
⇒ Df ≠ Dg
∴ Not Identical

Q. (xi) f ( x ) = x2 − 1 ; g ( x ) = x−1 x+1

A. Df : x2 – 1 ≥ 0
⇒ x ∈ (-∞, -1] ∪ [1, ∞)
Dg :x – 1 ≥ 0 ⇒ x ≥ 1
x + 1 ≥ 0 ⇒ x ≥ -1
⇒ Dg : x ≥ 1
⇒ Df ≠ Dg
∴ Not Identical
Function

49.
 Q. (xii) f ( x ) = 1 − x2 ; g ( x ) = 1 − x. 1 + x

A. Df : 1 – x2 ≥ 0
x ∈ [-1, 1]
Dg : 1- x ≥ 0 ⇒ x ≤ –1
1 + x ≥ 0 ⇒ x ≥ –1
⇒ Dg : x ∈ [-1, 1]
⇒ Df = Dg ⇒ Identical

Q. (xiii) f ( x ) = elne ; g ( x ) = ex
x

A. f(x) = ex = g(x)
(⸪ a(logaN) = N )
Df = Dg
∴ Identical

Q. (xiv) f(x) =
1 − cos2 x
; g ( x ) = sinx
2

A. f(x) ≥ 0 (Because of square root)

g(x) can be negative.
Range not same.
∴ Not Identical

Q. (xv) f(x) = log(x + 2) + log(x – 3); g(x)=log (x2 – x – 6)

A. Df : x + 2 > –0 ⇒x>-2
x – 3 > 0 ⇒x>3
∴ Df : (3 ∞)
Dg : x2 - x - 6 > 0
⇒ (x -3) (x + 2) > 0
⇒ x ∈ (-∞ , – 2) ∪ (3, ∞)
⇒ Df ≠ Dg
∴ Not identical

1 x
Q. (xvi) f(x) =
1
; g (x) =
1+x
1+
x

A. Df : x ≠ 0, – 1
Dg : x ≠ -1
⇒ Df ≠ Dg
Function

∴ Not Identical

50.
 Q. (xvii) f(x) = [{x}] ; g(x) = {[x]}

A. f(x) = 0 = g(x)
Df = Dg = (-∞, ∞)
∴ Identical.

Classification of Functions
1. Definition of One-one (injective Mapping)
f: A → B such that different elements of A have
different f images in B.
or x1, x2 ∈ A and f(x1), f(x2) ∈ B,
f(x1) = f(x2) ⇒ x1 = x2
or x1 ≠ x2 ⇒ f(x1) ≠ f(x2)

Ex.

(Every input has a different output)

Ex.

Not one-one (3 inputs have same output)

Q.1 Find if given functions are one-one or not

(i) y = x + 1

A. All horizontal lines can cut the graph at only 1

point.
⇒ One-one function
Function

51.
 Q. (ii) y = |x|

A. A horizontal line cuts this graph

at 2 points
⇒ f(x1) = f(x2) = K
⇒ Not one-one

Q. (iii) y = ex

A. dy
dx
= ex > 0

⇒ Increasing function ∀ x ∈ R
(continuous also)
⇒ at any two different x, y can’t be same
⇒ One-one function

Q. (iv) y = x3

A. dy
= 3x2
dx

dy
∴ ≥0
dx

⇒ Continuous and increasing.

Point to Remember!!!
⇒ Horizontal line cuts the graph at
y If a line parallel to x axis cuts the
only one-point.
graph of the function atleast at
two points, then f is many one.
⇒ One–one function
y A continuous function whose
derivative changes sign is a
many one function.

Q. (v) y = sinx

A. Many-one function
Horizontal line cuts at more than one point
Function

52.
 Find whether the given function are one one or not:
Q. 1 1
[x], {x}, , 2 cosx, tanx, sgn(x)
x x

A. [x] and {x} → different x but same y

→ Not one-one
1
⇒ y= ⇒ one-one
x
1
⇒ y= 2
x
f(2) = f(–2)
→ Not one-one
⇒ cosx and tanx ⇒ not one-one (similar to sinx)
⇒ (sgn(1) = sgn(2)) ⇒ sgn(x) is not one-one

2. Definition of Many-one function

f:A → B such that two or more elements of A have
the same f-image in B.
or There exist x1, x2 ∈ A such that
f(x1) = f(x2) but (x1) ≠ (x2)

Ex. y = (x-1) (x-2) (x-3)

Sol. f(1) = f(2) = f(3) = 0
⇒ Many-one function.

3. Definition of Onto (surjective mapping)

f: A → B such that each element in B is the f image
of at least one element in A.
In case of onto function, codomain is equal to
range.
So, to find if the function is onto, find range and
match it with codomain.

Ex. (i) f : R → R ; f(x) = 2x + 1

(ii) f : R+ → R ; f(x) = ln x
(iii) f : R → R+ ; f(x) = ex
Sol. (i) y = 2x + 1 ⇒ y ∈ R ⇒ Codomain = Range ⇒ Onto.
(ii) y = lnx ⇒ y ∈ R ⇒ Codomain = Range ⇒ Onto.
(iii) y = ex ⇒ y ∈ (0, ∞) or R+. ⇒ Codomain = Range ⇒ Onto function.
Function

53.
 4. Definition of Into function
f: A → B such that at least one element in B
(co-domain) is NOT the f image of any element
in domain A.

Ex. f: R → R; f(x) = {x}, [x], sgn x

Sol. Range: [0, 1), I, {-1, 0, 1} respectively Know the facts
But, codomain is R.
⇒ codomain ≠ Range ⇒ All are into function. For into function, codomain≠Range.

Q. Find whether the given functions are onto or into:

(i) f : R → [0, ∞); f(x) = x2
(ii) f : R → [0, ∞); f(x) = x2 +1
(iii) f : R → [0, ∞); f(x) = x100 + 100x98 + 3x2 + |x| + xsgn(x)

A. (i) x2 ∈ [0, ∞) ⇒ Codomain = Range ⇒ Onto function

(ii) x + 1 ∈ [1, ∞) ⇒ Codomain ≠ Range ⇒ Into function
2

(iii) y = x100 + 100 x98 + 3x2 + |x| + x sgn(x)

≥0 ≥0 ≥0 ≥0 |x|≥0
∴ y ≥ 0 (fmin = 0 at x = 0 and fmax =∞)
→ Range = Codomain.
→ Onto function.

One-one onto function (Bijective / invertible)

If function is both injective and surjective, then it is called a Bijective

function.

One one into function

This type of function is one-one but not onto.

Function

54.
Many one onto function

This type of function is both many-one and

onto.

Many one into function

This type of function is both many-one and

into.

Q. Classify as one one onto, one one into, many one onto or many one into:
f:[-1, 1] → [-1, 1] f(x) = sin2x

A. x ∈ [–1, 1]
y ∈ [–1, 1]
∴ Range = codomain ⇒ onto function
Horizontal line cuts at 2 points
⇒ Many-one function.
∴ f(x) = sin 2x is many one onto function.

Q. Classify as one one onto. one one into. many one onto or many one into:
2x2 − x + 5
f: R → R; f ( x ) =
7x2 + 2x + 10

A. Method-1:
dy
Find minimum and maximum values by finding critical points where =0
dx
Method-2:
Numerator and denominator are both positive ∀ x ∈ R as D < O
∴ f(x) > 0 ⇒ Range ≠ Codomain
∴ function is into
5 2x2 − x + 5
f (0) = =
10 7x2 + 2x + 10
⇒ 7x2 + 2x + 10 = 4x2 – 2x + 10
Function

55.
 ⇒ 3x2 + 4x = 0
4  −4 
⇒ x = 0 or − ⇒ f(0) = f  
3  3 
⇒ Many-one function.
∴ f(x) is Many-one into function.

Q. The function f [2, ∞) → Y defined by f(x) = x2 – 4x + 5

is both one and onto if:
(A) Y = R (B) Y = [1, ∞)
(C) Y = [4, ∞) (C) Y = [5, ∞)

A. f’(x) = 2x – 4 ≥ 0 ∀ x ≥ 2
Also, f(x) = (x-2)2 +1
⇒ Range: [1, ∞ )
⇒ f(x) is one-one for x ∈ [2, ∞)
⇒ Since f(x) is onto, Range = codomain.
So, option (B) is correct answer.

Permutation based problems:

f: A → B
Case-1:
When both the set A and B contain an equal
number of elements.

(i) Total number of functions.

Sol. ‘a’ has 3 options for image.
‘b’ has 3 options for image.
‘c’ also has 3 options i.e., 1,2 or 3.
∴ Total number of functions possible
= 3 × 3 × 3 = 33 Point to Remember!!!
Similarly, if no of elements is 5, then number
of functions = 55. If number of elements is ‘n’, then
Generalizing it, number of functions = nn.

(ii) Total number of one-one functions.

Sol. if number of elements is 5:
‘a’ has 5 option for image.
After that, ‘b’ will have 4 options, c → 3, d → 2,
e→1
Function

56.
∴ Total no of one-one function
= 5 × 4 × 3 × 2 × 1 = 5! Point to Remember!!!

For ‘n’ elements, number of one-

one functions = n!
(iii) Number of many one functions.
Sol. Number of many one functions = Total
Point to Remember!!!
number of functions – number of one-one
functions
Number of many-one functions
= nn – n!

(iv) Number of onto functions.

Sol. In case of equal number of elements in both Point to Remember!!!
domain and codomain, all the onto functions
are one-one function. Number of onto functions = n!

(v) Number of into functions.

Sol. Number of into functions = Total – Number
of onto functions = nn – n!

Case-2:
When number of elements in A (domain) is
more than B

(i) Total number of functions.

Sol. Let number of elements in A and B be 5 and
4 respectively.
Every input has 4 options for output.
So, total number of functions
= 4 × 4 × 4 × 4 × 4 = 45
Point to Remember!!!
(ii) Number of one-one (injective) functions
Sol. = 0 Total number of functions = nm
where, m is number of elements in
(iii) Number of many-one functions
A and n is number of elements in B.
Sol. = total – 0
= 45
= nm

(iv) Number of onto functions

Sol. It is calculated using group formation.
Function

57.
 Ex.
Number of ways to distribute 5 elements of A
 5! 1 
among 4 elements of B is  ×  × 4!
 2 ! ( 1!) 3
3 ! 


(v) Number of into functions

Sol. = Total – Number of onto functions.

Case-3:
Number of elements in codomain (B) is more
than that in A

(i) Number of total functions

Sol. It is same in this case also and is equal to nm,
where n and m are number of elements in ‘B’
and ‘A’ respectively.

(ii) Number of injective mapping

Sol. No of elements in A = 4
No of elements in B = 6
So, number of one-one function

=
6
 C4 × 4
!
selection of 4 output distribution

(iii) Number of many-one functions

Sol. = Total – one-one
= 64- 6C4 × 4!

(iv) Number of onto functions

=0

(v) Number of into functions

Sol. Number of into functions = nm

Composite Functions
Let f: A → B and 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.
 
g  f ( x )  = h(x)
Function


 input 

58.
Q.1 If f(x) = x2 and g(x) = x-7. Find: (i) gof (ii) fog

A. (i) gof = g(f(x)) = g(x2) = x2 – 7

(ii) fog = f(g(x)) = f(x-7) = (x-7)2

Q.2 If f(x) = x2 and g(x) = x – 7. Find: (iii) gog (iv) fof

A. (iii) gog = g(g(x)) = g(x-7) = (x-7)-7 = x – 14

(iv) fof = f(f(x)) = f(x2) = (x2)2 = x4

Q.3 Let f(x) = x ; g(x) = 2 − x . Find: (A) fog (B) gof

A. (A) fog = f ( 2 − x) = 2−x

(B) gof = g ( x ) = 2 − x

Q.4 Let f(x) = x : g ( x ) = 2 − x . Find: (C) fof (D) gog

A. ( x) = x = x
(C) fof = f 1/4

(D) gog = g ( 2 − x ) = 2 − 2−x

Domain of gog: 2- 2 − x ≥ 0 ⇒ 2 ≥ 2−x

⇒ 4 ≥ 2 – x ⇒ x ≥ − 2
 ⇒ x ∈ [ −2, 2]
Also 2 − x ≥ 0 ⇒ x ≤ 2 
For range: 2-x ∈ [0, 4] ⇒ 2 − x ∈ [0, 2]
⇒ 2- 2 – x ∈ [0, 2] ⇒ gog ∈ 0, 2 

Q.5 Prove that: f(x) = x2 if fofof … fof

 = x
2n

n times

A. Generally it becomes a sequence. So, find fof, fofof, fofofof(x).

Now observe the sequence and hence find the fofof … fof

n times

fof(x): ( x ) = x fofof ( x ) = ( x ) = x
2 2 4 4 2 8

fofofof = ( x ) = x = x ⇒ fofof... fof ( x ) = x(

8 2 16 24 2n)
Function

59.
 x
Q.6 f (x) = ; g(x)=fofofo…fof(x) (f written ‘n’ times), g(x) =?
(1 + x )
n 1/n

x
f (1 + x ) n 1/n
x
A. fof(x) = = =
(1 + f ) n 1/n
( 1 + 2x )
1/n n 1/n
 x  n

1 + 
 1 + xn 
x

f(f(f(x))) =
( 1 + 2x ) n 1/n

=
x
( 1 + 3x )
1/n n 1/n
 x  n

1 + n 
 1 + 2x 
By observation, it can be said that
x
fff…f(x) n times =
( 1 + nxn )
1/n

Non-uniformly defined functions

Q.7  1 + x if 0 ≤ x ≤ 2
f (x) =  . Find fof.
3 − x if 2 < x ≤ 3

A.  1 + f; 0 ≤ f ≤ 2
f(f(x)) = 
3 − f; 2 < f ≤ 3
 1 + (1 + x ) ; 0 ≤ x ≤ 2 0 ≤ 1 + x ≤ 2 ⇒ x ∈ [0, 1 ]

 1 + (3 − x ) ; 2 < x ≤ 3 0 ≤ 3 − x ≤ 2 ⇒ x ∈ (2, 3]
= 
 3 − (1 + x ) ; 0 ≤ x ≤ 2 2 < 1 + x ≤ 3 ⇒ x ∈ (1, 2]
3 − ( 3 − x ) ; 2<x≤3 2 < 3 − x ≤ 3 ⇒ x ∈φ
2 + x; x ∈ [0, 1]

∴ f(f(x)) = 4 − x; x ∈ ( 2, 3]
2 − x; x ∈ ( 1, 2]

Function

60.
 1 − x if x ≤ 0  −x if x < 1
Q.8 f(x) =  2
 x if x > 0
and g(x) = 
1 − x if x ≥ 1
.

Find (fog)(x)

1 − g ( x ) ; g ( x ) ≤ 0
A. f(g(x)) =  2
 g ( x ) ; g ( x ) > 0
 1 − ( −x ) ; x ∈ [0, 1)

= 1 − ( 1 − x ) ; x ∈ [ 1, ∞ )
 2
 ( − x ) ; x ∈ ( −∞, 0 )
(This time solved directly using the graph of g(x)).

1 + x3 if x < 0 ( x − 1)1/3 if x < 0

Q.9 f(x) =  2 and g(x)=  . Find g(f(x)).
 x − 1 if x ≥ 0
1/2
( x + 1) if x ≥ 0

1

( f − 1 ) 3 ; f < 0
A. g(f(x)) =  1
( f + 1) 2 ; f ≥ 0


( ( ) )
 1 + x3 − 1 1/3 ; x ∈ ( −∞, −1)
 2
(
 ( x − 1) − 1 ; )
1/3
x ∈ [0, 1)
= 
 ( 1 + x3 + 1) ;
1/2
x ∈ [ −1, 0 )

( )
 x2 − 1 + 1 ; 1/2
x ∈ [ 1, ∞ )
 x ; x ∈ ( −∞, −1)
 2
( x − 2)
1/3
; x ∈ [0, 1)
g(f(x)) = 
( x 3 + 2)
1/2
; x ∈ [ −1, 0 )

 x ; x ∈ [ 1, ∞ )
Function

61.
 Q.10 Find number of distinct real c satisfying f(f(f(c))) = 3 where f(x) = x2 – 2x

A. Let f (θ) = 3, θ = f(f(c))

θ2 – 2θ = 3 ⇒ θ2 – 2θ – 3 = 0
θ = 3 or − 1
⇒ f(f(c)) = 3 or -1
⇒ Let f(t) = 3 or -1, t = f(c)
t2 – 2t = 3 ⇒ t = 3 or – 1
and
t2 – 2t = -1
(t-1)2 = 0
t=1
∴ f(c) = 3 or 1 or -1
3
c2 – 2c – 3 = 0 
−1
c2 – 2c + 1 = 0 ⇒ c = 1
1 + 2
c2 – 2c – 1 = 0⇒ 
 1 − 2
∴ ‘c’ can be 1 ± 2 , 1, -1, 3.
Total 5 solutions.

Q.11 y = f(x); Domain ∈ [-3, 2]

Then domain of y = f(|[x]|) is? ([·] denotes greatest integer function)

A. |[x]| ∈ [-3,2]
⇒ |[x]| ∈ [0, 2] (⸪ |x| is always non-negative)
⇒ [x] ∈ [-2, 2]
⇒ x ∈ [-2, 3)
Function

62.
 Functional Equations

Q.1 2f(x) + f(1-x) = x2; f (4) = ??

A. 2f(x) + f(1 – x) = x2 ...(i)

In functional equations, if we replace ‘x’ by some ‘y’ in L.H.S, then we also

must replace ‘x’ by ‘y’ in R.H.S.
Applying this concept here, replace ‘x’ by ‘1-x’.
We get,
2f(1-x) + f(1-(1-x)) = (1-x)2
2f(1-x) +f(x) = 1-2x+x2 ...(ii)
Multiplying 1st equation by 2, we get
4f(x) + 2f(1-x) = 2x2 ...(iii)
subtracting (ii) from (iii), we get
3f(x) = x2 + 2x – 1
x2 + 2x − 1 23
⇒ f(x) = ⇒ f(4) =
3 3
1
 
Q.2 f ( x ) + 3x f   = 2 ( x + 1) ∀ x > 0. Find f(10099).
x
1
A. Replace x by
x
 1 3 1 
⇒ f   + f ( x ) = 2  + 1   × ( 3x ) and subtract the first equation.
 x x x 
⇒ 9f(x)-f(x)=6+6x – 2(x+1)
x+1
⇒ f(x)= ⇒ f(10099) = 5050
2
n

Q.3 Find natural number values of a for which ∑f (a + k ) = 16 ( 2

k =1
n
− 1)

f(x) satisfies f(x + y) = f(x) f(y) ∀ x, y ∈ N; f(1) = 2

x = y = 1 ⇒ f(2) = f(1) · f(1) = 2 · 2 = 22
A.
x = 2, y = 1 ⇒ f(3) = f(2) · f(1) = 22 · 2 = 23
x = 3, y = 1 ⇒ f(4) = f(3) · f(1) = 23 · 2 = 24
By observation, f(n) = 2n
n n

∑2 a +k
= 16 ( 2n − 1) ⇒ 2a ∑2k = 16 ( 2n − 1)
k =1 k =1

⇒ 2 .2 ( 2 − 1) = 16 ( 2 − 1)
a n n
⇒ 2 a+1
= 24

⇒ a=3
Function

63.
 Q.4 f (x) =
ax
, a > 0 . Find
2n− 1

∑
 r 
f .
a + a r = 1  2n 
x

A.  1   2   3 
(i) f   + f   + f   + ... + f 
 2n − 1 
=S
 2n   2n   2n   2n 
 2n − 1   2n − 2   1 
(ii) f  + f + ... + f   = S
 2n   2n   2n 
 
1 2
1− 1−
2n 2n

aα a 1−α
f( α ) + f(1- α ) =
+
a α + a a 1−α + a
aα a
= α +
a + a a + a aα
aα a
= + =1
a + a
α
a + aα
So, sum of terms equidistant from beginning and end is 1.
∴ 1 + 1 + 1 +…1(2n – 1 times) = 2S
2n − 1
∴ S=
2

Q.5 f(x) is a polynomial of degree 6 and leading coefficient 4.

f(1) = 6, f(2) = 5, f(3) = 4, f(4) = 3, f(5) = 2, f(6) = 1. Find f(7).

A. T1=6, T2 = 5, T3 = 4…., T6 = 1
From observation = f(x) = 7-x for x = 1, 2, 3, 4, 5 and 6
or f(x) – (7-x) = 0 ∀ x ∈ {1, 2, 3, 4, 5, 6}
⇒ f(x) – 7 + x = 4 (x -1) (x -2) (x -3) (x -4) (x -5) (x -6)
⇒ f(x) = 4(x -1) (x -2) (x -3) (x -4) (x -5) (x -6) + 7-x
⇒ f(7) = 4·6·5·4·3·2·1 + 7-7 = 2880

Inverse of function for bijective function

Let f: A → B be a one-one and onto function,
then there exists a unique function g: B → A
such that f(x) = y ⇔ g(y)=x, ∀ x ∈ A and y ∈ B.
Then g is said to be inverse of f.
Know the facts
Thus g= f-1 : B→ A {(f(x), x) |(x, f(x))∈ f|}
1 Condition for a function to have a
g = f-1 ≠
f inverse:
f ≡ {(a, 1), (b, 2), (c, 3)} (1) f must be bijective
⇒ g ≡ {(1, a), (2, b), (3, c)} (2) g is also bijective
Function

64.
Q. Find the inverse of the following:
(i) y = 2x (ii) y = x – 1 (iii) y = 4-x
x 1
(iv) y = (v) y =
4 x

A. (i) y = 2x; x → y and y → x

Point to Remember!!!
⇒ x = 2y
x
⇒y= Function given as y = f(x), try to
2
separate all terms of y from x and
x
⇒ f-1(x)= make the form x = g(y).
2
(ii) y = x -1
⇒ x = y – 1 ⇒ y = x + 1 ⇒ f-1(x) = x + 1
(iii) y = 4-x
⇒ x = 4-y ⇒ y = 4-x ⇒ f-1(x) = 4 – x Point to Remember!!!
x
(iv) y = f and f-1 are mirror image of each
4
y other about line y = x
⇒x= ⇒ y = 4x ⇒ f-1(x) = 4x
4
1
(v) y =
x
1 1 1
⇒x= ⇒y= ⇒ f-1(x) = = f(x)
y x x

Ex. y = f(x) = ex
f-1(x) = lnx
Graph can be understood by concept that (b, a) is
image of (a, b) about line y = x
Know the facts

Note: Since, image is unique, inverse of a

If y = f(x) and g(y) = x function is unique.
⇒ g(f(x)) = x and f(g(y)) = y
⇒ f-1(f(x)) = x
⇒ If f(g(x)) = x
⇒ f and g are inverse of each other.
Function

65.
 Q.1 Compute the inverse: f: R → R+, f(x) = 10x+1

A. y = 10x+1
x = 10y+1
⇒ log 10 x = y + 1
⇒ log 10 x − 1 = y
⇒ f −1 ( x ) = log 10 x − 1

Q.2 Compute the inverse: f: (-2, ∞) → R, f(x) = 1 + ln (x +2)

A. y = 1 + ln(x + 2)
⇒ x = 1 + ln (y + 2)
⇒ x -1 = ln (y + 2)
⇒ y + 2 = ex-1
⇒ y = f-1(x) = ex-1 – 2

2x
Q.3 Compute the inverse: f: R → (0, 1), f ( x ) =
1 + 2x

A. y =
2x
1 + 2x
2y 1
⇒ x= = 1−
1+2 y
1 + 2y
1
⇒ =1–x
1 + 2y
1
⇒ 1 + 2y =
1−x
1 x
⇒ 2y = −1=
1−x 1−x
 x 
⇒ y = f −1 ( x ) = log 2  
1− x

Q.4 Compute the inverse: ln x + x2 + 1( )

A. (
y = ln x + x2 + 1 )
(
⇒ x = ln y + y 2 + 1 )
⇒ y + y 2 + 1 = ex
Function

66.
 ⇒ y 2 + 1 = ex − y
⇒ y2 + 1 = y2 + e2x – 2y ex
e2x − 1
⇒ y= = f-1(x)
2ex

Q.5 If f: R → R ; f(x) = x3 + (a+2)x2 + 3ax + 5 is an invertible mapping. Find ‘a’.

A. f is one-one and onto

⇒ f´(x) = 3x2 + 2(a+2)x + 3a ≥ 0 or ≤ 0 always (Here f’(x) ≤ 0 is not possible)
⇒ 3x2 + 2(a+2) x+ 3a ≥ 0
⇒ D≤0
⇒ 4(a + 2)2 – 4 × 9a ≤ 0
⇒ a2 – 5a + 4 ≤ 0
⇒ a ∈ [1, 4]

ex + e− x
Q.6 f: [0, ∞) → [1, ∞); f(x) =
2
. Find f-1(x).

A. x=
ey + e− y
2
2xe = e2y + 1
y

e2y – 2xey + 1 = 0
2x ± 4x2 − 4
⇒ ey =
2
(
⇒ y = ln x ± x2 − 1 )
Range of f-1 will be [0, ∞)

(
∴ f-1(x) = ln x + x 2 − 1 )
 x if x<1
Q.7  2
f(x) =  x if 1 ≤ x ≤ 4 . Find f-1(x)
8 x if x>4


A. When
x < 1, f ( x ) ∈ ( −∞, 1) 

1 ≤ x ≤ 4, f ( x ) ∈ [ 1, 16] Range
x > 4, f ( x ) ∈ ( 16, ∞ ) 

Range become domain for f-1 (x)

Function

67.
 
 x, x ∈ ( −∞, 1 )

So, f-1(x) =  x , x ∈ [1, 16 ]
 2
 x , x ∈ ( 16, ∞)
 64

3  7
Q.8 
A function f :  , ∞  →  , ∞  defined as f(x) = x2 – 3x + 4.
2  4 
Solve the equation f(x) = f-1(x).

A. Solution of f(x) = x will also be solution of f(x) = f-1(x).

But there may be extra solution.
So, always check the graph before solving f(x)=x
Now, since there is only 1 solution on y = x (by graph)
f(x) = x ⇒ x2 – 3x + 4 = x
⇒ (x – 2)2 = 0 2

⇒x=2

Ex.
y = f(x) = 4-x
f-1(x) = 4 – x
⇒ f(x) = f-1(x) have infinite solution here.
While f(x) = x has only 1 solution.
So, always check the graph.

Properties of inverse function

(i) The inverse of a bijection is unique.
(ii) Domain Range interchange
(iii) Inverse of bijection is also a bijection
(iv) (gof)-1 = f-1(g-1(x)). (Reversal law)

Definition of Homogeneous Functions

A function is said to be homogeneous with
respect to any set of variable when each of its
terms is of the same degree with respect to
those variables.
f(tx, ty) = tnf(x, y) then f(x, y) is homogeneous
function of degree n.
Function

68.
Ex. f(x,y) = ax2 + 2hxy + by2
f(tx, ty) = a(tx)2 + 2h (tx) (ty) + b(ty)2

= t2 (ax2 + 2hxy + by2)
⇒ Homogeneous equation of degree 2.

x x
Q. Find if f ( x, y ) = ln   is homogeneous or not?
y y

tx  tx 
A. f ( tx, ty ) = ln  
ty  ty 
x x
= t0 ln   = t0 f ( x, y )
y y
⇒ Homogeneous function of degree 0.

Definition of Implicit / Explicit function

A function defined by an equation not solved for the
dependent variable (y) is called an Implicit function.
If y has been expressed in terms of x alone then it is
called an Explicit function.

Ex. y = ex + sinx + tanx + x2 → Explicit

tany
Ex. 2xy + x2 + siny + = 0 → Implicit
x2 + 1
(It is not solved for y)

Q. x = y2 – 2y. Find domain of expicit form.

A. y2 – 2y = x
⇒ y2 – 2y + 1 = 1 + x
⇒ (y - 1)2 = 1 + x
⇒ y -1 = ± 1+x
⇒y=1± 1 + x (2 branches ⇒ 2 functions)
⇒1+x≥0⇒x≥–1

⇒ Domain: x ∈ [-1, ∞)
Function

69.
 Note:
y2 = x is not a function
y=x
⇒  it represents 2 separate function.
y = −x 
Same is the case for x2 + y2 = 1
y2 = 1 – x2

⇒ y= 1 − x2

=– 1 − x2

Definition of Bounded Function Know the facts

A function is said to be bounded if |f(x)| ≤ M,
where M is a finite quantity. If graph of function can be bounded
between 2 horizontal lines, then
Ex. function is bounded.
(i) y = |sinx| ≤ 1 → Bounded
If range of function contain ∞ or -∞
(ii) y = |[x]| → No maximum or minimum value.
⇒ function is not bounded.
So unbounded.
(iii) y = ex → unbounded
(iv) y = lnx → unbounded
(v) y = {x} → bounded
(vi) y = sec x → unbounded.
(vii) y = x3 – 10x2 + 3x + 1 = f(x), f; [-100, 26] → R
→ bounded
(Since domain is finite, range will also be finite. So,
it is bounded)
Function

70.
(viii) y = sgnx → bounded
as range is finite {-1, 0, 1}

Definition of ODD and EVEN Functions

A function f(x) defined on the symmetric
interval (–a, a)
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) for all x in the domain of ‘f’ then
f is said to be an odd function.

Geometrical interpretation of odd and even

function

(i) y = f(x) ⇒ (x, y)

For even function, f(-x) = y = f(x)

⇒ Graph will be symmetric about y-axis.

(ii) For odd function f(–x) = –f(x)

 e can see that, for odd function,

W
Graph is symmetric about origin.

Ex. Common odd functions: f = sinx, tanx, x, x3 Know the facts

Ex. Common even functions: f = cos x, sin2x, x2,
(i) A function may neither be odd
|x|, |sinx|
nor even. eg. x2 + x, sinx + cosx.
(ii) Every function can be expressed
Ex.
as the sum of an even and an odd
(i) f(x) = x2 + x ⇒ f(–x) = x2 – x ≠ f(x) or -f(x)
function.
∴ x2 + x is neither odd nor even.
f ( x ) + f ( −x ) f ( x) + f (− x) f ( x) − f (− x)
(ii) (a) Let h(x) = f ( x) = +
2 2
 2

f ( −x ) + f ( x ) even odd
h(-x) = = h(x) so,
2 (iii) The only function which is
it is even function. defined on the entire number line
and is even and odd at the same
time is f(x) = 0.
Function

71.
 (b) Let h(x) = f(x) – f(–x)
h(–x) = f(–x) – f(x) = –h(x).
So, it is odd function.
Sum of both of above functions is f(x). Hence f(x) can be divided into
sum of even and odd function.

Q.1 Express ex as sum of an odd and even function

A.  ex + e− x 
ex =  +
 e x − e− x 
 
2
 2
 
even function odd function

Q.2 Identify given below functions as odd, even or neither odd nor even.
1− x
(i) f(x) = ln   = ln (1-x) – ln (1+x)
1+ x

A. 1+ x
f(-x) = ln   = ln (1+x) – ln (1-x) = –f(x)
1− x
So, it is odd function.

Q. (
(ii) f(x) = ln x + 1 + x2 )
A. f(–x) = ln (-x+ 1 + x2 )
f(x) + f(–x) = ln ((1+x2)- x2) = ln1 = 0
⇒ f(-x) = –f(x), so f(x) is odd function.

Q. (iii) f(x) = 1 + x + x2 − 1 − x + x2

A. f ( −x ) = 1 − x + x2 − 1 + x + x2 = − f ( x ) → odd

Q. (iv) f ( x ) = x
2x + 1
2x − 1
 1 + 2x 
 
A. 2 + 1 −x
f ( −x ) = −x  − x 2x 
 = −x 
2 − 1 1−2 
x

 
 2x 
 1 + 2x   2x + 1 
= − x x 
= x x = f(x) → even
1−2   2 − 1 
Function

72.
 (1 + 2 ) x 2

Q. (v) f ( x ) =
2x

A. f ( −x ) =
(1 + 2 )−x 2
1 + 2.2− x + 2−2x
=
2− x 2− x
22x + 2.2x + 1
= 2x + 2 + 2-x = = f (x)
2x
So, it is even function.

1 + 2x
Q. (vi) f ( x ) =
1 − 2x

1 + 2− x 2x + 1
A. f ( −x ) = =
1 − 2− x 2x − 1
= −f ( x ) → odd

f(x) = ([a]2-5[a] + 4) x3 + (6{a}2 – 5{a} +1) x + x tan x is even function.

Q. Find ∑ ai (ai are values of a for which function is even)
([.] is the greatest integer function and {.} is fractional part function)

A. f (x) = (
[a ] − 5 [a ] + 4 ) x + (6 {a} − 5 {a} + 1) x + xtanx
2 3 2

even part
odd part

For f(x) to be even, make odd part = 0

∴ [a]2 – 5[a] + 4 = 0
∴ [a] = 1 and 4
1 1
6{a}2 – 5{a} +1 = 0 ⇒ {a} = and
2 3
1 1 1 1
So possible values of a= [a] + {a} are 1 + ,1 + ,4 + ,4 +
2 3 2 3
2 2 35
∴ ∑ ai = 10 + 1 + = 11 =
3 3 3
Function

73.
 Note:

f(x) g(x) f(x) +g(x) f(x)-g(x) f(x)·g(x) f(x)/g(x) (gof)(x)

odd odd odd odd even even odd

even even even even even even even

odd even neither odd neither odd odd odd even

nor even nor even

even odd neither odd neither odd odd odd even

nor even nor even

Proof: Know the facts

Let’s prove the first:
h(x) = f(x) + g(x), f(x) and g(x) are odd (a) f(T) = f(0) = f(–T), where T is the
h(-x)= f(-x) + g(-x)=-f(x)-g(x)=-h(x) period.
∴ h(x) is odd
(b) Inverse of a periodic function
Let h(x) = g(f(x))
does not exist.
h(–x) = g(f(-x)) = g(-f(x)) = -g(f(x))
= – h(x) → odd. (c) Every constant function is always
Similarly, rest can be proved. periodic, with no fundamental
period.
Definition of Periodic Function (d) If f(x) has a period T and g(x)
A function f(x) is called periodic if there exists also has a period T then it does not
a positive number T (T>0) called the period of mean that f(x) + g(x) must have a
the function such that f(x + T) = f(x), for all period T. e.g., f(x) = |sin x| + |cos x|.
values of x within the domain of x.
(e) If f(x) is periodic and g(x) is also
f(x) = f(x + T) = f(x + 2T) = f(x + nT); n∈ I periodic then it does not mean that
Ex. f(x) + g(x) must be periodic.
f(x) = sinx → T1 = 2p → T can be 2p, 4p, 6p, 8p 1
g(x) = {x} → T2 = 1 → T can be 1, 2, 3, 4, 5 (f) If f(x) has a period p, then ,
f (x)
∴ No common T. ∴ f(x) + g(x) is not periodic.
and f ( x ) also has a period p
Ex. (i) y = cos x → Period = 2π
(g) If f(x) has a period T then
(ii) y = a+ b cos x → Period = 2π f(ax + b) has a period T/|a|.

Point to Remember!!!

Common periodic functions sinx, cosx, tanx, {x} , |sinx|, |cosx|, sin2x
Period → 2p 2p p 1 p p p
Function

74.
Ex. f(x) = sinx. Find period.
Know the facts
Sol. f(x + T) = f(x)
LCM of rational and irrational
⇒ sin (x + T) – sin(x) = 0 number is not defined.
p 
LCM of rational numbers , and
T  T q m
⇒ 2 sin cos  x +  = 0
2  2
r LCM of p,  and r
T T =
⇒ sin =0⇒ = np , n∈ I s HCF of q,mand s
2 2
⇒ T = 2n π but T should be smallest
Know the facts
positive value
In case of f(ax + b), let period of f(x)
⇒ T = 2π is T and f(ax + b) be T’.
f(a(x + T’) + b) = f(ax + b)
⇒ f(ax + b + aT’) = f(ax + b)
T
⇒ aT’ = T ⇒ T’ = .
a
But since period must be positive,
T
T’=
a
Examples on Periodic Functions

Find period of:

Q.1 2x 4x
(i) f(x) = cos − sin
3 5

A. Period of cosx = 2 π
2  2π
Period of cos  x  = = 3π
3  2/ 3
4  2π 5π
Similarly, period of sin  x  = =
 5  4 / 5 2
5π
LCM of 3 π and is 15 π
2
So, period is 15π.
Function

75.
 Q. (ii) f(x) = cos (sinx)

A. Period of sinx = 2 π
But f(x) is a composite in trigonometric function.
T T T
So, period may be , , .i.e., also.
2 4 8
So, we must check.
T
= π ⇒ f(x+π) = cos (sin (x+ π )) = cos (-sinx) = cos (sinx)
2
∴ π can be period.
T π
Now check for =
4 2
  π 
cos  sin  x +   = cos (cosx) ≠ f(x)
  2 
π
so, is not the period.
2
∴ π is the period.

Q. (iii) f(x) = sin (cosx)

A. Sol. Period of cosx = 2p = T

T
now check for = p
2
f(x + p) = sin (cos (x + p)) = sin (-cosx) = -sin (cosx) ≠ f(x)
∴ p is not the period.
So, T = 2π

Q. (iv) f(x) = sin4x + cos4x

A. f(x) = 1 −
1
( sin2x )
2

2
1  1 − cos4x 
f(x) = 1 −  
2 2 
2π π
Period of f(x) = period of cos (4x) = =
4 2
Function

76.
Q. (v) If period of sin ( πkx ) is 2, find k.

A. Period of sinx = 2 π
2π
∴ Period of sin ( πkx ) = =2
πk
⇒ |k| = 1
⇒ k = 1, -1

Q. (vi) f(x) = x – [x]

A. f(x) = {x}
∴ Period = 1

Q.2 f(x) = sinx + cos ax is a periodic function. Then prove that ‘a’ must be rational.

A. LCM of two numbers will only exist if either both numbers are rational, or
they are same type of irrational.
Period of sin x = 2π

Period of cos x = 2π
a
2π
LCM of 2π and is possible only when ‘a’ is rational.
a
Note:
(i) f(x) = cos x ; and sinx + {x} are aperiodic.
(ii) f(x) = xsinx is aperiodic
(iii) f(x) = sin (x + sinx) is periodic
Proof: x → x + 2 π
f (x + 2 π )= sin ((x + 2 π ) + sin (x + 2 π ))
= sin (2 π + x + sinx) = sin (x + sinx) = f(x)
∴ f(x) is periodic with period 2 π .

Q.3 Find period of:

(i) f(x) = {x} + {2x} + {3x}

1 1
A. T1 = 1 T2 =
2
T3 =
3
1 1
T = LCM of 1, and
2 3
T=1
Function

77.
 x  x 
Q. (ii) f ( x ) = {x} +   +  
2 3

A. T1 = 1; T2 = 2; T3 = 3
T = LCM of 1, 2 and 3
T=6
(n ) (n + 1 )
Q. (iii) f(x) = [x] + [2x] + [3x] + …+[nx] -
2
x

A. = [x] + [2x] + [3x] + …+ [nx] – (x + 2x + 3x + …+nx)

= – ((x-[x]) + (2x-[2x]) + (3x-[3x]) +…+ (nx–[nx]))
= – ({x} + {2x} + {3x} +…{nx})
1 1 1
Period of these small functions are 1, , , ...,
2 3 n
1 1 1
Period of their sum = LCM of 1, , , ..., = 1.
2 3 n

Q.4 f(x) + f(1 + x) = 0. Find T.

A. By functional Rule, x → x + 1
⇒ f(1 + x) + f(2 + x) = 0
Subtracting given equation
f(2 + x) – f(x) = 0 ⇒ f(x + 2) = f(x) ⇒ T = 2

Q.5 f(x + 2) + f(x – 2) = f(x) ∀ x ∈ Df. Find T.

A. x → x+2
⇒ f (x + 4) + f(x) = f(x + 2)
Add both equations.
⇒ f (x + 4) + f(x – 2) = 0
⇒ x → x+2
⇒ f(x + 6) = –f(x)
Now x → x + 6
⇒ f(x + 12) = -f(x+6)
⇒ f(x+12) = – (–f(x)) (from previous equation)
⇒ f(x) = f(x+12)
⇒ T = 12

Q.6 f ( x + λ ) = 1 + 2f − f 2 ∀ x ∈ R. Prove that f is periodic.

A. (f(x + λ)-1)2 = 2f – f2 = 1 – (f(x) – 1)2

x→x+λ
(f(x+2λ)-1)2 = 1– (f(x + λ)-1)2 = (f(x)-1)2
Function

⇒ f(x + 2λ) – 1 = f(x) – 1 or – (f(x)-1)

In 1st case, f(x+2 λ ) = f(x)
78.
 ⇒ f is periodic with T = 2λ
In 2nd case, f(x + 2λ) + f(x) = 2
But from original equation f(x) ≥ 1
∴ above equation can only be true, if f(x) = 1 always but f(x) = 1 does
not satisfy the original equation.
so, second case is rejected.
∴ f(x+2λ) = f(x) and T = 2λ

Q.7 f(x) = x ∀ x ∈ [0,2] and f(x) is even with period = 4.

Find f(5), f(7.1), f(-1), f(-7), f(2019).

A. f(x + 4) = f(x)
f(5) = f(1) = 1
f(7.1) = f(3.1) = f(-0.9) = f(0.9) = 0.9
f(-1) = f(1) = 1; f(-7) = f(-3) = f(1) = 1
f (2019) = f(-1 + 505 × 4) = f(-1) = 1

Q.8 f: R → R, f(x) = x3 -2x2 + 5x + 3 is

(A) one one, onto (B) one one, into
(C) many one, into (D) many one, onto

A. f’(x) = 3x2 – 4x + 5
a > 0, D < 0
⇒ f’(x) > 0 ∀ x ∈ R ⇒ f(x) is one-one.
Since f(x) is odd degree polynomial, its range is R which is codomain.
⇒ f is also onto. so (A) is correct.

Q.9 f : R → R, f(x) = 2x3 – 6x2 – 18x + 17 is:

(A) one-one, onto (B) one-one, into
(C) many-one, into (D) many-one, onto

A. f’x = 6x2 – 12x – 18 = 6 (x2 – 2x – 3)

f’(x) = 0 for x = -1 and 3.
graph will be like:
so, f is many one.
Range is R. So, f is onto.
⇒ (D) is correct answer.
Function

79.
 Q.10 f: [3, ∞) → [a, ∞), f(x) = 2x3 – 6x2 – 18x + 80 is an onto function, find a.

A. f’(x) = 6x2 – 12x – 18 = 6(x2 – 2x -3) = 6(x-3) (x + 1) > 0 ∀ x > 3

∴ fmin at x = 3. So a = f(3)
⇒ a = 2 × 33 -6 × 32 – 18 × 3 + 80
⇒ a = 26.

Q.11 f(x) = x2 + bx + 3 is not injective for x ∈ [0, 1], then the set of b is:
(A) (0, ∞) (B) (-2, 0) (C) (0, 2) (D)(2, ∞)

A. f is upward facing parabola. It will only be true if minima lies in interval (0, 1)
−b
x-coordinate of minima =
2
−b
⇒ 0< < 1 ⇒ b ∈ (-2, 0)
2
Function

80.
81.
82.