Nature of Chapter:
1. We will be studying all the standard functions in this session. It is
extremely important for you to understand significance of this topic.
2. You have to focus on two things. How a given function behaves and
how to solve equations and inequations involving this function.
3. Understanding and practice both will be required.
Weightage of Functions (Last 5 years)
2023
2022
2021
2020
2019
Average
JEE Main
5.6 %
3.6 %
4.4 %
3.8 %
4.2 %
4.32 %
Jee Advanced
6%
0%
0%
5%
0%
2.20 %
Standard Functions
❖ Modulus function
❖ Greatest Integer functions
❖ Fractional Part functions
❖ Exponential Functions
❖ Logarithmic Function
❖ Signum Function
Standard Functions
Critical Topics in the Chapter
●
Modulus Function
●
GIF
Modulus Function
Modulus Function
Basics of the Modulus Function
The general representation of the Modulus Function is f(x) = |x|
it is equal to x if x ≥ 0
|x|
it is equal to −x if x < 0
Ex: Consider the following examples.
(a) |−2| =
(b) |2| =
(c) |x2| =
Modulus Function
Basics of the Modulus Function
The general representation of the Modulus Function is f(x) = |x|
it is equal to x if x ≥ 0
|x|
it is equal to −x if x < 0
Ex: Consider the following examples.
(a) |−2| = 2
(b) |2| = 2
(c) |x2| = x2
Modulus Function
Basics of the Modulus Function
|x - 1|
Modulus Function
Basics of the Modulus Function
it is equal to x - 1 if x ≥ 1
|x - 1|
it is equal to 1 − x if x < -1
Modulus Function
Graph of Modulus Function
Y
O
X
Modulus Function
Graph of Modulus Function
Y
y=−x
y=|x|
y=x
O
X
Q
Solve the following:
(a) |x| = 5
(c) |3x - 5| = -3
(b) |2x - 1| = 2
Q
Solve the following:
(a) |x| = 5
Q
Solve the following:
(a) |x| = 5
Solution:
Q
Solve the following:
(b) |2x - 1| = 2
Q
Solve the following:
(b) |2x - 1| = 2
Solution:
Q
Solve the following:
(c) |3x - 5| = -3
Q
Solve the following:
(c) |3x - 5| = -3
Solution:
The output of the modulus function
cannot be negative.
So, the given equation has no solutions.
Modulus Function
Result
Q
Solve the following:
(a) |x|2 + 9|x| + 20 = 0
(b) x2 - 5|x| + 6 = 0
Q
Solve the following:
(a) |x|2 + 9|x| + 20 = 0
Q
Solve the following:
(a) |x|2 + 9|x| + 20 = 0
Solution:
Q
Solve the following:
(b) x2 - 5|x| + 6 = 0
Q
Solve the following:
(b) x2 - 5|x| + 6 = 0
Solution:
Modulus Function
Inequalities Related to the Modulus Functions
Q
Solve the following inequalities:
(a)
(b)
(c)
(d)
Q
Solve the following inequalities:
(a)
Q
Solve the following inequalities:
(a)
Solution:
−3
3
Q
Solve the following inequalities:
(b)
Q
Solve the following inequalities:
(b)
Solution:
Q
Solve the following inequalities:
(c)
Q
Solve the following inequalities:
(c)
Solution:
−5
5
Q
Solve the following inequalities:
(d)
Q
Solve the following inequalities:
(d)
Solution:
−𝜋
𝜋
Modulus Function
Inequalities Related to the Modulus Functions
Result
1.
2.
Q
Solve the following inequalities:
(a) |2x + 1| ≥ 2
(b) |4 - x| < 6
Q
Solve the following inequalities:
(a) |2x + 1| ≥ 2
Q
Solve the following inequalities:
(a) |2x + 1| ≥ 2
Solution:
Q
Solve the following inequalities:
(b) |4 - x| < 6
Q
Solve the following inequalities:
(b) |4 - x| < 6
Solution:
−2
10
Modulus Function
Observation
Modulus Function
Observation
Q
Solve the following inequalities:
(a) |8x + 4| > 0
(b) |5x + 3| ≤ 0 Q 7
(c) |x + 3| ≥ -2
(d)
Q
Solve the following inequalities:
(a) |8x + 4| > 0
Q7
Q
Solve the following inequalities:
(a) |8x + 4| > 0
Solution:
Q7
Q
Solve the following inequalities:
(b) |5x + 3| ≤ 0
Q7
Q
Solve the following inequalities:
(b) |5x + 3| ≤ 0
Solution:
Q7
Q
Solve the following inequalities:
(c) |x + 3| ≥ -2
Q7
Q
Solve the following inequalities:
(c) |x + 3| ≥ -2
Solution:
Q7
Q
Solve the following inequalities:
(d)
Q7
Q
Solve the following inequalities:
(d)
Solution:
Q7
Q
Solution:
Solve for x:
(a) |x| = x
(b) |x| > x
AIEEE 2011
Q
The domain of the function
is:
A
(0, ∞)
B
(- ∞, 0)
C
(- ∞, ∞) -{0}
D
(- ∞, ∞)
AIEEE 2011
Q
The domain of the function
is:
A
(0, ∞)
B
(- ∞, 0)
C
(- ∞, ∞) -{0}
D
(- ∞, ∞)
AIEEE 2011
Q
The domain of the function
Solution:
is:
Modulus Function
Inequalities Related to the Modulus Functions
Result
Q
Solve the following inequalities:
(a)
(b)
Q
Solve the following inequalities:
(a)
Q
Solve the following inequalities:
(a)
Solution:
Q
Solve the following inequalities:
(b)
Q
Solve the following inequalities:
(b)
Solution:
Modulus Function
Remark
Modulus Function
Observation
Try to observe the range of the following
(a) |a|, where a ∈ [-1, 1]
(b) |x - 1| + 2
Modulus Function
Now, let’s look at some properties of the modulus function.
Modulus Function
Now, let’s look at some properties of the modulus function.
(a) |a| ≥ a
(b) |ab| = |a| × |b|
(c)
Let’s do some examples based on these properties.
Q
Solve the following:
(a)
(b)
Q
Solve the following:
(a)
Q
Solve the following:
(a)
Solution:
Q
Solve the following:
(b)
Q
Solve the following:
(b)
Solution:
(Note that x ≠ 0)
Q
Solve the following:
(b)
Solution:
Q
Find the domain of
Q
Find the domain of
Solution:
or
or
For domain
Modulus Function
Triangle Inequality
Modulus Function
Triangle Inequality
(i) |a + b| ≤ |a| + |b|
(ii) |a - b| ≥ ||a| - |b||
Observation
Condition for equality is ab ≥ 0
Modulus Function
Observation
(a) |2x - 3| = |x - 1| + |x - 2| is possible if and only if
Modulus Function
Observation
(b) |2x - 3| < |x - 1| + |x - 2| is possible if and only if
Modulus Function
Observation
(a) |2x - 3| = |x - 1| + |x - 2| is possible if and only if
(x - 1) (x - 2) ≥ 0, that is x ∈ (-∞, 1] ∪ [2, ∞)
(b) |2x - 3| < |x - 1| + |x - 2| is possible if and only if
(x - 1) (x - 2) < 0, that is x ∈ (1, 2)
Q
Solution:
Modulus Function
Now, let’s do some examples where we need to make cases to handle
the modulus function.
Q
Solve the following: |x - 1| + |x - 3| ≤ 5
Solution:
|x - 1| + |x - 3| ≤ 5
Case 1 → x ∈ (-∞, 1)
(1)
−∞
(2)
1
(3)
3
Case 2 → x ∈ [1, 3)
∞
Case 3 → x ∈ [3, ∞)
Q
If |x - 1| + |x| + |x + 1| ≥ 6 , then x lies in
A
(- ∞, 2]
B
(- ∞, 2] ∪ [2, ∞)
C
R
D
ф
Q
If |x - 1| + |x| + |x + 1| ≥ 6 , then x lies in
A
(- ∞, 2]
B
(- ∞, 2] ∪ [2, ∞)
C
R
D
ф
Solution:
|x - 1 | + | x | + | x + 1 | ≥ 6;
Case 3: - 1 ≤ x < 0
1-x-x+x+1≥6
Case - 1: x ≥ 1
-x ≥ 4
x-1+x+x+1≥6
⇒ x ≤ -4
⇒ 3x ≥ 6
⇒ No solution
⇒x ≥2
Case 4: x ≤ -1
Case 2: 0 ≤ x < 1
1- x - x -1 - x ≥ 6
1-x+x+x+1≥6
⇒ x ≤ -2
x≥4
⇒ x ∊ (-∞, 2] ∪ [2, ∞)
⇒ No solution
Q
Solve the following: |2x - 1| + x ≤ 10
Solution:
(1)
(2)
|2x - 1| + x ≤ 10
−∞
Case 1 →
Case 2 →
In this case
In this case
…(1)
1/2
…(2)
∞
Modulus Function
Observation
Range of :
(a) |x| for x ∈ [-1 , 2] is _________
(b) |x2| for x ∈ [-1 , 2] is _________
(c) | 3 sinx | is __________
Modulus Function
Lets conclude modulus with one interesting observation:
Observation
|x - a| → distance of x from ‘a’ on real line.
Greatest Integer Function
Greatest Integer Function
The representation of the Greatest Integer Function (GIF) is f(x) = [x]
For any real number x :
[x] → is the greatest integer less than or equal to x.
Greatest Integer Function
Let’s see some examples to understand its behaviour.
Greatest Integer Function
Consider the following examples.
(a) [1.5] =
(b) [5.999] =
(c) [6] =
(d) [0.99] =
(e) [𝜋] =
(f) [-0.01] =
(g) [-2.8] =
(h)
Greatest Integer Function
Consider the following examples.
(a) [1.5] = 1
(b) [5.999] = 5
(c) [6] = 6
(d) [0.99] = 0
(e) [𝜋] = 3
(f) [-0.01] = -1
(g) [-2.8] = -3
(h)
Q
Find the value of
Where [.] is GIF.
Q
Find the value of
Where [.] is GIF.
Solution:
Q
Solve the following:
(a) [x] = 2
(b) [x] = -3
(c) [2x - 3] = 4
(d) [x] = 2.9
Q
Solve the following:
(a) [x] = 2
Q
Solve the following:
(a) [x] = 2
Solution:
Q
Solve the following:
(b) [x] = -3
Q
Solve the following:
(b) [x] = -3
Solution:
Q
Solve the following:
(c) [2x - 3] = 4
Q
Solve the following:
(c) [2x - 3] = 4
Solution:
Q
Solve the following:
(d) [x] = 2.9
Q
Solve the following:
(d) [x] = 2.9
Solution:
Greatest Integer Function
Result
Greatest Integer Function
Result
Q
Solve the following equations:
(a) [x] < 2
(b) [x] ≥ 2
(c) [x + 2] > 3
(d)
Q
Solve the following equations:
(a) [x] < 2
Q
Solve the following equations:
(a) [x] < 2
Solution:
[x] < 2
x<2
x ∈ (-∞, 2)
Q
Solve the following equations:
(b) [x] ≥ 2
Q
Solve the following equations:
(b) [x] ≥ 2
Solution:
[x] ≥ 2
x≥2
x ∈ [2, ∞)
Q
Solve the following equations.
(c) [x + 2] > 3
Q
Solve the following equations.
(c) [x + 2] > 3
Solution:
[x + 2] > 3
x+2≥4
x≥2
x ∈ [2, ∞)
Q
Solve the following equations.
(d)
Q
Solve the following equations.
(d)
Solution:
Q
Find domain:
Q
Find domain:
Q
Find domain:
Solution:
Q
Find domain:
Q
Find domain:
Solution:
It is valid for all real numbers except
integers.
As at integers [x] = x
Thus Domain of y is R - Z
JEE Main 11th April, 2023 S2
Q
The domain of the function
is:
(where[x] denotes the greatest integer ≤ x)
A
(-∞, -3] U [6, ∞)
B
(-∞, -2) U (5, ∞)
C
(-∞, -3] U (5, ∞)
D
(-∞, -2) U [6, ∞)
JEE Main 11th April, 2023 S2
Q
The domain of the function
is:
(where[x] denotes the greatest integer ≤ x)
A
(-∞, -3] U [6, ∞)
B
(-∞, -2) U (5, ∞)
C
(-∞, -3] U (5, ∞)
D
(-∞, -2) U [6, ∞)
Solution:
Q
Solve for x : [|x|] = 4
Q
Solve for x : [|x|] = 4
Solution:
[|x|] = 4
4 ≤ |x| < 5
|x| ≥ 4 and |x| < 5
x ≤ -4 or x ≥ 4 and -5 < x < 5
(-5, -4] ∪ [4, 5)
Greatest Integer Function
Result
1) [x + k] = [x] + k, for k ∈ Z
2) [-x] = -1 - [x] for x ∉ Z
NOTE
[kx] ≠ k [x]
Q
If y = 3[x] + 1 = 2[x - 3] + 5, then find the value of [x + y],
where [.] represents greatest integer function.
Q
If y = 3[x] + 1 = 2[x - 3] + 5, then find the value of [x + y],
where [.] represents greatest integer function.
Solution:
JEE Main 4th Sep, 2020
Q
Let [t] denote the greatest integer ≤ t. Then the
equation in x, [x]2 + 2 [x + 2] - 7 = 0 has:
A
exactly four integral
solutions
B
infinitely many solutions
C
no integral solution
D
exactly two solution
JEE Main 4th Sep, 2020
Q
Let [t] denote the greatest integer ≤ t. Then the
equation in x, [x]2 + 2 [x + 2] - 7 = 0 has:
A
exactly four integral
solutions
B
infinitely many solutions
C
no integral solution
D
exactly two solution
Solution:
[x]2 + 2[x + 2] - 7 = 0
[x]2 + 2[x] - 3 = 0
Let [x] = y
y2 + 3y - y - 3 = 0
(y - 1)(y + 3) = 0
[x] = 1 or [x] = - 3
X ∈ [1, 2) or x ∈ [-3, -2)
Greatest Integer Function
Graph of the Greatest Integer Function
Y
X
Greatest Integer Function
Graph of the Greatest Integer Function
Greatest Integer Function
Observation
Range of
(a) [ x2 ] for x ∈ [0, 2] is
(b)
for x ∈ [1, 4) is
Fractional Part Function
Fractional Part Function
The Fractional Part Function is denoted as
f (x) = {x} = x - [x]
NOTE
Every real number can be expressed as sum of two numbers
x = [x] + {x}
Fractional Part Function
Ex. Consider the following examples.
(a) {3.2} =
(b) {2.999} =
(c) {3} =
(d) {3.001} =
(e) {-3.2} =
(f) {-3} =
Fractional Part Function
Ex. Consider the following examples.
(a) {3.2} = 0.2
(b) {2.999} = 0.999
(c) {3} = 0
(d) {3.001} = 0.001
(e) {-3.2} = 0.8
(f) {-3} = 0
Q
If
, then {x} is ________
Q
If
Solution:
, then {x} is ________
Fractional Part Function
Observation
(a) If x belongs to
, then {sin x} = _______
(b) If x belongs to
, then {sin x} = _______
Fractional Part Function
Observation
(a)
(b)
Fractional Part Function
Remark
if x is an integer
otherwise
Q
Solve for x:
(a) [x] = 2{x} + 1
(b) 2x - 1 = 3[x] + 2{x}
Q
Solve for x:
(a) [x] = 2{x} + 1
Solution:
Q
Solve for x:
(b) 2x - 1 = 3[x] + 2{x}
Q
Solve for x:
(b) 2x - 1 = 3[x] + 2{x}
Solution:
2x - 1 = 3[x] + 2{x}
Putting
x = [x] + {x}
2([x] + {x}) - 1 = 3[x] + 2{x}
2[x] + 2{x} - 1 = 3[x] + 2{x}
[x] = -1
Thus, x ∈ [-1, 0)
Q
where [x] is the greatest integer less than or
equal to x, the number of possible values of x is
A
34
B
32
C
33
D
None of these
Q
where [x] is the greatest integer less than or
equal to x, the number of possible values of x is
A
34
B
32
C
33
D
None of these
Solution:
⇒ Total 33 values of x are possible
Fractional Part Function
Graph of the Fractional Part Function
Fractional Part Function
Graph of the Fractional Part Function
Exponential Function
Exponential Function
y = ax, where a > 0, a ≠ 1 ; is called exponential function
Exponential Function
Result
Q
Solve for x:
(a)
(b)
Q
Solve for x:
(a)
Q
Solve for x:
(a)
Solution:
Q
Solve for x:
(b)
Q
Solve for x:
(b)
Solution:
Exponential Function
Graph of Exponential Function
Exponential Function
Graph of Exponential Function
y = ax, a > 0, a ≠ 1
Logarithmic Function
Logarithmic Function
y = loga x, where a > 0, a ≠ 1, x > 0 ; is called logarithmic function
Logarithmic Function
y = loga x, where a > 0, a ≠ 1, x > 0 ; is called logarithmic function
Let’s look at some examples
(a) log28 =
(b) log10100 =
(c)
(d)
Logarithmic Function
Result
(a) loga x2 = logax1 ⇒
(b) loga x2 > logax1
(c) loga x > p
Logarithmic Function
Remark
While solving equations and inequations involving log,
be careful about domain.
Q
Solve for x:
(a) log10(x2 - 5x) < log106
(b)
(c)
, where [.] represents GIF.
Q
Solve for x:
(a) log10(x2 - 5x) < log106
Solution:
… (1)
Also,
x < 0 or x > 5
Now,
… (2)
Q
Solve for x:
(b)
, where [.] represents GIF.
Q
Solve for x:
(b)
, where [.] represents GIF.
Solution:
… (1)
Also,
… (2)
Q
Solve for x:
(c)
Solution:
Also,
… (2)
… (1)
Logarithmic Function
Graph of Logarithmic Function
Logarithmic Function
Graph of Logarithmic Function
y = logax, a > 0, a ≠ 1, x > 0
Y
Y
a>1
O
0<a<1
X
O
X
Q
Find the domain of the following functions:
(a)
(b) y = log10(log10(log10x))
Q
Find the domain of the following functions:
(a)
Solution:
Given function is
For domain,
and
I.e.
and
For
, we get
For
, we have
. . .(2)
. . .(1)
Taking intersection of (1) and (2),
we get
Q
Find the domain of the following functions:
(b) y = log10(log10(log10x))
Q
Find the domain of the following functions:
(b) y = log10(log10(log10x))
Solution:
y = log10(log10(log10 x))
For domain,
x > 0 and log10x > 0 and log10(log10x) > 0
x > 0 and x > 1 and x > 10
∴ x ∈ (10, ∞)
Q
Domain of the following function is :
f(x) = log2(log4(log2(log3(x2 + 4x - 23))))
A
(-8, 4)
B
(-∞, -8) ∪ (4, ∞)
C
(-4, 8)
D
(-∞, -4) ∪ (8, ∞)
Q
Domain of the following function is :
f(x) = log2(log4(log2(log3(x2 + 4x - 23))))
A
(-8, 4)
B
(-∞, -8) ∪ (4, ∞)
C
(-4, 8)
D
(-∞, -4) ∪ (8, ∞)
Q
Domain of the following function is :
f(x) = log2(log4(log2(log3(x2 + 4x - 23))))
Solution:
The given function is defined when
log2(log3(x2 + 4x - 23)) > 1
i.e., when log3(x2 + 4x - 23) > 2
i.e., when x2 + 4x - 23 > 32
i.e., when x2 + 4x - 32 > 0
i.e., when x < -8 or x > 4
Signum Function
Signum Function
For example
(1) sgn (x2 + 10) =
(2) sgn (x2 + x + 1) =
(3) sgn ({x} − 2) =