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Understanding Number Sets
Mathematics categorizes numbers into specific sets
based on their properties. Here are the three most
fundamental types :
(i) Natural Numbers (N)
These are the basic counting numbers we use in
everyday life. They begin at 1 and increase infinitely
by increments of 1.
- Notation : N = {1, 2, 3, 4,.......}
(ii) Whole Numbers (W)
Whole numbers are simply the set of natural numbers
with the addition of zero. Think of them as “Natural
Numbers + 0.”
- Notation : N = {0, 1, 2, 3, 4,....}
- Note : This set is sometimes denoted as N 0
(iii) Integers (Z or I)
Integers encompass the entire range of whole numbers
along with their negative counterparts. They do not
include fractions or decimals.
- Notation : Z = {...,–3, –2, –1, 0, 1, 2, 3,.....}
Important Notes on Integer Subsets
Understanding how we label groups of integers is
crucial for solving inequalities and advanced algebra
problems.
(1) Positive vs Non-Negative
or I
) : These are the standard
Natural Numbers. They start from 1 and go up.
Set : {1, 2, 3,......}
- Non-Negative Integers (W or Z 0
) : These are the
Whole Numbers. This set includes all positive
integers plus zero.
Set : {0, 1, 2, 3,......}
(2) Negative vs Non-Positive
Negative Integers (Z
or I - ) : These are all integers
strictly less than zero.
Set : {....,–3, –2, –1}
- Non-Negative Integers (Z 0
or I 0
) : This set
includes all negative integers plus zero.
Set : {....,–3, –2, –1, 0}
(3) Zero is neither positive nor negative but 0 is a
member of the set of non-negative integers as well
as of the set of non-positive integers.
(iv) Even Integers : Integers which are divisible by 2 are
called even integers. e.g. 0, 2, 4,....
(v) Odd Integers : Integers which are not divisible by
2 are called as odd integers. e.g. 1, 3, 5, 7,...
(vi) Prime Number : Natural number having exactly
two positive divisors i.e. 1 and itself are called prime
numbers.
e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,...
(vii) Composite Number : Let ‘a’ be a natural number,
‘a’ is said to be composite if it has atleast three
distinct positive divisors.
- 1 is the only natural number that is neither a prime
number nor a composite number.
- 2 is the only prime number which is even.
- Numbers which are not prime are composite
numbers (except 1).
- ‘4’ is the smallest composite number.
(viii) Co-prime number : Two natural numbers (not
necessarily prime) are co-prime, if their HCF (Highest
common factor) is 1.
e.g. (1,2), (1, 3), (3, 4), (3, 10), (3, 8), (5, 6), (7, 8),
(15, 16) etc.
These numbers are also called relatively prime
numbers.
- Two prime numbers are always co-prime but
converse need not be true.
- Two consecutive natural numbers are always
co-prime numbers.
- Two consecutive odd natural numbers are always
co-prime numbers.
(ix) Rational number : The numbers which can be
reduced in the form p/q where p,q, Zand q^ ^0 and
HCF (p, q) = 1, are called rational numbers. ‘Q’
represents their set.
- All integers are rational numbers with q = 1
- When numbers are expressed in reduced form of
p
,q 1
q
(^) , the rational numbers are called fractions.
- Rational numbers when represented in decimal
form are either ‘terminating’ or ‘non-terminating
but repeating’.
e.g. 5/4 = 1.25 (terminating)
5/3 = 1.6666 .... or (^) 1.6 or
1.6 (non-terminating but
repeating)
Expressing a Non-terminating but
Recurring Decimal to Rational Form
Let a non-terminating and recurring decimal can be
expressed as
m digits n digits
0 X Y such that X is the non-recurring
part having m digits and Y is the recurring part having n
digits.
Now, (^) R 0.X.....XY
m
10 R X.Y
....(i)
m n 10 R XY.Y
Subtracting (1) from (2), we have
m n m
10 10 R XY X
m n m
XY X
R
- Don’t learn the Formula, just memorize the process
Illustration 1 : Express the rational number as ratio
of two integers whose decimal expansion is
Solution : Let x = 0.12345454545......
Then 10
3 x = 123.454545...
5 x = 12345.4545....
Subtracting we get
5 3
10 10 x 12222 x
Illustration 2 : Express (^) 23.6245 as the ratio of two
integers.
Solution : Let R 23.
100R 2362.
10000R 236245.
9900R 236245 2362
R
Numbers to Remember
Number Square Cube Sq. Root
2 3 4 5 6 7 8 9
10
11
12
13
14
15
16
17
18
19
20
4
9
16
25
36
49
64
81
100
121
144
169
196
225
256
289
324
361
400
8
27
64
125
216
343
512
729
1000
1331
1728
2197
2744
3375
4096
4913
5832
6859
8000
2
3
Rounded
upto two
places
of
decimal
Continued Proportion and Power Means
If a series of ratios are equal, such as
a c e
b d f
(^) then
each individual ratio is also equal to :
1
n n n n
n n n
a c e .....
b d f .....
Example : Given the proportion
a c
b d
(^) , we can
derive several equivalent forms by applying the
rules above :
2 2
2 2
a c a c a c a c
b d b d b d b d
Understanding Intervals
Intervals are subsets of the set of real numbers (R). If we
have two real numbers a and b where a < b, we can define
the four primary types based on whether the end points (a
and b) are included.
- The Four types of Intervals
Type
Open
Closed
Half-
Open
(Left)
Notation
(a,b)
[a,b]
(a,b]
[a,b)
Set-Builder
Notation
{x R |
a < x < b}
{x R |
a x b}
{x R |
a x b}
{x R |
a x b}
Description
Half-
Open
(Right)
Neither a nor
b are included
Both a and b
are included
a is excluded;
b is included
a is included;
b is excluded
The infinite intervals are defined as follows
- (^) a, (^) (^) x : x a
- a,^ ^ ^ x : x^ a
- , b x : x b
- (^) , b (^) (^) x : x b
Intervals are particularly important in solving
inequalities or in finding domains etc.
Number Line Representations
Suppose we have to represent x 2,3 or 2 x 3
2 = Solid Dot 3 = Open Dot
Illustration 4 : Write Correct number line
representation for : (x < 8) or x(–, 8) or
Solution :
Illustration 5 : Write correct number line
representation for : x 1,9 (^) 10,11
Solution :
Shaded portion is the required interval.
Illustration 6 : Correct option representing
3 4 5 9 11
(a) x (^) 3,5 9,11
(b) x^ ^ 3,5^ 9,11 – {4}
(c) x 3, 4 4,5 – 3,11
(d) x (^) 3, 4 (^) 4,5 – 9,11
Solution : x^ ^ 3, 4^ 4,5 – 9,11
or
x (^) 3,5 {4} – 9,11
Hence (B), is the correct option.
Key Properties of Inequalities
Let a, b, c, d and m be real numbers.
(1) Basic Definitions and Transitivity
- Definition : a b means that either a < b or
a = b.
- Transitivity : If a < b and b < c, then it follows
that a < c.
(2) Addition and Subtraction
- Constant Addition : If a < b, then a + c < b
- Adding Inequalities : If a < b and c < d, then
a + c < b + d.
- Subtracting Inequalities : If a < b and c < d,
then a – d < b – c
- Be careful here! You subtract the larger value of the
second inequality from the smaller value of the first.
(3) Multiplication and Division
Multiplication is where things get interesting. The
direction of the inequality depends entirely on the
sign of the multiplier.
- Positive Multiplier ( m > 0) : If a < b, then ma <
mb. The sign stays the same.
- Negative Multiplier (m < 0) : If a < b, then ma >
mb. The inequality sign reverses,
- Example : Since 3 < 5, multiplying by –1 gives
(4) Powers and Exponents
For positive numbers 0 < a < b :
- If the exponent r > 0, then a
r < b
r .
- If the exponent r < 0, then a
r > b
r .
(5) Reciprocal Sums (Special Cases)
These are very common in optimization problems
and calculus :
- Positive Case : For a > 0, then
a 2
a
The equality holds only when a = 1.
- Negative Case : For a < 0, then
a 2
a
The equality holds only when a = –
Wavy Curve Method
The Wavy Curve Method (or Method of Intervals) is a
shortcut for solving inequalities involving polynomials or
rational functions. Here is the refined, simplified breakdown
to help you master it quickly.
(1) Factor & Find Critical Points : Ensure your inequality
is compared to zero (e.g., f(x) > 0). Factor everything
completely.
- Zeros : Values of x that make the numerator zero
(marked as solid dots • if the inequality is or )
- Discontinuities : Values of x that make the
denominator zero (always marked as open
circles o because you can’t divide by zero).
(2) Plot on Number Line : Place all critical points on
a number line in increasing order. This divides the
line into several intervals.
(3) Determine the Staring Sign : Look at the right -
most interval (greater than the largest critical point).
- If all x terms in your factors are positive
(e.g. (x–2)), the curve starts above the number line
(positive) from the far right.
(4) Draw the Wavy Curve (The “Bounce or Cross”
Rule) : Move from right to left through each critical
point. Look at the exponent (power) of the factor
associated with that point.
- Odd Power (Simple Point) : The curve crosses
the number line (sign changes).
- Even Power (Double Point) : The curve bounces
off the number line (sign stays the same).
(5) Select the Solutions :
- If the inequality is > 0 or (^) 0, pick the intervals
where the curve is above the line.
- If the inequality is < 0 or 0, pick the intervals
where the curve is below the line.
REMARKS :
- The point where denominator is zero or function
approaches infinity will never be included in the
answer.
- Points of discontinuity will never be included in
the answer.
- If you are asked to find the intervals where f(x) is
non-negative or non-positive then mark the intervals
closed corresponding to the roots of the numerator
and let it remain open corresponding to the roots
of denominator.
Polynomial Inequalities
Illustration 7 : Solve for x : (x – 1) (x – 2) > 0
Solution :
(1) Find the Critical Points :
Set each factor to zero to find the points where the
expression changes sign :
- x^ ^1 ^0 ^ x^ ^1 • x 2 0 x 2
(2) Test the Intervals : The critical points 1 and 2 divide
the number line into three regions. We check the sign
of (x – 1) (x – 2) in each :
x , (^2)
2
x x 2 0
x^ ^2 x^ ^1 ^ ^0
x , (^1) (^) 2,
Correct Answer : x 3, 2
Rational Inequalities
Illustration 14 :
2x 7
x 5
, Solve for x
Solution :
2x 7
x 5
2x 7 3x 15
x 5
x^8
x 5
x 8
x 5
Correct Answer : x ,5 (^) 8,
Illustration 15 : Values of ‘x’ satisfying (^)
2x
x 1
(a)
x , 1,
(b)
x , 2,
(c)
x , 2,
(d)
x , 2, {1}
Solution : •
2x
1
x 1
2x
x 1
x^ ^1
3x 1
x 1
x , 1,
....(i)
2x
4
x 1
2x
x 1
2x 4x 4
x 1
2x 4
x 1
2x^4
x 1
2x 4
0
x 1
x 1, x 2
x ,1 2, ....(ii)
Finally taking intersection (i) and (ii)
Correct Answer :
x , 2,
Option (C)
Modulus & its Graph
For any real numbers x, modulus or absolute value of x is
denoted by |x| and is defined as
x if x 0
x
x if x 0
^
y
x y
= (^) – x
- 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7
- x x 0
- Geometrically |x| is distance of real number x from
zero along the real number line
x – a
a (^0) x
- More generally |x – a| is distance between ‘x’ and
‘a’ on the number line
2
x x
Modulus Equations
Illustration 16 : Solve for x :
(a) | 2x + 5| = 2 (b) | x – 3| = –
Solution : (a) 2x^ ^5 ^2
2x 5 2 or 2x 5 2
x or x
Correct Answer :
x ,
(b) | x – 3| = – 1
A modulus quantity cannot be negative
So, x is our answer
Illustration 17 : Solve for x; 2x 3 = 3x + 5
Solution :
Case I :
x
2x 3 3x 5
2x 3 3x 5
x 8
(Accepted as
x
Case II :
x
2x 3 3x 5
5x 2
x Accepted as x
Case III :
x
2x 3 3x 5
x = –8 Rejected as x
Correct Answer :
x 8,
Illustration 18 : Solve for
2
x; x + 3 | x | + 2 = 0
Solution :
2 x 3 | x | 2 0
2
x 3 x 2 0
2
x 2 x | x | 2 0
x (^) | x | (^2) 1 | x | (^2) 0
(^) | x | (^2) | x | (^1) 0
Either | x | 2 0 or | x | 1
| x | 2 | x | 1
Modulus cannot be negative
So, x
Illustration 19 : Solve for |x| – 2 |x + 1| + 3 |x + 2| = 0
Solution :
Case I : When x < – 2
x 2 x (^1) 3 x (^2) 0
2x 4 0
x= – 2(Rejected as x < – 2)
Case II : When 2 x 1
x 2 x 1 3 x 2 0
4x 8 0 x 2
Case II : ^ ^ ^
x 4
0 x , 2 4,
x 2
...(ii)
Given inequation becomes
x 2 x 4
x 1 x 2
on solving we get
x 2, 4 / 5 1,
taking intersection with (ii)
x (^) 4, (^) ....(iii)
or
x 2 x 4
x 1 x 2
on solving we get
x 2,0 (^) 1,5 / 2
we get taking intersection with (ii) we get
x
Hence, solution of the original inequation :
x 2, {1}(taking union of (i) & (iii)
Illustration 24 : Find the set of all real values for x
that satisfy the following inequality :
x 1 + x 2 + x 3 6
(a) x 2or x 4 (b) x 0or x 4
(c) (^0) x 4 (d) x 0or x 6
Correct Answer : (b)
Solution :
Case I : (^) x 1 , then (^1) x 2 x 3 x 6
x 0 ....(i)
Hence (^) x 0
Case II : (^1) x 2 , then
x 1 2 x 3 x 6
x 2
But (^1) x 2 No solution ....(ii)
Case III : 2 x 3 , then
x 1 x 2 3 x 6
x 6
But 2 < x < 3 No solution .....(iii)
Case IV : x > 3, then
x 1 x 2 x 3 6 x 4
Hence x 4
From (i), (ii), (iii) and (iv) the given inequality holds
for (^) x 0 or x 4
Logarithm
(1) What is a Logarithm?
At its heart, a logarithm is just a different way of looking
at exponents. Every positive real number N can be
written in exponential form :
a
x = N
In this setup, a is the base and x is the exponent
We can translate this exact same relationship into
logarithmic form :
log a
N = x
The Simple Rule : The logarithm of a number is the
power to which you must raise the base to get that
specific number.
Relationship :
x
a
a N log N x
Essential Ground Rules (Limitations)
For a logarithm log a
N to exist in the real number system,
we have to follow three strict rules :
- The Number Rule : N must be greater than zero
(N > 0)
- The Base Rule : The base a must be positive
(a > 0)
- The Unity Rule : The base a cannot be equal to 1 a^ ^1
Important Things to Note
- Uniqueness : For any specific value of N, its logarithm
log a
N will always yield a single, unique result
- The Zero Barrier : You cannot take the logarithm of
zero; it is undefined.
- The Negative Barrier : In the system of real numbers,
logarithms of negative numbers are not defined.
Illustration 25 :
x
3 9
1
log log x + + 9 = 2x
2
, Solve for x
Solution : a = 3
N = a
L
x x 2x 2 x
9
log x 9 3 3 9
x x
9
log x 9 9
9
log x
1
2
x 9
Illustration 26 : The value of N, satisfying
a b c p
log 1 + log 1 + log 1 + log N = 0
(a) 4 (b) 3
(c) 2 (d) 1
Solution :
^
0
b c p
1 log 1 log 1 log N a 1
^ b c p
log 1 log 1 log N 0
c p
1 log 1 log N 1
c p
log 1 log N 0
p
1 log N 1
p
log N 0
N 1
Correct Answer : (d)
Illustration 27 : If 5
log p = a and 2
log q = a , then
prove that
4 4
2a-
p q
Solution :
a
5
log p a p 5
a
2
log q a q 2
4a 4 4 4a 4a
p q 5 .2^10
2a
2a 1
(2) Fundamental Logarithmic Identities
By applying the basic definition of a logarithm
x
a
a N log N x, we can derive three essential
properties that simplify complex equations :
(a) The Identity of Unity : The logarithm of 1 to any valid
base is always 0.
a
log 1 0
Reasoning : Because any non-zero base raised to the
power of 0 equals 1 (a
0 = 1)
(b) The Base Match Identity : The logarithm of a number
where the base is identical to the number itself is always 1
log N
N = 1
Reasoning : Because any number raised to the power
of 1 is itself (N
1 = N).
(c) The Reciprocal Identity : The logarithm of a number
where the base is the reciprocal of that number is
always –1.
1
N
log N 1 or^ N
log 1
N
Reasoning : Because raising a number to the power
of –1 flips it into its reciprocal (N
Pro-Tip : The power Cancellation Rule
There is a powerful “self-cancelling” property to keep in
mind :
a
log N
N a
When a base is raised to a logarithm with that same base,
they essentially “undo” each other, leaving you with just
the number N.
2
log 7
2 7
(3) The Principal Properties of Logarithms
For these rules to apply, ensure that m and n are positive
numbers, the base a is positive and not equal to
1 a 0,a 1 , and x is any real number..
(a) The product Rule : The logarithm of a product is equal
to the sum of the logarithms of the individual factors.
a a a
log mn log m log n
(b) The Quotient Rule : The logarithm of a fraction is
equal to the logarithm of the numerator minus the
logarithm of the denominator.
a a a
m
log log m log n
n
(c) The Power Rule : The logarithm of number raised
to an exponent is equal to the exponent multiplied by
the logarithm of that number.
log a
(m
x ) = x log a
m
(4) The Base Power Property
If the base of a logarithm is itself raised to an exponent,
that exponent comes out as a reciprocal (it goes into the
denominator) when moved in front of the log.
The Identity
k (^) a a
log m log m
k
Combining with the Number Power Rule
If both the base and the argument have exponents, you
can move them both to the front as a single fraction :
k
x
a a
x
log m log m
k
- Example : Suppose you need to solve log 8
(1) Recognize that the base 8 can be written as 2
3
(2) Apply the rule : (^3) 2 2
log 2 log 2
(3) Since log 2
2 = 1, the answer is simply 1/3.
Illustration 28 : If x = log 3
4 and y = log 5
3, then find
log 3
10 in terms of x and y.
Solution :
2
3
x log 2
3
x 2 log | 2 |
3
x 2 log 2
written as log x or log 10
x.
- Natural (Napierian) Logarithm : Uses base e
(where e (^) 2.718) and is denoted as ln x or log e
x.
Converting Between Bases
e 10
log a 2.303 log a
10 e
log a 0.434 log a
Illustration 32 : If a, b, c, are distinct positive real
numbers different from 1 such that
^ b c a a c b
log a .log a log a + log b.log b log b +
a b c
log c.log c - log c = 0 , then abc is equal to
(a) 0 (b) e
(c) 1 (d) None of these
Solution : b c
log a log a 1
a c a b
log b.log b 1 log clog c 1 0
log a log a log b log b
log b log c log a log c
log c log c
log a log b
3 3 3
log a log b log c
3log a log b log c
(^) log a log b log c (^) 0
[
3 3 3
If a b c abc 0, then a b c
0if a b c]
log abc log1 abc 1
Illustration 33 : Evaluate :
5 8 7
1/ log 3 log 36 4/ log 9
81 + 27 + 3
Solution :
log 5 3 3log 36 9 4log 7 9
3/2 (^2) 4log 5 3 log 3 36 log 7 3
3 3 3
Graph of Logarithmic Functions :
Graph of y = log a
x :
When a > 1
x
y
O (1,0)
O
y
x
When 0 < a < 1
Points to remember :
a a
x y if a 1
log x log y
x y if 0 a 1
^
(6) Characteristic and Mantissa
For any given number N, a logarithm can be expressed
as :
a
log N Integer Fraction
The integer part is called the Characteristic and the
fractional part is called the mantissa.
- The mantissa part of the logarithm of a number is
always positive (^) 0 m (^1)
- If the characteristic of log 10
N is n, then the
number of digits in N is (n + 1)
- If the characteristic of log 10
N is (–n), then there
exist (n – 1) zeros after the decimal in N.
Illustration 34 : Let x = (0.15)
20
. Find the
characteristic and mantissa in the logarithm of x, to
the base 10. Assume log 10
2 = 0.301 and log 10
Solution :
log x log 0.15 20 log
20 log15 (^2) 20 log 3 log 5 (^2)
20 log 3 1 log 2 (^2)
log 5 log
(^20) 1 log 3 log 2
(^20) 1 0.477 0.301
Hence characteristic = – 17
Mantissa = 0.
Illustration 35 : Determine the total number of digits
in the integer expansion of 2
75 .
Solution : Computing log 2
75 , we have log 2
75
- log 2 75. (0.3010) = 22.5750.
Consequently, the characteristic of this
common logarithm is equal to 22.
Therefore, 2
75 = a. 10
22 , where (^1) a 10 , a
is an integer, and hence the number 2
75 has
23 digits.
Illustration 36 : Number of cyphers after decimal
before a significant figure starts in
5
is equal to
[Use : log
10
2 = 0.3010]
Solution : Let
100
N
10 10
log N 100 log
10 10
100 log log 4
10
100 1 3log 2
100 1 (^) 3 0.3010
100 1 0.9030 100 0.
Number of zeroes = 9
Logarithmic Inequalities
Understanding logarithmic inequalities requires paying close
attention to the base (a). The direction of the inequality
depends entirely on whether the base in greater than 1 or
between 0 and 1.
Points to Remember
(1) Comparison of Logs :
a a
0 x y if a 1
log x log y
x y 0 if 0 a 1
^ ^ ^
(2) When Base a > 1 (Inequality sign remains the
same) :
(a)
p
a
log x p 0 x a
(b)
p
a
log x p x a
(3) When Base 0 < a < 1 (Inequality sign flips) :
(a)
p
a
log x p x a
(b)
p
a
log x p 0 x a
Illustration 37 : Solve for x :
(a) (^)
2
log x 5x 6 1
Solution : (^)
2
log x 5x 6 1
1 2 0 x 5x 6 0.
2 0 x 5x 6 2
2
2
x 5x 6 0
x 1, 2 3, 4
x 5x 6 2
Hence, solution set of original inequation :
x 1, 2 3,4
(b) (^)
2
1/3 4
log log x 5 0
Solution : (^)
2
1/3 4
log log x 5 0
2
4
0 log x 5 1
2 2
4
2 2
4
0 log x 5 x 5 1
log x 5 1 0 x 5 4
2
1 x 5 4
2
6 x 9 x 3, 6 6,
Hence, solution set of original inequation :
x 3, 6 6,
Illustration 38 : Determine the set of all real values
for x that satisfy the following logarithmic inequality
: (^) (^)
2
e
log x 2x 2 0
Solution : The values of x satisfying the inequality
2
e
log x 2x 2 0 must be such that
2 0 x 2x 2 1
we have,
2 2 x 2x 2 0 x 1 3
2
x 1 3 x 1 3
x 1 3 or^ x^ ^1 ^3
x 1 3 or x 1 3 ....(i)
Again
2 2
x 2x 2 1 x 2x 3
(^2 )
x 1 4 | x 1| 4
| x 1| 2 2 x 1 2
(1) The Golden Rule : Don't Cross-Multiply Variables
In a normal equation like 1/x = 1, you can move x to the other side to get 1 = x, but in an inequality
(like, or or < or >), you cannot do this.
(2) Mis-applying the Power Rule : The power rule states that log b
(m
n ) = n. log b
(m). A common mistake is
moving the exponent when it applies to the entire log function rather than just the argument.
The Wrong : (log
x
3 = 3 log x
Correct : Only log
x
3 ) = 3 log x
Note : (log
x
3 is simply the value of the log cubed; it cannot be simplified using the power rule.
(3) Inventing "Distributive" Laws : Students often try to distribute the log across addition or subtraction
within the argument. This is perhaps the most frequent error.
Wrong : log(x + y) = log x + log y
Wrong : log(x – y) = log x – log y
Correct : There is no simplified identity for log (m n). The product and quotient rules only work
in reverse : • log x + log y = log(xy) • log x – log y = log(x/y)
(4) Confusing the Quotient Rule with Division : Students often confuse the log of a fraction with the
fraction of two logs.
Wrong :
log x
log x y
log y
Correct :
x
log log x log y
y
Correction :
log x
log y
is actually the Change of Base formula, which equals log y
x.
(5) Including Roots of the Denominator : This is a classic ''domain'' trap in rational inequalities
like
P x
Q x
The Mistake : Using a closed circle (bracket) for a value that makes the denominator zero just
because the inequality is or
The Fix : Even if the inequality is inclusive (^) (^) , the roots of the denominator must always be
excluded (open circle/parentheses) because division by zero is undefined.
Fractional Part Function
y; = f(x) = {x} = x – [x]
Domain : (^) x R; Range : (^) 0,1
Eg. 2.3 = 2 + 0.3 fractional part
Integer part
–1 –2 0 1 2 3 x
y
Remarks
(i) Fractional part of any integer is zero
(ii) (^) x n {x}, n I
(iii)
0 ; x I
{x} (–x}
1 ; otherwise
^
"1729 is a very interesting number; it is the smallest number expressible
as the sum of two cubes in two different ways.".......S.Ramanujan
Part - 1 (Objective Type Questions)
Only One Option Correct Type
A1. Complete solution set of inequality
x x
x x
is
(a)
,0 (b)
,0 2,3 (c)
2,3 (^) (d)
A2. The solution of the inequality
x x
x
is
(a)
1,3 5, (b)
1,3 5, (c)
,1 5, (d) none of these
A3. Sum of solutions of the equation is
3 2
x | 4 x | 3 x 0
(a) 4 (b) 3 (c) 0 (d) 1
A4. The solution set of the inequality
x x
x
(^) is
(a) (0, 1) (b) [0, 2 ] (c)
,0 1, (d) none of these
A5. Number of integral values of x satisfying the inequality
2 6 7
x x
x x
is
(a) 5 (b) 6 (c) 7 (d) 8
A6. Which of the following equations has maximum number of real roots?
(a)
2
x x 2 0 (b)
2
x 2 x 3 0 (c)
2
x 3 x 2 0 (d)
2
x 3 x 2 0
A7. The complete solution set of the equation
2 2
x 5 x 6 x 12 x 27 17 x 21 is
(a)
x 9,3 (^) (b)
x 3,2 2,3 (c)
x 9, 3 2,
(d)
x 2,
A8. Solutions of 4 x^ ^3 ^3 x^4 ^12 are
(a)
x ,
(b)
x (c)
x (d)
x
A9. If (^) 4 x 2 , then (^) x 2 3 lies in the interval
(a) (1, 3] (b) [1, 3] (c) [0, 3] (d)
A10. The number of real solutions of 2 2
x
x x
is
(a) 0 (b) 1 (c) 2 (d) infinite
A4. Find all possible values of the following expressions :
(a) (^2)
x 2
(b) (^2)
x 2 x 3
(c) (^2)
x x 1
A5. Determine the resulting interval for
2
x given the following intervals for^ x^.
(a) If (^) x is in the range [–5, –1] (b) If (^) x is in the range (3, 6)
(c) If x is in the range (–2, 3] (d) If x is in the range (–3, )
(e) If x is in the range (– , 4)
Section - B (Topic Name : Based on Wavy Curve Method )
B1. Determine the set of all real numbers x that satisfy the following quadratic inequality. Provide your final answer in
interval notation
2
x x – 2 0
B2. Solve:
x 1 x 2 1 2 x 0
B3. Solve :
x x
x x
B4. Find the solution set for the following rational expression. State your final answer in interval notation.
2 3
x
x x x x
B5. Solve:
x x x
x
B6. Solve:
x x x
x x
B7. Solve:
4
2
x
x
B8. Solve:
2
2
x x
x x
B9. Solve : ^
x x
x x
B10. Number of integers satisfying the inequality,
4 2
x 29 x 100 0 is
Section - C (Topic Name : Based on Absolute Value Function)
C1. Find all real values of x that satisfy the following equations:
(i) Find x such that the absolute value of x equals 5.
(ii) Solve the quadratic equation involving the absolute value of x:
2
x x 2 0
C2. Identify which of the following statements is guaranteed to be true for all real numbers a and b where a b:
(a) If a^ ^ b, it follows that
2 2 a b
(b) If a^ ^ b, it follows that
a b
(c) If a b, it follows that a b
C3. Find all values of
f x for which
2
f x x x
C4. Find the set of all real values of x for which the following equations hold true.
(Hint: Consider the definition a aif a^ ^0 and a a if a^ ^0 ).
(a) x 2 x 2 (b)
x 3 x 3
(c)
2 2
x x x x (d)
2 2
x x 2 x x 2
C5. Solve:
x
x
C6. Find all real values of x that satisfy the following absolute value equation.
x 3 x 2 1
Section - D (Topic Name : Based on Modulus Inequalities)
D1. Solve the x and state the solution set in interval notation.
2
x 4 x 3 0
D2. Solve: 3 x 2 4
D3. Find all numbers x whose distance from 3 is at least 2 units.
D4. Solve for x: 1 x 2 3
D5. Find the interval of x satisfying the following inequality:
x
D6. Solve for (^) x : (^) x 3 2
D7. Find the interval of x satisfying the following inequality: x^ ^ x^2 2.
D8. Solve for x:
2 2
x x 4 x 4 x.
Section - E (Topic Name : Logarithms)
E1. Find the value of each of the following:
(a) 9
log 81 (b) 2
log (^4) (c) 2 3
log 1728
(d)
tan 40 ^
log cot
(e)
log 0.4 (f)^
2 3
log 5 2 6
E2. Find the value of
9
2log 3
3
E3. Find the value of x in each of the following cases:
(a) (^81)
log
x (^) (b) 2
log x (^4) (c) 2
log 4 x 5
