IIT JEE
BEST STUDY
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NOTES
SAVE YOUR TIME!
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NO NEED OF
TAKING FULL NOTES
NOW!
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JUST PRINTOUT THESE
AND USE THEM IN
YOUR LECTURES :-)
INDIA’S FIRST NOTES WITH MOST
IMPORTANT SUBTOPICS MARKED
ACCORDING TO JEE MAINS
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SAMPLE PROBLEMS FOR
UNDERSTANDING WHICH ARE
HIGHLY EXPECTED IN JEE MAINS
EXAM
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1
BASIC MATHEMATICS
1.1. NUMBER SYSTEM
Some important sets of numbers are given below:
1.1. Natural numbers: The numbers 1, 2, 3, ...., are called natural or counting numbers. The set of
natural numbers is denoted by N. i.e. N = {1, 2, 3, 4, ....}.
1.2. Integers: The numbers 0, ± 1, ± 2, ..., are called integers. The set of integers is denoted by Z or I.
i.e. Z = {..., –3, –2, –1, 0, 1, 2, 3, ...}.
The set of positive integers is denoted by Z+ = {1, 2, 3, ...} = N and Z– = {–1, –2, –3, ...}.
It should be noted that 0 is neither positive nor a negative integer. An integer x is positive if x > 0
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and negative if x < 0.
The set {0, 1, 2, ...} is called the set of whole numbers and is denoted by W.
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1.3. Rational numbers: A number of the form p/q where p and q are integers and q  0 is called a
rational number. A set of rational numbers is denoted by Q.
i.e. Q = {p/q : p, q are integers and q  0}
For example,
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Rational numbers can also be represented as terminating or non-terminating recurring decimals.
x = 1/3, 5/2, 3, –5, ..., 2.135, 5.123, where 5.123 = 5.1232323 ... are rationals.
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1.4. Irrational numbers: The real numbers which are not rational are called irrational numbers.
e.g. 2, 5, 31/5, , e, log2 10 log10 5 etc.
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1.5. Real numbers: A number which is either rational or irrational is called a real number. Thus a set
obtained by taking all rational and irrational numbers is called a set of real numbers and it is
denoted by R. Real numbers can also be treated as points on a line (also called the real line).
1.6. Complex numbers: The numbers of the form a + ib where i = (–1) and a, b are real numbers, are
known to be complex numbers.
Sample Problem-1:
Let a, b  odd integers a > b then prove that a2 – b2 is divisible by 8. Can we say a2 – b2 is +ve?
Solution:
Let a = 2m + 1 and b = 2n + 1 where m > n, then a2 – b2 = 4 (m + n + 1) (m – n).
Since (m + n + 1) (m – n) is always even so a2 – b2 is divisible by 8.
(a2 – b2) may also be negative.
1.1
Basic Mathematics
Sample Problem-2:
In (9990001408)2009, find digit at the unit place.
Solution:
Since 81 = 8
83 = 512
82 = 64
84 = 4096
85 = 32768
Thus, the digit at unit place is 8, 4, 2, 6, 8, 4, 2, 6, ....
so in (9990001408)2009 the digit a unit place is same as in (8)2009 which is 8.
Sample Problem-3:
Prove that product of three consecutive numbers is divisible by 6.
Solution:
Let n, n + 1 and n + 2 be consecutive numbers, then
n(n + 1) (n + 2) = n3 + 3n2 + n
In 3 consecutive numbers one is always 3 type so
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putting n = 3 we get
n (n + 1) (n + 2) = 27 2 ( + 1) + 6 
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Since  ( + 1) is divisible by 2 so above expression is divisible by 6.
1.2. SOME PROPERTIES OF INTEGERS
Even integers: The integers which are divisible by 2 are called even integers. The set of even
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integers is {..., –6, –4, –2, 0, 2, 4, 6, ...}.
Note that 0 (zero) is an even integer. Usually, 2n, where n is any integer represents an even integer.
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Odd integers: The integers which are not divisible by 2 are called odd integers.
{..., –5, –3, –1, 1, 3, 5, ...} is the set of all odd integers. Usually an odd integer is taken as 2n + 1,
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where n is any integer.
Also note that
(i)
(even integer) +, –, × (even integer) always give an even integer.
(ii) (odd integer) +, – (odd integer) give an even integer while (odd integer) × (odd integer) results
in odd integer.
(iii) (even integer) × any natural number is an even integer.
(iv) 0/0, 0° are not integers. Actually they do not exist.
Prime and composite integers: A positive integer which has no divisors other than 1 and itself is
called a prime number and others are called composite integers. By convention, unity is neither a
prime nor a composite number. 2, 3, 5, 7, 11, 13, 17, ... are examples of some primes in ascending
order. The number 2 is the least prime and the only even prime. All other primes are odd. Any
composite number can be represented as a product of primes (called prime factorization).
- 1.2-
Basic Mathematics
)
Examples :
(i)
520 = 23 . 5 . 13
(ii) 2500 = 2254
(iii) 3610 = 2 . 5 . 192
Co-prime numbers: The numbers p and q are called co-prime if h.c.f. or g.c.d. of p and q is 1. For
Ex. 2 and 9 are co-prime as h.c.f. of 2 and 9 is 1.
Ex : 5 and 18 are co-prime, as h.c.f. of 5 and 18 is 1.
1.3. INTERVAL NOTATIONS
The various intervals are represented as follows:
Closed interval [a, b] = {x : a  x  b}.
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Open interval ]a, b[ or (a, b) = {x : a < x < b}, semi closed interval [a, b[ or [a, b) = {x, a  x < b},
which is also called closed on left.
The set of all real x such that x  a can be written as [a, [, while the set of x values for which x < b
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can be represented as ]–, b[. For example,
{x : x  7} =[7, [, {x : x > – 2} = ]–2, [,
{x : x  5} = ], –5[, etc.
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1.3.1. THE SYMBOLS <, >, , 
Let a, b be real numbers. Then a < b iff (a – b) is negative and a > b iff a – b is positive.
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The symbol a  b (read as a is less than or equal to b) is true when a is less than b or a = b. For Ex.
2  3 and 2  2 both are true.
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Similarly a  b read as a is greater than or equal to b means a is not less than b.
Note that if a > b then ka > kb iff k is positive. If k is negative than ka < kb. Also a/b > c  a > bc.
It depends on b.


a/b > c 
 a > bc if b > 0
a < bc if b < 0.
1.4. ABSOLUTE VALUE OF REAL NUMBER
The absolute value of a real number x also called the modulus of x is denoted by |x| and is defined as
follows.
 x if
 x if
|x| = 
Thus |x| =
x0
x0
( x2 ) .
1.3
Basic Mathematics
For example.
(i)
|5| = 5, |0| = 0, |–5| = –(–5) = 5
(ii) |x – 5| = x – 5 if x  5 and |x – 5| = – (x – 5) = 5 – x, if x < 5.
1.5. PROPERTIES OF ABSOLUTE VALUE
The following properties are very useful in the solution of many problems. For any real numbers a
and b, the following properties hold
(i)
|a|  0
(iv) |a – b|  ||a| – |b||
(ii) |a| = |–a|
(iii) |a + b|  |a| + |b|
(v) |ab| = |a||b|
(vi)
a |a|

, b  0.
b |b|
(vii)
|a2 | = a2
Important note: We often find students writing.
 f ( x), if
 f ( x), if
|f (x)| = 
x0
x0
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which is quite incorrect. |f (x)| is defined as
f ( x)  0
f ( x)  0
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 f ( x), if
 f ( x), if
|f (x)| = 
For example, consider the function
f (x) = x – |x – x2|, –1  x  1.
i.e. if x (1 – x)  0 i.e. if 0  x  1.
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Now |x – x2| = x – x2, if x – x2  0
and |x – x2| = – (x – x2) = x2 – x if x – x2 < 0
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i.e. if x (1 – x) < 0 i.e. if x < 0 or x > 1.
i.e. if – 1  x < 0, taking the values of x from the given domain –1  x  1.

x  ( x  x2 )  x2 ,
if
2
2
 x  ( x  x)  2 x  x , if
f (x) = 
0  x 1
1  x  0
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Hence, we have
Simplify : f (x) = |x –1| + |x + 2|.
x – 1  0 when x  1 and x –1 < 0 when x < 1. Thus change point for |x –1| is x = 1. Similarly
change point for |x + 2| is –2. Now, –2 and 1 divide the number line in three parts:
x < –2, –2  x < 1, x  1.
when x < – 2, x + 2 < 0, x – 1 < 0 so
|x + 2| = – (x + 2) and |x – 1| = – (x –1).


f (x) = – x – 2 – x + 1 = –2x –1 for x < –2.
when –2  x < 1, x + 2  0, x – 1 < 0 so
- 1.4-
Basic Mathematics
)
|x + 2| = x + 2 and |x – 1| = – (x –1) and so f (x) = x + 2 – x + 1 = 3.
Finally for x  1, x + 2 > 0, x – 1  0 so f (x) = x + 2 + x – 1 = 2x + 1.
x  2
2 x  1 :

Hence f (x) = |x – 1| + |x + 2| =  3
: 2  x  1
 2x 1 :
x 1

OBJECTIVE QUESTIONS-1.1 TO 1.5:
1.
2.
(a)
–3 < x < –1
If
| x + 3| + x
> 1 then x lies in
x+ 2
(a)
(–5, )
(b) (–5, –1)  (–1, )
(c)
(–5, –2)  (–1, )
(d) none of these
Solution of x 
(2 –
(c)
R – (–2 –
3,2+
3 )  (–2 –
3 , –2 +
(c) (5, )
(d)
1

 ,  (5, )
3

3)
3 , –2 +
3 ) (b) R –(2 –
3)
(d) none of these
(d) none of these
(c) (2, )
(d) (–, )
(b) (–, –2)  [4, ) (c) (–, 0]  [4, )

1
 in which x lies, is
1 | x | 3
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(–, –2)
3,2+
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[0, 4]
If 1 
(b) (–2, 2)
| x|
1
 , then the largest interval in which x lies, is
1 | x | 3
–2  x  –1
(b) –1 < x  1
(c) –2  x  2
(d) 1  x  3
(c) (–, 1/2)
(d) (1/2, 0)  (0, )
The solution set of the inequality | x | < | x – 1| is
(a)
9.
(d) x < –3 or x > 3.5
1
 4 is
x
The solution set of the in equation
(a)
8.
1 
 ,5 
3 
Solution of |x –1| + |x –2| + |x –3| 6 is
(a)
7.
(b)
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(a)
(a)
6.
1

 , 
3

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5.
(c) –1 < x < 1
Solution of |2x –3| < |x + 2| is
(a)
4.
(b) –3 < x < – 1.5
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3.
If x  R, then | x + 1| < 2 and | x – 2| > 3.5 are together equivalent to
(0, 1)
The equation |x| +
(a)
(1, )  {0}
(b) (1, )
x
x2
=
will be always true if x belongs to
x - 1 | x - 1|
(b) (1, )
(c) (–1, 1)
1.5
(d) (– , )
Basic Mathematics
HINTS/SOLUTIONS OBJECTIVE QUESTIONS-1.1 TO 1.5:
1.
(b): |x + 1| < 2
 –2 < x + 1 < 2  –3 < x < – 1
Again, |x –2| > 3.5, x  (–, –1.5)  (5.5, )
The intersection of intervals implies that x  (–3, –1.5)
2.
(b):

| x  3|  x  x  2
0
x2
| x  3 | 2
 0 , its roots are –2, –1 and–5
x2
 x  (–5, –2)  (–1, )
3.
3
 3
(b): Make 3 cases, x  –2, x   2,  and x 
2
 2
4.
(a): Case 1: x +
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Case 2: x + >
1
– 4  x  (2  3,  2  3)  (0, )
x
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5.
1
< 4  x  (–, 0)  (2  3,2  3)
x
(c): Make cases x  1, x  [1, 2]
x  [2, 3] and x  3 and solve.
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Draw the graph to get the answer.
(b): Since 1 + |x| > 0, so cross multiply and solve.
7.
(c): Simplify inside mod and solve.
8.
(c): Make three cases x  0, x  (0, 1) and x  1, then solve.
9.
(a): Use |x| + |y| = |x + y|
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6.
if and only if x and y are of same sign.
1.6. THE SOLUTION OF (x – a) (x – b) > 0
Let (x – a) (x –b) > 0, a < b, a, b  R.
Then there arise two cases:
Case I: x – a > 0 and x – b > 0, so x > a and x > b i.e. x > b.
Case II: x – a < 0 and x – b < 0, so x < a and x < b i.e. x < a.
combining these facts we get x < a or x > b.
If we consider weak inequality (x – a) (x – b)  0 then in the same fashion we have x  a or x  b.
Similarly, the solution of (x – a) (x – b) < 0, a < b, will be found to be a < x < b.
And for (x – a) (x – b)  0, we have a  x  b.
- 1.6-
Basic Mathematics
)
Thus for a < b
(x – a) (x – b) < 0  a < x < b, (x – a) (x – b) > 0  x < a or x > b.
Aliter: Consider f (x) = (x – a) (x – b), a < b.
Then put the change points a and b on the real line.
a and b divide the real line in three parts x < a, a < x < b and x > b.
Start with +ve sign (+) for extreme right and then alternately –, +, –, ...,
Thus
(x – a) (x – b) > 0  x < a or x > b and (x – a) (x – b) < 0  a < x < b.
Also (x – a) (x – b)  0  x  a or x  b, and (x – a) (x – b)  0  a  x  b.
Similarly for positive a
x2 < a2  – a < x < a and x2 > a2  x < – a or x > a.
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Also |x| < a  – a < x < a and |x| > a  x < – a or x > a.
Sample Problem-4:
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Solve the in equation
( x - 1)( x - 2)
< 0.
( x + 3)( x - 4)
( x - 1)( x - 2)
. Then change points are –3, 1, 2 and 4. Then
( x + 3)( x - 4)
Using wavy curve
f (x) =
( x - 1)( x - 2)
< 0  –3 < x < 1 or 2 < x < 4.
( x + 3)( x - 4)
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Solution: Let f (x) =
Also f (x) > 0  x < –3 or 1 < x < 2 or x > 4
1.7. EQUATION AND IDENTITY
It two functions of unknown x are connected by equality sign, e.g. f (x) = g (x), then it is called an
equation if it holds for some values of x in the domain of definition and an identity if it holds for
all values of x in the domain of definition. For Ex. x2 + x – 2 = 0 is an equation as it is satisfied only
by x = –2 and 1, on the other hand (x + 1)2 = x2 + 2x + 1 is an identity as
L.H.S. = R.H.S. for all values of x.
If there exists no x which satisfy an equation, equation is called inconsistent. An example,
x + 2 = x + 3 is an inconsistent equation.
1.7
Basic Mathematics
OBJECTIVE QUESTIONS-1.6 TO 1.7:
2
1.
2.
(2 x - 1)( x - 1) ( x - 2)
( x - 4)4
If
3
> 0 then x lies in
(a)
(–, 4)
(b) (–, ½)  (2, ) – {4}
(c)
æ1 ÷
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çç , 2÷
çè 2 ÷
ø
(d) none of these
( x - 2)(1- x)( x - 3)3 ( x - 4) 2
If
 0 then x lies in
( x + 1)
(a)
(–1, 1]  [3, )
(c) (–, –1)  [2, 3]
(b) (–1, 3]
(d) none of these
2
3.
The interval in which
(a)
(c) (–, 3)
(d) (–, 1) (2, 3)
(b) [2, 4]
(c) (2, 4)
(d) (–, 1)  (2, )

(– , –3)
1
 2 is
( x  3)

(b) R
(a)
[3, )
If
x3 ( x  1) 23 ( x  2)333
0
888( x  3)( x  33)2 ( x  333) 4
(a)
(c)
(–33, –3)  (–2, –1)  (0, )
(–2, 0)
(c) (–1, ) – {1}
(d) (–1, ) – {1, –8}
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| x  1| ( x  3) 2
 0 , then x
( x  2)6
(b) [–3.5, –3)
(d) (– , –3.5)  (–3, )
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If
(–1, )
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10.
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9.
(1, 4)
( x 2  3)( x  1)4
 0 then x
If
( x  1)
(a)
8.
(b) (2, 3)
The solution set of the equation
(a)
(c)
7.
(–, 1)
x2
 4 is
x 1
Solution of (5x –1) < (x + 1)2 < (7x –3) is
(a)
6.
(d) none of these
The complete set of values of x which satisfy the in equations: 5x + 2 < 3x + 8 and
(a)
5.
(b) (–, 1) (2, ) (c) (–1, )
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4.
(–, 2)
x + 2x + 3
is positive is
x 2 - 3x + 2
(b) (–1, 3)
(c) [–1, 3]
(d) none of these
(b) (–333, –33)  (–3, –2)  (–1, 0)
(d) (–3, –2)  (–1, 0)
(2 x  1)( x  1) 2 ( x  2)3
 0 then x lies in
If
( x  4)4
(a)
(–, 4)
(b) (–, ½)  (2, ) – {4}
(c)
1 
 , 2
2 
(d) none of these
- 1.8-
Basic Mathematics
)
1.
HINTS/SOLUTIONS OBJECTIVE QUESTIONS-1.6 TO 1.7:
(b): The wavy curve is
2.
(d): The wavy curve is
–
–1
3.
–
+
2
1
(b): The wavy curve is
3
–
+
6.
( x  2)
x2
0
4 
( x  1)
x 1
(c): Solve three inequalities.
1
(b):
20
x3
+
–1
–3
+
1
(d): The wavy curve is
+
+
+
+
–2
–1
(d): The wavy curve is
3
+
+
–
+
–33
(b): Make the wavy curve
–3
–
+
–
+
2
1
–1
+
0
+
4
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1/2
–2
–
+
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–333
10.
+
–7/2
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9.
–
+
ES
2x  7

0
x3
(a): The wavy curve is
–
8.
2
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7.
4
(d): 5x + 2 < 3x + 8  x < 3
and
5.
–
+
1
4.
–
+
1.8. SETS
1.8.1. A Set is a well defined collection of distinct objects.
Each object is called an element of the set.
If ‘a’ is an element of set ‘A’ then we write a  A (a belongs to A)
Sample Problem-5:
Write the set {x : x is a positive integer and x2 < 40} in the roster form.
Solution: The required numbers are 1, 2, 3, 4, 5, 6. So, the given set in the roster form is {1, 2, 3, 4, 5, 6}.
Sample Problem-6:
1 2 3 4 5 6 
2 3 4 5 6 7 
Write the set  , , , , ,  in the set-builder form.
1.9
Basic Mathematics
Solution: We see that each member in the given set has the numerator one less than the denominator.
Also, the numerator begin from 1 and do not exceed 6. Hence, in the set-builder form the


given set is x : x 
n

, where n is a natural number and 1  n  6
n 1

1.8.2. Representation of a set
(a) Roster Form: In this form elements are written in curly brackets and repeated elements are
dropped out. e.g., {a, b, c}.
(b) Set builder Form: In this form an element ‘x’ is written with a specific property. e.g., {x : x is
an even natural number < 20}.
1.8.3. Null Set or Empty Set : A set containing no elements is called an empty set. It is denoted by ‘’ or
{}.
1.8.4. Singleton Set: Set containing one element is called a singleton set e.g., {1}, {a}.
1.8.5. Finite Set : A set is called a finite set if it contains no element or its elements can be counted till a
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certain natural number n.
e.g., {2, 4, 3, 5}; {a, b, c, d, e, f}.
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1.8.6. Cardinal Number of a Finite Set: The number of elements belonging to a finite set A is called
cardinal number of that set and is denoted by n(A). e.g., cardinal number for set {2, 3, 4, 5} is 4.
1.8.7. Infinite Set: A set, whose elements can not be counted upto any natural number n, is called an
infinite set e.g.,{2, 4, 6, 8, ........}.
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1.8.8. Equivalent Sets: Two finite sets A and B are said to be equivalent if they have same number of
elements i.e., their cardinal numbers are same, e.g., {3, 5, 7} and {a, b, c} are equivalent sets.
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1.8.9. Equal Sets: Two sets A and B are said to be equal if all the elements of set A are in set B and all the
elements of set B are in set A. We can also define sets A and B to be equal if
(i)
they are equivalent sets
(ii) have the same elements.
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e.g., {1, 2, 3} and {3, 1, 2} are equal sets.
1.8.10. Subset: Set A is said to be a subset of set B if all the elements of set A are in B and we write it as
A  B. Also have if a  A then a  B.
e.g., {1, 2, 3, 4} is a subset of a set of natural numbers.
Note: Null set and the set itself are always subsets of a given set.
1.8.11. Proper Subset: Set A is said to be a proper subset of B if all the elements of set A are in set B but
there is at least one element in set B which is not in set A and we write as A  B. e.g., {a, b, c} is a
proper subset of {a, b, c, d}.
We can also write B  A i.e., set B is called super set of set A.
1.8.12. Power Set: A set of all the subsets of given set is called its power set. If a set contains n elements
then the number of elements in the power set will be 2n.
- 1.10-
Basic Mathematics
)
1.8.13. Universal Set: A set which contains all the sets and elements under consideration is called
universal set e.g., set of students in a school. School acts as a universal set for set of students in
L.K.G., U.K.G., I, II, III, IV, V, VI, VII, VIII, IX, X, XI and XII classes.
1.8.14. Venn Diagrams: Sets are also represented by using diagrams known as Venn diagrams. A circle
or ellipse represents a set and a rectangle represents a universal set.
1.8.15. Operations on Sets
(a) Union of Sets: Union of two given sets A and B is the set of
all the elements which either belong to set A or to set B or to
both. We represent union as A  B.

A  B = {x : x A or B or both}
(b) Intersection of Sets: Intersection of two given sets A and B is
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the set of all the elements common to both A and B. We
represent intersection of set A and set B as A  B.

A  B = {x : x  A and B}.
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(c) Disjoint Sets: Two sets A and B are said to be disjoint if they
have no common elements or their intersection
is a null set. i.e., A  B = .
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(d) Difference of Sets: Let A and B be two given sets. Then the
difference of set A and set B, denoted as A –B, is a set of all
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the elements which belong to set A
but not to set B. i.e., A – B = {x : x A and x  B}.
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(e) Symmetric Difference of Two Sets: Given two sets A and B
then symmetric difference of two sets is a set of elements
which belongs to A or B but are not common to both the sets.
It is denoted by A B.
 AB = (A – B)  (B – A).
(f)
Complement of a set : Given a universal set U and a set A,
then complement of set A is the set of all the elements which
do not belong to A. It is denoted by A or AC or A or U – A.
 A = {x : x  A}
1.8.16. Algebra of Sets
(a) Idempotent law
(i)
AA=A
(ii) A  A = A
1.11
Basic Mathematics
(b) Identity law
(i)
A=A
(ii) U  A = A
(c) Commutative law
(i)
AB=BA
(ii) A  B = B  A
(d) Associative law
(i)
(A  B)  C = A  (B  C)
(ii) A  (B  C) = (A  B)  C
(e) Distributive law
(i)
(f)
A  (B  C) = (A  B)  (A C)
(ii) A  (B  C) = (A  B)  (A  C)
De-Morgan’s law
(i)
(A  B) = A  B
(ii) (A  B) = A  B.
1.8.17. Results on cardinal numbers (number of elements in a set) of finite sets.
n(A  B) = n (A) + n (B) – n (A  B)
(ii)
n(A  B) = n (A) + n (B), if A and B are disjoint.
(iii)
n(A – B) = n(A) – n (A  B).
(iv)
n(A  B) = n(A – B) + n (B – A) + n (A  B).
(v)
n(A  B  C) = n(A) + n (B) + n (C) – n (A  B) – n (B  C) – n (C  A) + n (A  B  C).
(vi)
n (A B) = n {(A  B)} = n(U) – n (A  B).
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(i)
Sample Problem-7:
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(vii) n (A B) = n {(A  B)} = n(U) – n (A  B).
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Which of the following pairs of sets are equal? Justify your answer.
(i) X, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”.
(ii) A = {n : n  Z and n2  4} and B = {x : x R and x2 – 3x + 2 = 0}.
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Solution: (i) We have, X = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then X and B are equal sets as
repetition of elements in a set do not change a set. Thus, X = {A, L, O, Y} = B
(ii) A = {–2, –1, 0, 1, 2}, B = {1, 2}. Since 0  A and 0  B, A and B are not equal sets.
Sample Problem-8:
Are the following pair of sets equal ? Give reason.
(i) A = {2, 3}, B = {x : x is a solution of x2 + 5x + 6 = 0}
(ii) A = {x : x is a letter in the word FOLLOW}
B = {y : y is a letter in the word WOLF}.
Solution: We have,
(i) A = {2, 3}, B = {x : x is a solution of x2 + 5x + 6 = 0}
Now, x2 + 5x + 6 = 0
 x2 + 3x + 2x + 6 = 0
- 1.12-
Basic Mathematics

)
 x (x + 3) + 2 (x + 3) = 0
 (x + 3) (x + 2) = 0  x = – 2, – 3
Therefore, B = {–2, –3}
Here, we observe that the elements of set A are not exactly the same to that of set B, hence A
and B are not equal sets.
(ii) We have, A = {x : x is a letter in the word FOLLOW}
 A = {F, O, L, W}
And B = {x : x is a letter in the word WOLF}
 B = {W, O, L, F}
Here, we observe that the elements of both sets are exactly same, hence the sets are equal.
OBJECTIVE QUESTIONS-1.8:
1.
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C.
JE
If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B
and C and 8 liked all the three products. How many liked product C only?
(a)
(b) 12
(c) 13
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2.
11
(d) 10
In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the
students knows either Hindi or English. How many students are there in the group?
3.
(b) 115
(b) B = C
(c) B = B (B C)
(d) None of these
(b) A C
(c) A C
(d) None of these
(b) x  B
(c) x  A
(d) None of these
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B C
AC
If A  B and x  B , then
(a)
BA
If X and Y are two sets such that n ( X ) = 17, n ( Y ) = 23 and n ( X  Y ) = 38, then n ( X  Y )
(a)
8.
(d) 30
(c) 15
If A  B and B  C , then
(a)
7.
(b) 20
Let A, B, and C be the sets such that A  B = A  C and A  B = A  C then
(a)
6.
25
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(a)
5.
(d) 125
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only
and not cricket?
4.
(c) 110
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(a) 100
2
(b) 1
(c) 3
(d) 4
If X and Y are two sets such that X  Y has 18 elements, X has 8 elements and Y has 15 elements ;
how many elements does X  Y have?
(a)
1
(b) 3
(c) 4
1.13
(d) 5
Basic Mathematics
9.
In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two
drinks. How many people like both coffee and tea?
(a)
1.
29
(b) 22
(a): n(A) = 21
(c) 25
(d) 19
HINTS/SOLUTIONS OBJECTIVE QUESTIONS-1.8:
n(A  B) = 14
n (A  B  C) = 8
n(B) = 26
n(A  C) = 12
n(C) = 29
n(B  C) = 14
n (only C) = n (C) – n(A  C) – n (B  C) + n (A  B  C)
= 29 – 12 – 14 + 8 = 11
(d): n (A  B) = 100 + 50 – 25 = 125
3.
(a): n (C  T) = n(C) + (n (T) – n(C T)
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2.
n(T) = 65 – 40 + 10 = 35
n (T only) = 35 – 10 = 25
(b): We can prove
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4.
B = (B  C) C and C = (B C)  (A  C)
(a): All the elements of A will be inside B and inside C also as B is subset of C.
6.
(c): Since all the elements of A are inside B so x  A.
7.
(a): n (X Y) = 17 + 23 – 38 = 2
8.
(d): n(X Y) = 8 + 15 – 18 = 5
9.
(d): 37 + 52 – 70 = 19
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5.
1.9. LOGARITHMS
1.9.1. If ‘a’ ( 1) is a positive real number and x is a rational number such that ax = t, then we say that x is
logarithm of t to the base ‘a’ and we write it as loga t = x.
Thus ax = t  loga t = x.
e.g., 23 = 8  log2 8 = 3; log10 100 = 2  102 = 100
1.9.2. Laws of logarithm
(a) If m and n are positive rational numbers, then loga (mn) = loga m + logan.
This result can be generalized to the multiplication of n numbers.
æ ö
çè n ø
m
(b) If m and n are positive rational numbers, then loga çç ÷
÷
÷= logam – logan.
(c) If m and n are positive rational numbers, then loga (mn) = n logam.
- 1.14-
Basic Mathematics
)
(d) loga 1 = 0 as a0 = 1 i.e., logarithm of 1 to any non zero base is zero.
(e) loga a = 1 as a1 = a i.e., logarithm of a non zero number to the same base is always 1.
(f)
If m is a positive rational number and ‘a’ and ‘b’ are positive real numbers (a  1, b  1), then
(i) loga m =
log b m
log b a
(ii) loga m =
1
.
log m a
(g) logb x > y
 x > by (if b > 1)
 x < by (if 0 < b < 1)
(h) logb a = logby (ay)
Sample Problem-9:
JE
Show that : 2 log (8/45) + 3 log (25/8)  4 log (5/6) = log 2.
 8 
 25 
5
Solution: 2log    3log    4log  
 45 
 8 
6
Sample Problem-10:
Prove that
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82
(25)3 (6) 4


45  45 (8)3 (5) 4
= log 2.
 log
Solution: Permissible values are
N > 0, a  (0, 1)  (1, )
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log a N
= 1 + logab & indicate the permissible values of the letters.
log a b N
and ab  (0, 1)  (1, )
b  (0, 1)  (1, )
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Sample Problem-11:
(a) Given : log1034.56 = 1.5386, find log103.456 ; log100.3456 & log100.003456.
(b) Find the number of positive integers which have the characteristic 3, when the base of the
logarithm is 7.
Solution: (a)
log103.456
= log10
34.56
 log34.56  1
10
= 1.5386 – 1 = 0.5386.
log10 0.3456 = 1.5386 – 2 = – 0.4614
log10 0.003456 = log10 34.56 – 4
(b)
= 1.5386 – 4 = –2.4614
log7x = 3  x = 73
so numbers will be (74 – 73).
1.15
Basic Mathematics
Sample Problem-12:
(a) If log10 (x2  12x + 36) = 2
(b) 91+logx  31+logx  210 = 0 ; where base of log is 3.
Solution: (a) log10(x2 – 12 x + 36) = 2
 x2 – 12x + 36 = 100
 x2 – 12x – 64 = 0
 x2 – 16x + 4x – 64 = 0
 (x – 16) (x + 4) = 0
 x = 16, x = – 4.
(b)
91 + log x – 31 + log x – 210 = 0
 9x2 – 3x – 210 = 0
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 3x2 – x – 70 = 0
 (3x + 14) (x – 5) = 0
Sample Problem-13:
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x=5
Simplify : log1/3 4 729 . 3 91 . 27  4/3
= log1/3 (36 . 3–2/3. 3–4/3)1/4
OBJECTIVE QUESTIONS- 1.9:
1
1
1
+
+
has the value equal to
log bc abc log ca abc log ab abc
(a)
2.
1/21
If log (x + y) = log 2 +
(b) 1
(c) 4
4.
(d) none of these
1
1
log x +
log y, then
2
2
(c) x2 + xy + y2 = 0
(a)
3.
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1.
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= log1/3 3 = –1.
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Solution: log1/3 4 729 . 3 91 . 27  4/3
x+y=0
(b) xy = 1
1 x 
If f ( x)  ln 
 , then
1 x 
(d) x – y = 0
(a)
f ( x1 ). f ( x2 )  f ( x1  x2 )
(b)
f ( x  2)  2 f ( x  1)  f ( x)  0
(c)
f ( x)  f ( x  1)  f ( x2  x)
(d)
 x x 
f ( x1 )  f ( x2 )  f  1 2 
 1  x1 x2 
log10 tan1  log10 tan 2  ...  log10 tan 89 is equal to
(a)
0
(b) 1
(c) 27
- 1.16-
(d) 81
Basic Mathematics
5.
If
(a)
6.
(–, 1)
(c) (2, )
(b) (1, 2)
1
(b) 3
3
(c) 4
(b) 1
1
(d) none of these
(d) 5
(c) 2
(b) 2
(d) 0
(c) 3
JE
x=0
(d) 4
(b) x = 2
(c) x = log102
(d)
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HINTS/SOLUTIONS OBJECTIVE QUESTIONS- 1.9:
1
1
 log abc bc 
 log abc ac
(b):
log bc abc
log ac abc
log ab abc
 log abc ab
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Adding we get log abc abc  1
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1
2.
(d) 1
If log102, log10 (2x + 1), log10 (2x + 3) are in AP, then
(a)
1.
(c) 0
If log5 (x + 23) – 2 log5 (1 – 5x –3) = – 2 log5(0.2 – 5x –4), then x is
(a)
10.
(b) abc
The number of solution of log 4 ( x  1)  log 2 ( x  3) is
(a)
9.
xyz
The solution of the equation log7 log5 ( x  5  x )  0 is
(a)
8.
log x log y log z
, then xa yb z c is equal to


bc c a a b
If log 0.3 ( x  1)  log 0.09 ( x  1) then x lies in the interval
(a)
7.
)
(d): log (x + y) = log 2 xy
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(x + y)2 = 4xy 

 (x – y)2 = 0  x = y
3.
 1  x1  1  x2 
 1  x1  x2  x1 x2 
 x1  x2 
(d): f (x1) + f (x2) = log 

 = log 
 = f

 1  x1  1  x2 
 1  x1 x2  x1  x2 
 1  x1 x2 
4.
(a): log10 tan 1° tan 2° ..... tan 89°
= log10 1 = 0
5.
(d): x = 10b –c, y = 10c – a and z = 10a – b
xa ybzc = 10a(b – c) + b(c –a) + c (a – b) = 1
6.
(b): log0.3 (x –1) > log(0.3)2(x –1)
 (x –1) <
x 1
 (x – 1) (x –2) < 0  x  (1, 2)
1.17
1
log 2 5
2
Basic Mathematics
7.
(a): log7 log5 ( x  5  x ) = 0
 ( x  5  x)  5
 x = 4 is solution.
8.
(b): log4 (x –1) = log2(x –3)
 (x –1) = (x –3)2  x2 – 7x + 10 = 0
x = 2 and x = 5 (But x = 2 gives (x – 3) negative)
9.
(b): log5(x + 23) – 2 log5(1 – 5x –3) = –2log5(0.2 – 5x – 4)
 x = 2 is answer
10.
(d): 2log10 (2x + 1) = log10 2 + log10(2x + 3)
1
 (2x + 1)2 = 2(2x + 3) x = log 2 5
2
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- 1.18-
ADD NOTES HERE :-)
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ADD NOTES HERE :-)
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ADD NOTES HERE :-)
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ADD NOTES HERE :-)
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ADD NOTES HERE :-)
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ADD NOTES HERE :-)
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ADD NOTES HERE :-)
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ADD NOTES HERE :-)
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ADD NOTES HERE :-)
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ADD NOTES HERE :-)
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