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Mathematics
Statistics
Session Objectives
Session Objectives
1. Intervals
2. Basic properties of inequalities
3. Definition and solution of linear inequation
4. Solution of modulus inequations
5. Solution of two variable inequations
6. Inequalities related to AM, GM and HM
Interval
(i) Open interval:
]a, b[ or (a, b) =x  R:a  x  b
a
b
(ii) Closed interval:
[a, b] =x  R : a  x  b
a
b
Interval
(iii) Open closed interval:
(a, b] =x  R : a  x  b
a
b
(iv) Closed open interval:
[a, b) =x  R : a  x  b
a
b
Note: (iii) and (iv) are
also called semi-closed
or semi-open intervals.
Basic Properties of Inequalities
(i) If a > b, b > c, then a > c.
(ii) If a > b, then a + m > b + m.
(iii)If a > b, then am > bm for m > 0 and
am < bm for m < 0.
(iv) If a > b > 0, then
1 1
 .
a b
Basic Properties of Inequalities
(v) If a1  b1, a2  b2 , ..., an  bn, then
a1  a2  a3  ...  an  b1  b2  ...  bn for
all positive numbers ai and bi for i = 1, 2, ... n.
(vi) If a1  b1, a2  b2 , a3  b3 , ...an  bn , then
a1 a2 ...an  b1 b2 ... bn for all positive
numbers ai and bi for i = 1, 2, ... n.
(vii) If a > b > 0 and n > 0, then an  bn and
1
an

1
bn .
Definition of Inequation
A statement involving variable(s) and
the sign of inequality, i.e. >, <,  or 
is called an inequation.
For example:
f  x   0 or f  x   0 or f  x   0 or f  x   0
Illustrative Example
Solve for x where x is non-negative integer.
(i) 2x + 8 = 20
(ii) 2x + 8 < 20
(iii) 2x  8  20
Solution:
(i) 2x + 8 = 20
(ii) 2x + 8 < 20
 2x  20  8  12
 2x  12
 x6
x  6
Possible values of x
are 0, 1, 2, 3, 4, 5.
Illustrative Example
(iii) 2x  8  20
 2x  12
 x6
 Possible values of x are 0, 1, 2, 3, 4, 5, 6.
An inequation may be linear or quadratic or cubic,
etc., containing one or more variables.
Solutions of Linear Inequations in
One Variable
It is the process of obtaining all possible
solutions of an inequation.
Solution set
The set of all possible solutions of an
inequation is known as its solution
set.
For example:
The solution set of the inequation x2 + 2 > 0 is
the set of all real numbers whereas the solution
set of the inequation x2 + 2 < 0 is the null set.
Rules for Solving Inequations
1. Adding or subtracting the same number
or expression from each side of an
inequation does not change the
inequality.
2. Multiplying or dividing each side of an
inequation by the same positive number
does not change the inequality.
3. Multiplying each side of an inequation by
the same negative number reverse the
inequality.
Rules for Solving Inequations
Example :
Solve the inequations
12  5x  3  3x
5x  3x  3  12
2x  15
x  7.5
 Answer is ]  ,  7.5]
–
 –
7
.
5
0
Rules for Solving Absolute Value
Function Inequations
1.
Factorize the expression.
2.
Get the roots of the expression or say critical point (critical
points are the points where the expression becomes zero or  ).
Expression only changes its sign at critical value.
3.
Make various interval on number line.
4.
Assign the sign of each bracket in these intervals and check the
sign of expression.
5.
List out the intervals where expression is positive or negative
separately.
6.
Left as it is, when expression is positive and multiply with –1
when function is negative to make
it positive.
Rules for Solving Absolute Value
Function Inequations
Example
|x + 1| > 4
Step 1: Already in factorized form, i.e x + 1
Step 2:
x+1=0
x = –1
Step 3:
–

–
1

0
So intervals are  ,  1 and  1,  .
Rules for Solving Absolute Value
Function Inequations
Step 4:
Sign of x + 1 in their interval  ,  1
is negative. To get the sign of x + 1 in
the interval we generally a value of x
which is less than –1. Say it as –1.008.
Then obviously x + 1 will be negative
and in the interval  1,   expression x + 1
is positive. This can be realized by
taking the value of x = –0.98 which
lies in the interval  1,   .
Rules for Solving Absolute Value
Function Inequations
Step 5: So in  ,  1 , (x + 1) is negative.
and in  1,   , (x + 1) is positive.
1  x  
 x 1
Step 6: x  1  
  x  1   x  1
Minus sign is added here as x + 1 is negative
in this interval and we are interested in positive
value of the expression.
Rules for Solving Absolute Value
Function Inequations
Now solve the equation.
For  1  x  
|x + 1| > 4
But the value of |x + 1| in this interval is x + 1.
So x + 1 > 4
x>3
So the solution is 3  x  
For    x  1 the value of |x + 1| is –(x + 1).
   x  1  4 – x – 1 > 4
Rules for Solving Absolute Value
Function Inequations
 x 5
 x  5
 x   ,  5
 Possible solutions are
 ,  5
–

3, 
–
5
3

Some Important Results on Modulus
Inequations or Absolute Value function
Result 1: If ‘a’ is a positive real number,
then
(i) x  a   a  x  a
(ii) x  a   a  x  a
i.e. x   a, a 
i.e. x  a, a
(iii) x  a  x  a or x  a
(iv) x  a  x  a or x  a
Some Important Results on Modulus
Inequations or Absolute Value function
Result 2: Let r be a positive real number
and ‘a’ be a fixed real number. Then
(i) x  a  r  a  r  x  a  r i.e. x  a  r, a  r 
(ii) x  a  r 
ar  x  ar
i.e. x  a  r, a  r 
(iii) x  a  r  x  a  r or x  a  r
(iv) x  a  r  x  a  r or x  a  r
Graphical Solution of One or Two
Variable Equations
Let the equation is Ax + By + C < 0.
Algorithm
Step 1: Convert the inequation into equation, i.e. Ax +
By + C = 0
Step 2: Draw the graph of Ax + By + C = 0.
Step 3: Take any point [generally (0, 0)] or any point not on the
line whose position is known to you with respect to line.
Step 4: Put this point in the given inequation and check the
validity of the inequation. If inequation satisfied then the
corresponding side of the line where lies the chosen point is the
graph of the inequation.
Illustrative Example
Draw the graph of the inequation
2x – y < 4.
Solution:
Step 1: 2x – y = 4
Step 2:
y
x
0
(2, 0)
(0, – 4)
y
x
Illustrative Example
Step 3: Take the point as (0, 0).
Step 4: 2 × 0 – 0 < 4
0 < 4, so it is valid. Hence,
the shaded side is the required
solution.
y
x
0
(2, 0)
(0, – 4)
y
x
Graphical Solution of Two Variable
Algorithm
(i) Draw the graph of two inequations.
(ii) Get the required common region
bounded by the two lines.
Illustrative Example
Get the required region.
x – 2y < 3
2x + y >1
Solution:
(0, 1)
x
Dotted part is for x – 2y < 3.
y=
–3
2
(3, 0)
+
1
So the required region is that
region where both cross and
dots are present, i.e. the circled
region.
1
–, 0
2
2x
Crossed part is for 2x + y > 1.
0
–
=
2y
3
Inequality related to AM, GM and HM
Let ‘a’ and ‘b’ be two real positive and
unequal numbers and A, G, H are
arithmetic, geometric and harmonic
means respectively between them.
ab
2ab
 A
, G  ab and H 
2
ab
 a  b   2ab 
2
Now AH  

ab

G
 ab
2



A G


G H
...(i)
ab
Again A  G 
 ab
2
Inequality related to AM, GM and HM
2
a  b  2 ab  a  b 


  0

2
2 

 A G  0
 AG
...(ii)
A
Again from (ii), A  G 
1
G

G  0
A G
 From  i  ,
 1
G H
 GH

H  0
...(iii)
From (ii) and (iii), A > G > H, i.e. AM > GM > HM
Inequality related to AM, GM and HM
Cor: If the two numbers are equal,
i.e. a = b, then
2
 a b
A G  
  0

2


A G
Again from (i),
 1
G H
 AG
 GH
Hence, A = G = H
Therefore, we can write AM  GM  HM equally holds
when a = b.
Inequality related to AM, GM and HM
Note: This inequality holds for n numbers
also.
a1  a2  a3 ...  an
AM of 'n' numbers 
n
GM of 'n' numbers  
1
a1 a2 a3 ...an n

n
and HM of 'n' numbers 
1
1
1
1


 ... 
a1 a2 a3
an
1
a1  a2  ...  an
n
AM  GM  HM 
  a1 a2 ...an  n 
.
1
1
1
n

 ... 
a1 a2
an
Class Test
Class Exercise - 1
x 3x  2 5x  3
Solve


.
5
4
5
Solution
5  3x  2  4  5x  3 
x
We have, 
5
20

x 15x  10  20x  12

5
20
 20x  25x  10
 x

x 5x  2

5
20
 45x  10
10
2
 x
45
9
 Solution set of the given inequality is  , 2  .
9


Class Exercise - 2
x
1
Solve
 .
x5 2
Solution
We have

x
1

x5 2
x
1
 0
x 5 2
2x   x  5 

0
2  x  5

2x  x  5
0
2  x  5
x5

0
 x  5
...(i)
Solution contd..
Here coefficient of x is positive. Now equating
numerator and denominator to zero, we get
x = –5 and 5 respectively.
– 5
+
+
–
5
x5
From i,
0
 x  5
 Required solution region   ,  5 5, 
Class Exercise - 3
Solve the following system of inequations
5x 3x 39 2x  1 x  1 3x  1


,


.
4
8
8
12
3
4
Solution
The given system of inequation is
5x 3x 39


4
8
8
2x  1 x  1 3x  1


12
3
4
 From (i),

...(i)
...(ii)
5x 3x 39


4
8
8
10x  3x 39

8
8
 13x  39
 x3
...(iii)
Solution contd...
From (ii),
2x  1 x  1 3x  1


12
3
4

2x  1  4  x  1
12
3x  1

4
  2x  3  3 3x  1
2x  3 3x  1


12
4
  2x  9x  3  3
  11x  0
x>0
...(iv)
Combining the solution (ii)
and (iii) on the number line
–

0
3
 Solution set  3, 

Class Exercise - 4
Solve the following system of inequations
x  1  5, x  2.
Solution
The given system of inequation is
x 1  5
...(i)
x 2
..(ii)
From (i), x  1  5
  5   x  1  5 
x  a   a  x  a 
  5  1  x  1 1  5  1
 4 x6
 Solution of inequation (i) is [–4, 6].
...(iii)
Solution contd...
From (ii), x  2
 x2
or
 x   ,  2
x  2
2, 
...(iv)
Combining the solution (iii) and (iv) on the
number line, we get
–

–
4–
2
2 4 6
The combined solution is
4,  2 2, 6
Class Exercise - 5
Solve
x 1
x2
 1.
Solution
We have

x 1
x2
x  1   x  2
x2
1 0.
0
...(ii)
Now the following cases arise:
Case I: When x  1  0, i. e. x  1 .
In this case, we have |x – 1| = x – 1.
 From (i),

x  1   x  2
x2
x  1 x  2
0
x2
0
Solution contd..
3

0
x2

 x20 

a

 0 and a  0  b  0 
b

 x  2
But in this case x  1.
Solution is 1, .
...(i)
Case II: When x – 1 < 0, i.e. x < 1.
In this case |x – 1| = – (x – 1).
Solution contd..
 Inequation (i) can be written as
  x  1   x  2 
x2
2x  1

0
x2

0
2x  1
 
0
x2
2x  1
0
x2
Equating numerator and denominator to zero,
we get x  
1
and –2 respectively.
2
Solution contd..
Therefore,
+ –
–
2
+
1
–
–
2
 Solution is  ,  2 
 1


,


.
 2

But x < 1, therefore
 1 
x   ,  2    , 1
 2 
...(ii)
Combining the solution (i) and (ii) on
the number line, we get
–

–
2

01
1
–
–
2
 The solution is
 ,
 2
 1

  2 ,  .


Class Exercise - 6
In the first four papers each of 100
marks, Mukesh got 83, 73, 72, 95
marks. If he wants an average of
greater than or equal to 75 marks
and less than 80 marks, find the
range of marks he should score in
the fifth paper.
Solution
Let Mukesh scores x marks in the fifth paper.
According to the given condition,
75 
83  73  72  95  x
 80
5
 75 
323  x
 80
5
 375  323  x  400
 52  x  77
Hence, Mukesh must score between 52 and 77 marks.
Class Exercise - 7
Draw the diagram of the solution set
of linear inequations,
x  y  1, x  2y  8, 2x  y  2, x  0, y  0.
Solution
Here we have to draw the graph of
the corresponding equations, i.e.
x – y = 1, x + 2y = 8, 2x + y = 2.
We get
Putting (0, 0) into each
inequation, we get the
shaded region as the
required solution.
Y
4
–
–
–
–
–
–
3
2
–
–
–
–
–
–
1
X´
–5 –4 –3 –2 –1
O
–1
–4
Y´
3
=2
–
1 –3
2
+y
x
y=
1
2x
–2
X
4
5
6
7
8 x +9
2y
=
8
Class Exercise - 8
If x, y and z are three positive
numbers, prove that (x + y + z)
 1 1 1
     9.
 x y z
Solution
x, y and z are positive quantities.
 AM of x, y and z > GM of x, y, z.
1
xyz

  xyz  3
3
 x  y  z  3  xyz 
Again AM of
1
3
...(i)
1 1 1
1 1 1
, ,  GM of , ,
x y z
x y z
1 1 1
1
 
x y z  1 1 1 3

 . . 
3
 x y z
Solution contd..
1
3
 1
1 1 1

  3

x y z
xyz


...(ii)
Multiplying of corresponding sides of
(i) and (ii), we get
1
 1 1 1
 x  y  z       9  xyz  3 
 x y z
 1 1 1
  x  y  z      9
 x y z
1
 xyz 
1
3
Proved.
Class Exercise - 9
If a, b, c and d be four distinct
positive quantities in HP, then
show that
(i) a + d > b + c
(ii) ad > bc
Solution
(i)
a, b, c and d are in HP.
 For first three terms AM > HM
 a, b and c are in HP
ac

b 

2

b
is
HM
of
a
and
c


 a  c  2b
...(i)
and for last three terms
AM > HM
 Again b, c and d are in HP
bd
c 

2

c
is
HM
of
b
and
d


 b  d  2c
...(ii)
Solution contd..
From (i) and (ii),
a + c + b + d > 2b + 2c
 adbc
Proved
(ii) For first three terms, GM > HM
ac  b
 ac  b2
...(iii)
and for the last three terms
 bd  c 2
...(iv)
From (iii) and (iv)
 ac  bd  b2c 2
bd  c
 ad  bc
Pr oved.
Class Exercise - 10
If a > 0, b > 0, c > 0, prove that
1
1
1
9


 , where
s  a s  b s  c 2s
s = a + b + c.
Solution
As we know that for three positive
quantities, x, y and z, we have
xyz
3
and A > H.
A
, H
1
1
1
3
 
x y z
Here (s – a), (s – b), (s – c) are positive quantities.

s  a  s  b  s  c  
3
3
1
1
1


sa sb sc
Solution contd..

3s   a  b  c 
2s


3
3

3
1
1
1


sa sb sc
3
1
1
1


sa sb sc
1
1
1
9




s  a s  b s  c 2s

s  a  b  c
Pr oved.
Thank you