mathematics

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

(iv) Closed open interval:

[a, b) = x R : a x b

a 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 andam < bm for m < 0.

(iv) If a > b > 0, then 1 1

.a b

Basic Properties of Inequalities

(v) If , then

for

all positive numbers ai and bi for i = 1, 2, ... n.

1 1 2 2 n na b , a b , ..., a b

1 2 3 n 1 2 na a a ... a b b ... b

(vi) If , then for all positive numbers ai and bi for i = 1, 2, ... n.

1 1 2 2 3 3 n na b , a b , a b , ...a b

1 2 n 1 2 na a ...a b b ... b

(vii) If a > b > 0 and n > 0, then 1 1

n n n na b and a b .

Definition of Inequation

A statement involving variable(s) andthe sign of inequality, i.e. >, <,is called an inequation.

or

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

2x 20 8 12

x 6

(ii) 2x + 8 < 20

2x 12

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 possiblesolutions 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 Inequations1. 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 makeit 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 3:0– 1–

So intervals are . , 1 and 1,

Rules for Solving Absolute Value Function Inequations

Step 4:

Sign of x + 1 in their intervalis negative. To get the sign of x + 1 inthe interval we generally a value of xwhich is less than –1. Say it as –1.008.Then obviously x + 1 will be negativeand in the interval expression x + 1is positive. This can be realized bytaking the value of x = –0.98 whichlies in the interval

, 1

1,

1, .

Rules for Solving Absolute Value Function Inequations

Step 5: So in , (x + 1) is negative. , 1

and in , (x + 1) is positive. 1,

Step 6:

x 1 1 xx 1

x 1 x 1

Minus sign is added here as x + 1 is negativein this interval and we are interested in positivevalue 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

3– 5–

, 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 i.e. x a, a

(ii) x a a x 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 numberand ‘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 inequation2x – y < 4.

Solution:

Step 1: 2x – y = 4

Step 2:

(2, 0)

(0, – 4)

0

y

y

x 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.

(2, 0)

(0, – 4)

0

y

y

x 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 < 32x + y >1

Solution:

12–, 0 (3, 0)

32

(0, 1)

2x + y = 1

x – 2y = 3

0

–

Dotted part is for x – 2y < 3.

Crossed part is for 2x + y > 1.

So the required region is that region where both cross and dots are present, i.e. the circled region.

Inequality related to AM, GM and HM

Let ‘a’ and ‘b’ be two real positive andunequal numbers and A, G, H arearithmetic, geometric and harmonicmeans respectively between them.

a b 2ab

A , G ab and H2 a b

2a b 2ab

Now AH ab G2 a b

A G

...(i)G H

a bAgain A G ab

2

Inequality related to AM, GM and HM

2a b 2 ab a b

02 2

A G 0

A G ...(ii)

A

Again from (ii), A G 1 G 0G

A G

From i , 1G 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

2a b

A G 02

A G

A G

Again from (i), 1G H

G H

Hence, A = G = H

Therefore, we can write equally holdswhen a = b.

AM GM HM

Inequality related to AM, GM and HM

Note: This inequality holds for n numbers also.

1 2 3 na a a ... a

AM of 'n' numbers n

1

n1 2 3 nGM of 'n' numbers a a a ...a

1 2 3 n

nand HM of 'n' numbers

1 1 1 1...

a a a a

11 2 n n1 2 n

1 2 n

a a ... a nAM GM HM a a ...a .

1 1 1n ...a a a

Class Test

Class Exercise - 1

x 3x 2 5x 3Solve .

5 4 5

Solution

We have,

5 3x 2 4 5x 3x5 20

x 15x 10 20x 12

5 20

x 5x 25 20

20x 25x 10 45x 10

10 2

x x45 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

0x 5 2

2x x 50

2 x 5

2x x 50

2 x 5

x 50 ...(i)

x 5

Solution contd..

x 5From i , 0

x 5

Required solution region , 5 5,

Here coefficient of x is positive. Now equating numerator and denominator to zero, we getx = –5 and 5 respectively.

+ – 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

...(i)4 8 8

2x 1 x 1 3x 1...(ii)

12 3 4

5x 3x 39

From (i),4 8 8

10x 3x 398 8

13x 39

x 3 ...(iii)

Solution contd...From (ii),

2x 1 x 1 3x 112 3 4

2x 1 4 x 1 3x 112 4

2x 3 3x 112 4

2x 3 3 3x 1 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 thenumber line, we get

– – 4 – 2 2 4 6

The combined solution is

4, 2 2, 6

Class Exercise - 5

x 1Solve 1.

x 2

Solution

We have .

x 11 0

x 2

x 1 x 2

0 ...(ii)x 2

Now the following cases arise:

Case I: When . x 1 0, i. e. x 1

In this case, we have |x – 1| = x – 1.

x 1 x 2

From (i), 0x 2

x 1 x 2

0x 2

Solution contd..

3

0x 2

a

x 2 0 0 and a 0 b 0b

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

0x 2

2x 1

0x 2

2x 10

x 2

2x 1

0x 2

Equating numerator and denominator to zero,

we get and –2 respectively.1

x2

Solution contd..

Therefore,

+ – +

– 2 12

– –

1Solution is , 2 , .

2

But x < 1, therefore

...(ii)

1x , 2 , 1

2

Combining the solution (i) and (ii) onthe number line, we get

0– 2 12

– – 1

–

The solution is

1, 2 , .

2

Class Exercise - 6

In the first four papers each of 100marks, Mukesh got 83, 73, 72, 95marks. If he wants an average ofgreater than or equal to 75 marksand less than 80 marks, find therange of marks he should score inthe fifth paper.

Solution

Let Mukesh scores x marks in the fifth paper.

According to the given condition,

83 73 72 95 x75 80

5

323 x75 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 setof 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

1 2 3 4 5 6 7 8 95 4 3 2 1– – – – –

4

3

2

1

1

2

3

4

–

–

–

–

Y

Y´

X´ X

x – y =

1

2x + y = 2

x + 2y = 8

O

––––––––––

––––––––––

Putting (0, 0) into eachinequation, we get theshaded region as therequired solution.

Class Exercise - 8

If x, y and z are three positivenumbers, prove that (x + y + z)

1 1 19.

x y z

Solution

x, y and z are positive quantities.

AM of x, y and z > GM of x, y, z.

1

3x y z

xyz3

1

3x y z 3 xyz ...(i)

1 1 1 1 1 1

Again AM of , , GM of , ,x y z x y z

1

3

1 1 11 1 1x y z

. .3 x y z

Solution contd..

1

31 1 1 13 ...(ii)

x y z xyz

Multiplying of corresponding sides of(i) and (ii), we get

1

31

3

1 1 1 1x y z 9 xyz

x y zxyz

1 1 1x y z 9

x y zProved.

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 HPa cb

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 HPb dc

2 c is HM of b and d

b d 2c ...(ii)

Solution contd..

From (i) and (ii),

a + c + b + d > 2b + 2c

Proved a d b c

(ii) For first three terms, GM > HM

ac b

2ac b ...(iii)

and for the last three terms bd c

2bd c ...(iv)

From (iii) and (iv)

2 2ac bd b c

ad bc Pr oved.

Class Exercise - 10

If a > 0, b > 0, c > 0, prove that

where

s = a + b + c.

1 1 1 9

,s a s b s c 2s

Solution

As we know that for three positivequantities, x, y and z, we have

x y z 3A , H

1 1 13x y z

and A > H.

Here (s – a), (s – b), (s – c) are positive quantities.

s a s b s c 31 1 13

s a s b s c

Solution contd..

3s a b c 31 1 13

s a s b s c

2s 3s a b c

1 1 13s a s b s c

1 1 1 9

Pr oved.s a s b s c 2s

Thank you