478 (1).pdf

Lines & Slopes:

Standard equation of a Linear Polynomial is y = mx + c, where m is the slope and c is the constant

or it is also known as y-intercept.

If the slope of the line i.e. m is positive, then the line will be an Increasing Line.

While if the slope is negative, then the line will be decreasing.

Examples of Increasing line:

If we want to plot y = (x/2) + 3. Many students find it difficult to plot the graph. It can be done in just

two steps. First of all the line given is an increasing line since the slope m is positive.

We just need to find the two points on X-axis and Y-axis respectively.

If x = 0, then y = 3, (3 is the y-intercept which we learnt in the equation y = mx + c).

And if y = 0, then x = 6.

We just need to plot these two points on the X-axis and Y-axis, join these two points.

(x = 0, y =3) will be point on positive Y-axis.

(x = 6, y =0) will be point on negative X-axis. On, joining the graph will look like this.

Compiled by Anil Nair Classes www.anilnair.in


Similarly, if we want to plot the graph of y = x + 3. Again, this is an increasing line but with a slope of

+1. We need just two points to plot the graph.

If x = 0, then y = 3, (again the y-intercept is same).

And if y = 0, then x = 3. Let us plot it on the same graph on which previous line was plotted.

After looking at the two graphs, we should be able to conclude that if the slope increases, the line

will be steeper.

In the graph, the line y = x + 3 is more steep than y = x/2 + 3 because the slope of the first line is

greater than the second line.

Now, if we want to plot y = 3x + 3.

If x = 0, then y = 3, (again the y-intercept is +3).

And if y = 0, then x = 1. Let us plot these two points and join them to get the graph.



After observing the third line, our conclusion should be justified that if the slope increases, the line

will be steeper than the lines with the lesser slope.

Slope: Slope or Gradient of a line is the inclination or the angle of the line. A higher slope value

indicates a steeper line. Slope can be calculated as the change in the y-coordinate divided by change

in the x-coordinate.

2 1

2 1

y y ySlopex x x

= =

.

So, if we know the co-ordinates of two points, we can find out the slope of the line joining them. For

e.g. if P = (5, 6) and Q = (1, 6).

We can find out the slope of the line PQ as 2 1

2 1

( 6 6) 12 2[1 ( 5)] 6y yx x

= = =

. So, the slope of the

line is negative that means the line will be decreasing one.



Also, the relevance of the slope can be understood by the formula only,ySlopex

=

. Let us take the

example of the line for which we graph was drawn.

Y = x + 3;

X Y

0 3

1 4

2 5

3 6

By, observing the table, we can say that when there is a change of 1 unit in x, y also increases by 1

unit. Thats why the slope of the line is 1. We can conclude some properties of increasing line.

a) When x increases, then y will also increase depending on the slope.

b) And if x decreases, then y will also decrease depending on the slope.

Let us see another example of increasing line i.e. y = 3x + 3.

X Y

0 3

1 6

2 9

3 12

Again in this line, for one unit increase in x, change in y-coordinate is of 3 units, thus the slope of

the line is +3.

Conclusion: If an increase in x-co-ordinate leads to increase in y-co-ordinate, the slope will be

positive and the line will be INCREASING.

And If an increase in x-co-ordinate leads to decrease in y-co-ordinate, the slope will be negative

and the line will be DECREASING.

Examples of Decreasing Line: We have already seen how to identify whether a line is increasing or

decreasing. If the slope is negative, line will always be decreasing.

For e.g. 2x + y = 4.

On seeing this line, many students will say that this is an increasing line with a slope of +2. But,

actually, this is an example of decreasing line of slope 2. Whenever, we need to find the slope of a

line, we will have to compare it with the standard equation of straight line y = mx + c.

If we rewrite the equation given in the question as y = 2x + 4, we get a decreasing line.



Properties of Decreasing Line: If x increases, then y will decrease depending on the slope and if x

decreases, then y will increase depending on the slope.

X Y

0 4

1 2

2 0

3 2

As there is a change of +1 unit in x-coordinate, y co-ordinate decreases by 2 units, thus a negative

line with slope of 2.

Let us plot the graph of this line. Again we just need two points.

If x = 0, then y = 4 (4 is the y-intercept, so graph will cut the y-axis at +4) and when y = 0, then x = 2.

If we decrease the magnitude of the slope i.e. make it 1, and then try to plot it.

Let say the line y = x + 4, a decreasing line with a slope of 1.

When x =0, then y = 4, and when y = 0, then x = 4.

X Y

0 4

1 3

2 2

3 1

As x increases by 1 unit, the y co-ordinate decreases by 1 unit, thus a slope of 1.



Again, we can see as the magnitude of slope of a line decreases, we get a flatter line. The line y = x

+ 4 is a decreasing line with slope of (1), while the line y = 2x + 4 is a decreasing line with a slope of

2. As the magnitude of the slope increases, we get a steeper line and vice-versa.

ROOTS & SOLUTIONS

Roots: Roots are the point where the graph of that equation cuts the x-axis. Or roots are the point

where y-coordinate is equal to 0 which should be quite obvious. As roots are the points on X-axis

and at any point on the X-axis, y-co-ordinate is equal to 0.

Also, a linear polynomial can have only one root. Root is dependent on the highest power of the

polynomial.

So, y = 2x + 4, highest power of polynomial here is 1, thats why it can have one root only.

Solutions: Solutions are the values which will satisfy the equation. But you will think that on plug-in

value of the root also, the equation will be satisfied. Yes, its true. Roots are a part of Solutions. In

other words, we can say that for every corresponding value of x, we will get a corresponding value of

y and all of those will be solutions.

For e.g. if x =1, then y = 2. This is a solution.

If x = 2, then y = 0. This is also a solution and root.

If x = 3, then y = 2 , this is also a solution.



Like this, we can assume infinite values of x, and for all those infinite values of x, we will get

corresponding values for y.

So, any linear equation has maximum of one root and infinite solutions.

If we want to interpret this logic graphically, then all the points on the line are solutions of that

equation. And there are infinite points on a line.

In the graph, the line cuts the X-axis at 2 which is the root and the line cuts the y-axis at 4 which is

the y-intercept.



In the graph shown above, the point (x , y) as (1, 2) also lies on the graph and is a solution. That was

just one point on the graph, like that we can get infinite points on the graph and all of them will be

the solutions.

Parallel Lines: If the slopes are same, then the lines will be parallel.

For example, y = 2x + 4 and y = 2x 4 are increasing lines with slope of (+2) and y-intercepts of +4

and 4 respectively. Let us plot them.



Theory of Quadratic Equation

Standard equation of a Quadratic Polynomial is ax2 + bx + c, and since the greatest degree of the

polynomial is 2, it can have maximum of two roots. We will discuss all the possibilities of quadratic

equation having two roots, one root or no root.

The graph of a quadratic polynomial depends on the coefficient of x2 i.e. a. If a is positive, graph of

quadratic polynomial will be a upright parabola (U-shaped) and if a is negative, graph will be of the

shape of inverted parabola. The logic is same as that of a linear polynomial as the coefficient of x (y

= mx + c) determines whether the line will be increasing or decreasing.

Let us plot a quadratic polynomial y = x2 3x 4; While plotting the graph we should take care of

few points. First of all, we should find out the roots of the polynomial as the roots will help us in

determining the points where the graph will cut the X-axis. And secondly, we can also find out the y-

intercept easily which will be constant part in the equation given. In this equation given, roots are (

1 &4) and y-intercept will be 4. Also, the graph of a quadratic polynomial will increase rapidly when

we increase the value of x or decrease the value of x.

X 3 2 1 0 1 2 3 4

Y 14 6 0 4 6 6 4 0

On observing the table, x = 1 and x =4 are roots of the polynomial as value of y-coordinate is equal

to 0. Also, y-intercept is 4 as the value of x-coordinate is equal to 0. And also as the value of x

starts increasing, value of y increases rapidly. And when value of x decreases, then also y

increases rapidly.



Let us plot one more graph of a quadratic polynomial with negative sign of a.

Y = x2 x + 6; Again we need to find out the roots of the polynomial, we need to equate the value

of polynomial to zero.

So, we get x2 x + 6 = 0, which is same as x

2 + x 6 = 0. So, the roots are (2) & (3) and the y-

intercept is +6. (y = x2 x + 6).



After observing both the graphs, we would have got an idea that whenever the coefficient of x2 is

positive, the graph will be a Upright-Parabola and when the coefficient of x2 is negative, the graph

will be an Inverted-Parabola.

Factor Theorem: If we know roots of a quadratic expression, we can frame the factors and if we

know the factors of a quadratic expression, we can get the roots of the equation. So, both factors

and roots are interrelated.

For e.g. if a quadratic expression is given as y = x2 10x + 24. We can factorize the expression as y =

(x 6) (x 4).

So, (x 6) and (x 4) are factors of the expression and to find the value of roots, we need to equate

the expression to 0.

So, (x 6)(x 4) = 0 which gives the value of x as 4 or 6. These two are the roots. So, if we know the

factors of an expression, we can equate it to 0 to get the roots. And if we know the roots, we can get

the factors by subtracting the root from x.

For e.g., If its given that 2 and 3 are the roots of a quadratic expression. So, we can frame the

factors as (x 2) and [x (3)]= (x + 3) are factors of the expression .

So, Factor Theorem states that if (x ) is a factor of polynomial f(x), then on plug-in value of x = in

f(x), f(x) will be equal to 0.



For e.g: If its given that (x 3) is a factor of 2x2 + 5x + k, find the value of k.

If (x 3) is given as the factor, then applying the factor theorem, we can say that x = 3 will be root of

the given expression and on substituting x = 3 in expression, the value of expression must be equal

to zero.

So, on plug-in 2 32 + 5 3 + k = 0; K is equal to 33.

We will learn some more applications of Factor theorem later on.

Sum of the roots and Product of the roots:

Roots of a quadratic expression ax2 + bx + c can be calculated in two ways:

i) By factorisation i.e. splitting the middle term.

2 4)2

b b acii Rootsa

= .

Now, the question arises if we want to find the sum of roots or product of roots of quadratic

expression is it mandatory to find out the roots first?

Answer is no. We can directly find out sum of the roots and product of roots and use it as a formula.

We will see the derivation now.

Let us assume the two roots of quadratic expression ax2 + bx + c as and .

Now using FACTOR THEOREM, if and are the roots then (x ) and (x ) will be the factors of

the quadratic expression.

So, we can express as (x ) (x ) = ax2 + bx + c.

On multiplying, we get [x2 ( + )x + ] = ax

2 + bx + c.

The equation is not balanced as the coefficient of x2 on the L.H.S is 1, but the coefficient of x

2 on

R.H.S is a. So, to make it balanced we will divide the expression on the R.H.S by a.

2 2( ) b cx x x xa a

+ + = + + .

On observing the above equation, we can conclude one more point about quadratic expression.

A new quadratic equation can be expressed as x2 (sum of the roots)x + product of roots = 0.

On equating the coefficient of x and constant on both sides, we get

Sum of the roots = ( ) ba

+ =

Product of roots =c

a = .



Now, we can use these two formulas directly in any quadratic expression. Let us see some

applications.

E.g. 2: Construct a quadratic equation if its two roots are 4 and 5.

To frame quadratic equation, we need to find sum of the roots and product of the roots.

Sum of the roots = 4 + (5) = 1.

And product of the roots = (4) (5) = 20.

So, the equation can be framed by substituting the values in x2 (sum of the roots)x + product of

roots = 0.

On plug-in values, we get x2 (1)x + (20) = 0 or, x

2 + x 20 = 0.

E.g. 3: If p and q are the roots of equation x2 5x + 7 = 0, frame a new equation whose roots are (p +

5) and (q + 5).

To frame a new equation, we need to find the sum of roots and product of roots of the required

equation.

Our required equation is x2 (p + 5 + q + 5)x + (p + 5) (q + 5) = 0.

So, we need to know the value of (p + 5) + (q + 5) = p + q + 10. We already know the value of (p + q)

from the first equation as p & q are roots of the first equation.

So, p + q = b/a = (5)/1 = 5, then value of (p + q + 10) = 5 + 10 = 15.

Now, we should also know the value of the product of roots of the required equation which is (p + 5)

(q + 5).

So, (p + 5) (q + 5) = pq + 5(p + q) + 25 = 7 + 5 5 + 25 = 57 [pq = product of roots of first equation

whose value is c/a = 7].

So, now we can plug these values in equation x2 (sum of the roots)x + product of roots = 0.

So, answer is x2 15x + 57 = 0.

E.g. 4: If (p + 3) and (q + 3) are roots of x2 + 8x 10, then frame a equation whose roots are p & q.

The new equation with the roots p & q will be of the form x2 (p + q)x + pq = 0.

That means we need to find the value of (p + q) and pq.

We know sum of roots of equation given in question i.e. (p + 3 + q + 3) = 8 which gives us the value

of (p + q) = 14.

Similarly, we can find the product of roots from the equation given.

(p + 3) (q + 3) = 10, On multiplying we get [pq + 3(p + q) + 9] = 10.



On substituting the value of (p + q) obtained earlier, we can find the value of pq.

So, [pq 42 + 9] = 10, pq = 23

Now, we know the value of (p + q) = 14 and pq = 23.

So, the new equation framed is x2 + 14x + 23 =0.

E.g. 5: If two roots of equation 3x2 7x + 2k + 5 = 0 are reciprocal of each other, then find the value

of k.

This is a pretty easy question which can be solved orally if we know that if the roots are reciprocal of

each other leads to product of the roots being equal to 1.

For e.g. if one root is p, second root will be 1/p since roots are reciprocal and their product will

always be 1.

We know the product of roots = c/a = (2k + 5)/3 which is equal to 1.

So, on solving k = 1.

E.g. 6: If two roots of equation 4x2 (5p + 15)x 4 = 0 are equal in magnitude but opposite in sign,

find the value of p.

Again, a straightforward question if we know the logic that if the roots are equal in magnitude and

opposite in sign, then the sum of the roots is always equal to zero.

If we assume one root as z, then the second root will be z. And sum of those two roots will be 0.

We know sum of the roots given in the question is (5p + 15)/4 and equate it to 0.

We get (5p + 15)/4 = 0. On solving, we get value of p = 3.

Determinant:

Determinant is denoted by D whose value is equal to b2 4ac. It is called Determinant as it helps in

determining the nature of roots of quadratic equation. We will take some examples which will make

us understand the logic.



1st

case: X2 5X + 6 = 0 2

nd case: X

2 6X + 9 = 0 3

rd case: X

2 + X + 1 =0

Roots of the equation can be found out

by the formula

2 42

b b aca

.

( 5) 25 24 5 12 2

Roots = = .

Similarly, roots of this

equation can be found out.

6 02

Roots =

And roots of this

equation,

1 52

Roots = .

First case: if Determinant is positive, then the roots will be real & unequal.

Second case: If determinant is 0, then the roots will be real & equal. (Also pay attention the

quadratic equation in 2nd

case is a perfect square. So, we should keep in mind that determinant of a

perfect square will always be zero which will give us real and equal roots).

Third case: If determinant is negative, then the roots will be imaginary as square root of a negative

number is imaginary.

Now, we look at the positioning of the graphs in all these different cases.



E.g. 7: If the equation x2 15 m(x 4) = 0 has equal roots, find the value/s of m.

Before solving this problem, we need to rearrange the equation in standard form which is ax2 + bx +

c = 0.

On rearranging, we get x2 mx + (4m 15) = 0.

Since, the equation has equal roots, determinant must be equal to zero.

Thus, on equating b2 4ac = 0, we get m

2 4(4m 15 ) = 0.

On solving, we get values of m as 6 or 10.

Common Roots: Sometimes two quadratic equations have one root in common or two roots in

common. In such scenarios we will go back to basics of the definition of root. Roots are the point

which intersects the X-axis or roots are the values which on substituting in quadratic expression, the

value of expression becomes zero. So, if a pair of quadratic equation has one root in common, then

on substituting the value of root in both the equations, value becomes zero.

For example, the expressions (x2 3x + 2) and (2x

2 3x 2) have (x 2 ) as the common factor which

means x = 2 is the common root of both the expressions. And if we substitute x = 2 in both the

expressions, values of both the expressions will become zero.



E.g. 8: If (x + 3) is common factor of x2 5x + k = 0 and x

2 6x p = 0, then find the values of k and p.

This is a very easy problem based on common root. If (x + 3) is common factor to both the equations,

then x = 3 will be the common root to both the equations and on substituting x =3, the value of

the equations will become zero.

So, plug-in value of x in both the equations and find out respective values of k & p.

In first equation, on substituting, we get 9 + 15 + k = 0; so, k = -24.

And in second equation, on substituting, we get 9 + 18 p = 0; we get p = 27.

What if both the roots of two quadratic equations are common?

We can handle this situation in two ways:

First way: If both the roots of two equations are common, then we can equate the sum of roots of

the first equation with sum of roots of second equation as both will be equal. Similarly, we can also

equate product of roots of the first equation with the product of roots of the second equation as

both will be equal.

Second way: If two equations are ax2 + bx + c =0 and dx

2 + ex + f =0 having two roots in common,

then the ratio of the coefficients of x2, x and constant will be equal.

a b cd e f= = . On plug-in values of a, b, c, d, e and f, we can solve the question.

For e.g. (x2 + 3x + 2 = 0) and (2x

2 + 6x + 4= 0) are two equations with both the roots common and the

ratio just discussed earlier will be same.

Cubic Equation: Cubic equation can have maximum of 3 roots and as we saw there is relationship

between roots of a quadratic equation, similarly there exists certain relationship among roots of a

cubic equation.

If p, q and r are the roots of cubic equation ax3 + bx

2 + cx + d =0.

Then sum of roots taken one at time i.e. (p + q + r) = b/a.

Sum of roots taken 2 at a time i.e. (pq + pr + qr) = c/a.

And sum of roots taken 3 at a time or product of roots i.e. (pqr) = d/a.

Also, we can notice the pattern that the negative & positive sign changes alternately.

E.g. 9: If p, q and r are the roots of cubic equation 2x3 4x

2 + 6x 10 = 0, then find value of p

3 + q

3 +

r3.

Before doing anything, we should see what is the requirement to find out value of p3 + q

3 + r

3?

We know, p3 + q

3 + r

3 = (p + q+ r)[p

2 + q

2 + r

2 (pq + pr + qr)] + 3pqr.



We know the value of (p + q + r) which is sum of roots of cubic equation given equal to 2. Similarly

we also know the value of (pq + pr + qr) which is sum of roots taken 2 at a time equal to 3. And

finally, we also know value of pqr = 5.

Now, to find out value of p3 + q

3 + r

3, we should also know the value of p

2 + q

2 + r

2.

No, the value of p2 + q

2 + r

2 can be found out with the help of an identity.

So, p2 + q

2 + r

2 = [p + q + r]

2 2(pq + pr + qr] = 4 6 = 2.

Now, we can plug-in values of all of them in formula of p3 + q

3 + r

3 to obtain its value.

So, p3 + q

3 + r

3 = (2) [ 2 (3)] + 3 5 = 10 + 15 = 5.

So, this problem is an example of application of identities and relations between the roots of a cubic

equation. If we are comfortable with these formulas, we can solve these problems orally.

Exercise:

1. If p and q are the roots of x2 cx + b = 0, find the values of

a) p2 + q

2.

b) p3 + q

3.

2. If p and q are the roots of equation x2 5x + 7 =0, then frame a new equation whose roots are p/q

and q/p.

3. For what values of p will the equation x2 2x(1 + 3p) + 7(3 + 2p) = 0 have real & equal roots?

4. For what value of k will the equation x2 5k(x + 2) 2(x + 3) = 0 have roots equal in magnitude

but opposite in sign?

5. If p and q are the roots of equation 2x2 6x + 3 = 0, then frame a equation whose roots are (p q)

2

and (p + q)2.

6. If p, q and r are the roots of the cubic equation 5x3 3x

2 + 4x 3 =0, find the value of

p q rqr pr pq

+ + .

7. If the difference between the roots of the equation 2x2 mx + 15 = 0 is 1.5, find m.

8. If one of the root of the equation x2 2x 15 = 0 is the same as one root of the equation x

2 10x +

p =0, find p.

9. P is a positive integer satisfying P2 16. How many equations of the form x2 + Px + 4 = 0 exist

such that the roots are real and unequal?

10. If the equation 2x2 px + 8 = 0 has imaginary roots, then how many positive integral values can

p take?



11. Which is the least positive integral value that p can assume if the roots of the equation 4x2 px

+ 5 = 0 are real and unequal?

12. How many integral values of p satisfy the relation x2 5px + 4p

2 + 1> 0 for all real values of x?

13. If both the roots of the equation x2 6px + 2 2p =0 exceed 5, then find the range of p.

14. Find the range of p if the roots of the equation 2x2 5x + p

2 6p + 8 = 0 are opposite in sign?

15. What is the condition for one root of the quadratic equation px2 + qx + r =0 to be thrice the

other? (Answer in terms of p, q and r).

16. While solving a quadratic equation, Ram made a mistake in copying the constant term and got

the roots as 5 and 1. While Shyam made a mistake in copying the coefficient of x and got the roots

as 3 and 4. Find the correct equation and its roots.

17. If one of the roots of the equation x2 + px 28 = 0 is 4 and roots of the equation x

2 + px + r = 0

are real and equal, then find the value of r.

18. If two quadratic equations px2 + x + 5 = 0 and 2x

2 + x + q = 0 have a common root x = 2, then

which of the following statements hold true?

a) 4p + q = 17 b) p + q = 8.25 c) p q = 11.75 d) 4p q = 3

Answer:

1.) (a) c2 2b 1. (b) c

3 3bc.

2. 7x2 11x + 7 = 0. 3. 2 or 10/9 4. 0. 4 5. X

2 12x + 27 = 0

6. 31/15 7. 129 8. 25 or 39 9. None

10. 7 positive integral values 11. 9 12. 1 13. P > 1.66 or P < 11.5.

14. 2 < P < 4 15. 3q2 16pr = 0 16. The correct equation is x

2 + 4x 12 = 0, and

roots are 2, 6. 17. r = 2.25 18. (a)



Decoding Inequalities

What is basic difference between solving an Inequality and Equality problem?

Answer is pretty simple. While solving an equality problem, we get a fixed value/s of the variable but

in case of an inequality, we get a range of values.

For e.g., On solving 2x 5 = x 3, we get value of the variable x as 2.

But the same problem with inequality will give us range of values. 2x 5 > x 3; On solving this, we

get x > 2. The range signifies that x can take all the values which are greater than 2 and uptill +

infinity.

Let us see pictorial understanding of > and

So, throughout the inequality chapter we will use the notation right of or left of particular value

instead of greater or less than that particular value. This concept helps us in visualizing the number

line and gives us our required region.

AND/ OR Funda:

We use funda of AND/OR quite extensively without paying attention to its details.

For e.g., if there are two sets given, lets say.

{Classical Batsmen} = [Dravid, Kallis, Sachin, Mahela, Ponting, Clarke].

{Hardhitting batsmen} = [Hayden, Sehwag, Sachin, Kallis, Ponting].

If now, the question arises which batsmen will comprise the set who are Classical AND hardhitting

batsmen. I am sure all of us will answer this. We will find out such names in two sets which are

common to both sets. [Kallis, Sachin, Ponting].



Similarly, if the question asks which batsmen will comprise the set who are classical OR hardhitting?

Answer will be the set of all the batsmen who are uniquely present in both the sets. [Hayden,

Sehwag, Sachin, Kallis, Ponting, Mahela, Dravid, Clarke].

One more pointer in AND/OR case is that OR set will always be greater than AND set.

OR (Union): In case of OR, either of the conditions has to be satisfied. So, even if one condition is

satisfied, that element will be included in OR. In the above example, if any of the batsmen satisfy

condition of hardhitting or classical, they will be included in the set.

AND (Intersection): In case of AND, both the conditions have to be satisfied simultaneously. If only

one of the conditions is being satisfied, then that case will not be considered. In the above example,

if we are looking for set of classical batsmen AND hardhitting batsmen, then both the conditions

have to be met at same time.

Application of AND/OR in Inequality:

AND/OR is used quite frequently in inequality problems, so we will revisit this funda again. But this

time we will play with inequality signs.

E.g. 1: If x > 2 AND x < 3.



E.g. 2: If x > 2 AND x > 3.

E.g. 3: If x > 2 OR x > 3.



E.g. 4: If x < 2 AND x > 3.

E.g. 5: If x < 2 OR x > 3.

These many examples should be enough to understand the logic of AND/OR. Best way to remove

any doubts related with this logic is again OR set will be greater than AND set.



SOME RULES OF INEQUALITY:

A) Addition or subtraction: Addition or subtraction in case of inequality is exactly same as equality

problems. There is no change. We can add/subtract without any worry.

For e.g., if a < b, if we add k (k can be a positive number or a negative number) on both sides,

inequality sign will remain same.

I want to say that a + k < b + k.

Take values and verify it for yourself.

We know 5 < 8, if we add 4 on both sides we get (5 4 < 8 4) = 1 < 4.

Inequality sign still holds even if we add a negative number on both sides.

B) Multiplication or Division: We have to tread this path carefully as there are some restrictions in

this area. First of all, when we are multiplying or dividing by a number in inequality problems, we

should know the polarity (sign) of that particular number. If we dont know the sign, we cant

multiply or divide blindly. If we are multiplying or dividing by a positive number, then inequality sign

will hold true. But if we are multiplying or dividing by a negative number, then inequality sign will

reverse. We have to be utmost careful here as this is an errorprone area.

For e.g., (i) if a < b and we are multiplying by a number k on both the sides and k being a positive

number, inequality sign will not change.

It means we will get a k < b k.

We can verify it by taking values, If 5 < 7 then multiply by 2 on both sides. ( 2 is a positive number).

Inequality sign will hold true as we get 10 < 14.

(ii) if a < b and we are multiplying by a number k on both the sides with k being a negative

number, inequality sign will reverse.

It means we will get a k > b k.

If we take values like 3 < 5 and multiply by 2 on both sides, inequality sign will reverse as we will get

6 > 10 . (2 is a negative number).

Same logic holds true for division also.

C) Squaring and Cubing: We do not have issues with cubing. Cubing both sides with any inequality

sign is tensionfree. So, we dont need to worry when we need to cube both sides as inequality sign

remains the same.

For e.g. if a > b, then a3 > b

3 (irrespective of the sign of a & b).

In case of squaring, we need to be wary of signs. We will take three cases to check all the

possibilities.



(i) When both the numbers are positive; on squaring inequality sign remains same.

For e.g., If a > b, then a2 > b

2 (if a & b are +ve numbers).

If 4 > 3, then on squaring sign will hold true as we will get 16 > 9.

(ii) When both numbers are negative; on squaring inequality sign will reverse.

If a > b, then a2 < b

2 (if a & b are negative numbers).

For e.g., if 3 > 5, then on squaring inequality sign will reverse as we will get 9 < 25.

(iii) When one number is positive and other one in negative, nothing can be said of sign. It will

depend on the magnitude.

For e.g., if 5 > 3, then on squaring we get 25 > 9. In this case the inequality sign holds true as the

magnitude of 5 is greater than 3.

But if 5 > 6, then we will get 25 < 36 on squaring as the magnitude of 6 is greater than 5.

So, when one of the numbers is +ve and another one is ve, we will have to look out for magnitude

and accordingly proceed.

There are many other rules in inequality which we will come to know when we solve different

problems. We will learn those theories when we encounter those problems.

Applications of Inequalities:

E.g. 6: Solve: 2x + 5 > 2(x + 1).

This is a simple problem based on the rules of addition/subtraction.

We can rewrite this problem as 2x + 5 2x > 2.

On solving we get, 5 > 2.

We are getting an answer independent of x. Whenever this happens, we need to check the

condition whether its true or false.

Condition we are getting is 5 > 2. Is it true?

Yes, it is true. So, answer should be x can take all real values.

We could have also approached this problem in a different way.

Question is 2x + 5 > 2x + 2. Now, 2x is on both sides and 2x added to 5 will always be greater than 2x

added to 2 for any real value of x. Thats why the answer is all real values of x.

Now, we will look at how to tackle higherdegree inequalities like quadratic, cubic and so on.



E.g. 7: Solve: x2 x 6 > 0.

There are two ways of solving this problem. First way is the graphical method which we will see later

and the second method is the polarity method which we will learn now. All higherdegree

inequalities can be solved using POLARITY METHOD. Lets see how to do

POLARITY METHOD:

x2 x 6 > 0 can factorised by splitting the middle terms as (x 3) (x + 2) > 0.

Two numbers i.e. (x 3) and ( x + 2) are being multiplied to result in a positive number. (Start

reading > 0 as positive). Think what are the cases possible for two numbers being multiplied and

resulting into a positive number.

First case: Both numbers are positive i.e. both(x 3) and (x + 2) are positive.

Second Case: Both (x 3) and (x + 2) are negative.

Now, next step is to plot points on the number line where (x 3) and (x + 2) will become zero. (x 3)

will become zero at x = 3 and (x + 2) will become zero at x = 2.

So, entire number line is divided into threeregions. Those regions are x > 3, 2 < x < 3 and x < 2.

If we take any value in the first region which is x > 3, then both the brackets (x 3) and (x + 2) will be

positive and two positive numbers multiplied will be a positive number.

Take a value for x in the region x > 3. If x = 5, then (x 3) (x + 2) > 0 is true as both the brackets will

be positive and when multiplied will give us the required result. So, we can say that x > 3 is one of

the regions which satisfy our requirement.

If we take the middle region which is 2 < x < 3, then one of the brackets will be positive and other

will be negative; and multiplication of positive number with a negative number results in a negative

number. Take a value for x in the region 2 < x < 3. If x = 0, then value of (x 3) becomes 3 and

value of (x + 2) becomes 2. And when we multiply these two, we get a negative number. So, this

middle region does not suit our requirement.

If we take the leftmostregion which is x < 2 , then both the brackets will be positive and two

positive numbers multiplied always give a positive resultant.

Take a value in the region x < 2. If x = 4, then value of (x 3) becomes 7 and value of (x + 2)

becomes 2; when these two multiplied, we get a positive number which fulfils our need.

This pattern will always follow. That is, in the rightmost region, we got positive result; in the

middleregion, we got negative result; and in the leftmost region we got positive result. Observe

that the polarity sign changes alternately if we start from rightmost region to leftmost region. We

will use this funda throughout the chapter. But, what is the logic behind this funda?



The polarity funda is based on this underlying logic of leftright of a point. This logic is very useful

and will be used in subsequent chapters also. And in all the inequality problems, we will always start

from the rightmost region because we will always get positive region. And after that we dont need

to do anything as sign changes alternately.

So, you should have got the idea that for a quadratic inequality, if the question is asking for > 0,

then answer will always lie beyond the boundaries and if the question is asking < 0, then our

required region will be within the boundaries. See the picture above to get the logic.

And if we have understood this logic, we can solve these questions orally by just finding out the

limiting points (points where that particular bracket will be zero).

E.g. 8: Solve: x2 x 6 < 0.

Same question as the previous one, only difference is that we need to find range of values for x for

which the quadratic expression given in the question will be negative.

Again, we will factorize the expression, we get (x 3) ( x + 2) < 0.

We will find out the limitingpoints or boundaries where the two brackets will become zeros.

The points are same as the previous one and since the question is asking for the negative result, our

required region will lie between the boundaries.

So, answer is 2 < x < 3.



E.g. 9: Solve x2 x 6 0.

This question is similar to x2 x 6 > 0, only difference being that in previous question the value of

the quadratic expression had to be greater than zero, but in this question the value of quadratic

expression has to be either equal to zero or greater than zero.

So, the boundaries will still remain the same, means the boundaries are still 3 and 2. But, in this

question the points 3 and 2 will also be included in the answer. Because when x takes the value of

2, the value of the quadratic expression (x 3) ( x + 2) 0, will be zero and when x takes the value of

3, the value of the expression will be zero.

So, answer will be x 3 or x 2.

When there is an equal to sign with an inequality sign (i.e. or ), we dont need to do anything

different in finding out the required region. Just plot the points and if equal to sign is there, then

those points will also be included in our answer.

E.g. 10: Solve (x + 5) (x + 7) 0.

First of all, we will find out the boundaries or points where the expression given becomes zero.

Those points are 5 and 7 respectively.

Since, we are looking for the value of the expression to be negative; our answer will be within the

region. Since, in the question equal to sign is also included the boundaries (i.e. 5 & 7 will also be

included).

So, answer is 7 x 5.

So, all the expressions including two brackets or quadratic expression can be solved orally like we

have learnt just now. Let us see a variation in this type.

E.g. 11: Solve x2 x +12 > 0.

First of all, this problem is not in the form of the standard way. Whenever, the coefficient of x2 is

negative, first thing that we should do is to make the coefficient positive by multiplying the entire

expression by 1. And we have learnt that when we are multiplying by a negative number, inequality

sign reverses.

So, after multiplying, we get x2 + x 12 < 0, now this problem is in standard format. We can go ahead

and solve this question with the logic learnt.



On factorizing, we get(x + 4) ( x 3) < 0. So, the boundaries are (4) and (3) respectively and

whenever the quadratic expression is negative (less than zero), required answer lies within the

region.

So, answer is [4 < x < 3].

So, dont forget to convert the coefficient of x2 if its negative.



E.g. 12: Solve (x)(x 2) ( x + 3) > 0.

This is a cubic inequality which can be solved using the polarity logic. Just plot the points on the

number line where each of the brackets becomes zero. After that, we will start from the rightmost

region since we always get a positive result in that region. After that, we know that the sign changes

alternately in the adjacent regions.

So, answer is x > 2 OR 3 < x < 0.

E.g. 13: Solve (2x + 3) ( x 5) (x + 4) 0.

Again a cubic inequality, so we should find out the points where each of the brackets will become

zero. First bracket will be zero when x = 3/2 or 1.5. Second bracket will become zero when x = 5

and third bracket will become zero at x = 4. Plot these points on the number line.



So, answer is 1.5 x 5 OR x 4. Notice, equal to sign is also included in the answer.

E.g. 14: For how many integral values of x, is the expression x(x 5) (2x 6) (x + 3) 0?

Most of the students get confused in these types of questions despite being based on the same

logic. Only difference in this question compared to those which we have done is that in this question

range is not being asked, but number of integral values of x is being asked. If we could find the

range for which the expression is less than or equal to zero, we can also count the number of

integral values of x for which this expression will be true.

Let us find out the range first.

The boundaries are 0, 5, 3 and 3 respectively. Plot them on number line and start from the right

most region because we always get positive there.



If range was being asked, answer is quite visible which is [3 x 5 OR 3 x 0].

But, since the number of integral values is being asked, we will count integers in the range found

above. So, answer is 7 integral values. [Be careful, the boundaries will also be included in the answer

as the value of the expression can be zero also].

E.g. 15: Solve: 2 05

x

x

>

.

Now, in this problem expression is being divided. How to solve this?

If the problem was (x 2)(x 5) > 0, we would have solved it orally and got the answer as x > 5 OR x

< 2.

You would be surprised that answer for 2 05

x

x

>

is also same. Why?

For, (x 2) (x 5) > 0, how many scenarios are possible to multiply two numbers to get a positive

result.

There are only two cases possible: a) Both numbers are negative or b) Both numbers are positive.

If you observe carefully, polarity logic will also hold true for division. For, 2 05

x

x

>

, which cases are

possible to divide two numbers to get a positive result. Again same two cases two possible; either

both numbers are positive or negative.

Thats why the answer for both expressions is same.



Even if the question was asking to solve 2 05

x

x

2 OR 4 x 1].

Remember, denominator becomes zero at x =2, and since the question is asking for values for which

the expression becomes greater than or equal to zero, we will have to exclude 2 from our answer.

Inequalities involving Perfect squares.

E.g. 18: Solve x2 > 0.

We need to find the range of values of x for which x2 will be positive. We know x

2 is a perfect

square and irrespective of whichever value of x, x2 will always be positive. Even if we substitute the

value of x as negative number, x2 will be positive.

So, answer should be x can take all real numbers except x = 0. x cannot take value 0 because if we

substitute x = 0 in x2, value of x

2 will become 0 and the question is asking for greater than 0.

E.g. 19: Solve (x 2)2 0.

Again (x 2)2 is a perfect square and it will be positive for all real values of x. And in the question,

we need to find values of x for which the expression (x 2)2 will be greater than or equal to zero.

So, x can take value of 2 also as in question equality sign is included.

E.g. 20: Solve (x + 7)2 < 0

Answer should be no solution as (x + 7)2 is a perfect square and it cannot be negative for any real

values of x.



E.g. 21: Solve ( x + 5)2 (x 5) 0.

Now, this problem consists of two terms in which one of the terms is a perfect square. We can solve

all such problems by ignoring the term involving perfect square as the perfect squares are always

positive. So, answer is not dependent on them. Still if you are not clear with logic, we will go back to

original logic of polarity.

Logic: When two terms are being multiplied to result into positive, only two cases are possible.

First case: Both the terms are negative. This case is not valid since the term (x + 5)2 cannot be

negative as it is a perfect square.

Second case: Both the terms are positive. This case is valid since (x + 5)2 is positive. And since (x + 5)

2

is positive, second term also has to be positive.

So, our question reduces to (x 5) 0. [Thats why i stated initially that we can ignore the perfect

square as it is always positive, so answer is dependent on the second term only].

So, on solving answer is x 5, but we need to check that the term which we ignored becomes zero at

what point. And the question asks for expression to be greater than or equal to zero, so we will have

to include that point in answer.

(x + 5)2 becomes zero when x = 5 and we got our answer as x 5. X = 5 does not lie in the second

region, so we will have to write it separately.

So, our final answer is x 5 and x = 5.

E.g. 22: Solve (x 3)2 ( x 5) < 0.

Again the first term is a positive square, so we can ignore it initially as it is always positive.

So, question reduces to ( x 5) < 0.

And answer is x < 5, again we need to check for the term which we ignored where it becomes zero.

The term ( x 3)2 becomes zero at x = 3 and since the question is asking for all values of x for which

the value of expression is less than zero and not equal to zero.

So, if x = 3 comes in our answer region, we will have to exclude it. On checking, we can see that x = 3

lies in the region x < 5, so we will have to exclude it.

So, final answer is x < 5 and x 3.

E.g. 23: Solve: ( )( )2

20

3x

x

.

Again, the denominator is a perfect square, so we can ignore it and work on the numerator.

Our question reduces to (x 2) 0. So, the answer is x 2. But, again we need to check where the

denominator becomes zero. The denominator ( x 3)2 becomes zero x = 3. And the question is

asking for values of x for which the expression is greater than or equal to zero. But, we have to also



take in mind that the (x 3) 2 is in denominator and it cannot be equal to zero as if the denominator

becomes zero, the entire expression will become undefined.

So, if x = 3 comes in answer region x 2, we will have to exclude it. And x = 3 lies in the answer

region. So, our final answer is x 2 and x 3.

So all the questions involving perfect squares can be solved pretty easily, but we just need to be

careful in writing our final answer depending on whether the expression is asking greater than or

equal to or less than zero.

E.g. 24: ( )( )

220

3x

x

In this question, numerator is a perfect square and we can ignore it and go ahead with our question

and solve it.

So, our question reduces to ( )1 0

3x

, for which the answer is x > 3. [Be careful, there will be no

equality sign in answer as the term (x 3) is in denominator].

Now, we need to check for the perfect square which we ignored. The term ( x 2)2 becomes zero at

x = 2 and since the question is asking foe values of x for which the expression is greater than or

equal to zero, we will have to include x = 2 in our answer.

But, x = 2 does not lie in the answer region x > 3, so we will have to write it separately.

Final answer is x > 3 and x = 2.

E.g. 25: Solve (x 5)3 (x + 3) 0.

Now, cube comes in picture, but dont worry we can handle it by manipulating the question.

We can rewrite the question as (x 5 )2 (x 5) ( x + 3) 0.

Now, we know how to handle inequality with perfect square, we can ignore the perfect square and

solve the remaining inequality.

Again, our question reduces to (x 5) ( x + 3) 0.

We know how to find the answer of such inequalities, answer always lies outside the boundaries and

the boundaries are + 5 and 3.

So, answer is x 5 OR x 3.



E.g. 26: Solve x2 + x + 1 > 0.

There are two ways of solving this problem.

First Method: For any real value of x, x2 will be always positive. And even if x is negative, x

2 will

always positive and when x is added to x2, x

2 + x will be still a positive number. And when we add 1

to x2 + x, the resultant x

2 + x + 1 will always be positive.

So, we reached the conclusion that x2 + x + 1 will always be positive for any real value of x and

question is also asking for positive value. So, answer is all real values of x.

Second method: Whenever we encounter a quadratic expression, we try to split the middle term to

find the roots. But, in this case we cannot split the middle term. So, here should immediately find

out the determinant of the expression.

D = b2 4ac= 1 4 = 3.

If D is negative and coefficient of x2 is positive, the graph of the expression y = x

2 + x + 1 will always

be above the xaxis. It would not cut or touch the xaxis because if it touches or cuts the Xaxis,

then the roots become real which is a contradiction to the result obtained.

So, whenever, the quadratic expression cannot be factorized, we need to find the determinant

immediately. If the determinant is negative and the coefficient of x2 is positive, value of that

particular quadratic expression is always positive.

E.g. 27: Solve (x2 + 2) (x

2 10x 24) < 0.

Again the first term is a quadratic expression which cannot be factorized and if we find out its

determinant, it is negative. So, x2 + 2 is always positive as the roots are imaginary and the graph will



always lie above the Xaxis. So, we can ignore it completely as it would not have any effect on

inequality.

So, our question reduces to x2 10 x 24 < 0 which can be further split into (x 12) ( x + 2) < 0.

We know how to solve such inequalities orally and answer is 2 < x < 12.

When there is some other number on right hand side of inequality apart from zero:

E.g. 28: Solve ( x 2)2 >49.

Best way to solve such problems is to bring whatever is on the righthand side to lefthand side

because that is our standard format. If we do not do that, we will have to two cases to solve it which

will consume slightly more time.

So, we can rewrite the question as (x 2)2 49 > 0.

Or (x 2)2 7

2 > 0; ( x 2 + 7) ( x 2 7) > 0.

We get (x + 5) ( x 9) > 0.

So, the answer is x > 9 OR x < 5.

E.g. 29: Solve 1 1

8x> .

Again we will bring whatever on the righthand side to LHS.

So, we get 1 1 80 0

8 8x

x x

> = > .

We have learnt that whenever the coefficient of x is negative, we need to make it positive and for

that we will have to multiply the expression by (1).

After multiplying, we get 8 0

8x

x

< .

We know how to solve it as boundaries are 8 and 0 respectively.

So, the answer is 0 < x < 8.

E.g. 30: Solve 3 21

x

x

+>

.

Again we will have to get 2 on LHS and solve it thereafter.

3 5 52 0 0 0 1 51 1 1

x x xx

x x x

+ + > > < < 0.

4. (x2 4) (x

2 9) 0.

5.

2

23 0

3x

x x

+

.

6.

2

23 0

6 9x

x x

+ +

.

7. (x2 16) (x + 5)

2 0.

8. (x + 8)3 (x

2 25) 0.

9.

2

24 0

8 16x x

x x

+ +

.

10. (2x + 3)2 < 49.

11. 7x 2x2 + 15 0.

12. (x 9)2 (x 7)

3 (x

2 2x 143) 0.

13.

3 2

2( 3) ( 7) 0( 4) ( 9)x x

x x

+ +

.



14. 211 3( 2)

x

x

+.

15.

2

212 2( 3)

x

x

0

19. (x2 4) (x + 5) 0.

20. (x2 + 5x + 7) ( x + 5) 0.

21. 1 1

5x x , then also the graph would had remained same, but only difference would had been that the V-shaped line would be excluded since equality sign is

missing. So, the graph would have been the area denoted in the previous graph by red arrows.

And if the question was to plot the graph of 2 3y x< , the graph would had remained same, but this time the required area will be below the V-shaped line excluding the V-shaped line since the

question is asking for less than.



We learnt how to plot the graphs of modulus with vertical and horizontal movements. Now, we will

look at some of the application of those graphs.

Type I:

E.g. 1: Find the area of the region enclosed by the graphs of 4 5y x= and the X-axis.

Now herein lays the advantage of learning how to plot the graph immediately and then finding out

the area of the required region.

The graph will be V-shaped made at (4, 5). And we do not need an exact graph. We just want to

know the points where the graph will cut the X-axis.

There are many ways of finding the area of the required region. We will look at two of those

methods and both those methods are pretty easy.

First Method:

First of all, to find the area of the triangle, we should know the base and height of that triangle.

Height is clearly visible as it will be the y-intercept. So, the height of the given triangle is (5 units),

and our objective should be to find out the base of the triangle. Base of the triangle is the X-axis line

enclosed between the V-shaped modulus graphs. So, if we can find out the co-ordinates of the end-

points of V-shaped modulus graph intersecting the X-axis, we can easily find out the distance

between the two points which will serve our purpose.



All of us know what is a slope and what does it mean. We also know thaty

slopex

=

. This means

that slope is the ratio of change between the Y-co-ordinates and X-xo-ordinates. And also the slope

of the line given in the question is 1. What can we conclude if the slope of a line 1?

If the slope of line is 1, it means whatever is the change in Y-coordinate, same magnitude of change

will also take place in X-coordinate.

Now, the value of x-coordinate is 1 and +9. Can we find out the distance between these two points

which will give us the length of the triangle?

Yes, the distance is 9 (1) = 10 units and the height is 5 units.

So, area of the required area is = (1/2) 10 5 = 25 square units.



Alternative method:

This is the far better efficient way of solving the question related with area problems. We just need

to find out how much is the slope of the line, the perpendicular will bisect the base into two equal

parts. Depending on what is the change in Y-coordinate and slope of the line, we can find out the

two parts of base and add them to find the length of base of triangle.



E.g. 2: Find the area of the region enclosed by the graphs of1 2 42

y x= + .

Again, we need to plot the graph immediately. The graph will be inverted V-shaped since negative

sign attached to modulus and the origin has shifted to +4. [1 22

x Will become zero at x = 4] and

the y-intercept is +4.

So, area of the required area is [1/2 16 4] = 32 square units.



E.g. 3: Find the area of the region enclosed by the graphs of 2 4 4y x= + and the X-axis.

After solving, two examples we should understand that drawing graph is not necessary to find the

area. We can find out the area orally. To find the area we need the height and base of the triangle.

Height is already given in the question as the y-intercept is 4.

And base can be found out with the help of slope and y-intercept. Slope in this case is 2 and change

in y-intercept is 4 {as y-co-ordinate changes from 4 to 0}.

So, 42 2yslope x

x x

= = =

Remember the triangle in previous question, half of the base is 2, so length of base will be 4. Now

the base is 4 and the height is also 4, so area of the triangle is 8 square units.

See the question can be solved orally without plotting the graph. So, once you get a hang of plotting

graphs, you can solve many questions orally.

E.g. 4: Find the area of region enclosed by the graphs of 3 4 52

y x= and the X-axis.

Again, we should not waste our time in plotting the graph. We can say that height of the triangle is 5

units and to find out the base, we will take help of slope. Slope of the line is 3/2 and change in the y-

intercept is 5 {as y-coordinate changes from 5 to 0}.

So, 3 5 102 3

yslope x

x x

= = =

Again half of the base of the triangle is 10/3, so the length of total base will be 20/3.

1 20 5052 3 3

Area =

So, all such questions can be handled without any difficulty if the basic understanding of graph is

clear.



Type II:

E.g. 5: How many integral points will satisfy 3 4y x + and y 0.

This is a different type of question than what we have learnt. But the logic is still hidden in the funda

of graph only.

Let us plot the graph first; we have already seen that whenever there is an inequality sign involved,

we first plot the graph for equal to and then determine which region will be applicable if > or 0 above X-axis. So, the region satisfying y 0 will be X-axis and region

above X-axis.

The graph will be an inverted V-shaped made at (3, 4).

We will have to count all integral points on X-axis, points on inverted V-shaped and all the points

lying in the region enclosed between them since equality sign is included in question.



E.g. 6: How many integral points will satisfy 3 4y x + and y > 0.

This is also the same question, only difference being that we will not take any point on X-axis as y >

0. Apart from that remaining all points will be counted. So, the simplest logic for finding out integral

points will be to count integral points on X-axis which will help us in determining points on lines

above X-axis. And then in final answer we will ignore points on X-axis.



E.g. 7: How many integral points will satisfy 3 4y x< + and y 0.

Again same type of question, only difference being that equality sign is missing. The logic still

remains same; we will count all the integral points on X-axis and points lying inside the region

enclosed. We cannot take any points lying on Inverted V-shaped sign as those points will be valid

when equality sign was included. So, we will have to exclude them.



E.g. 8: How many integral points will satisfy 3 4y x< + and y > 0.

This is the fourth variation of this problem. In this case, we would not count any points on X-axis as y

> 0 and as well as any points lying on inverted V-shaped graph. We will just have to count integral

points lying in the enclosed region.



E.g. 9: How many integral points will satisfy 2 2 4y x + and y 0.

In this question slope of the line is 2, but no need to worry. We will still get the answer by following

the learnt logic. We will have to count all the integral point on inverted V-shaped graph, as well as all

the points on X-axis and also all the points enclosed between region.

The change in y-co-ordinate is 4, so change in x-coordinate will be 2.

42 2 2y xx x

= = =

.



Some Practice questions:

1. How many integral points will satisfy 2 5y x and y 0.

2. How many integral points will satisfy 3 9 6y x and y 0.

3. How many integral points will satisfy 1 2 32

y x + and y > 0.

4. Find the area of region enclosed by the graphs of 2 4 53

y x= + and the X-axis.

5. Find the area of region enclosed by the graphs of 1 4 42

y x= and the X-axis.



Type III:

E.g. 1: Find the area and perimeter of the region enclosed by the graphs of 5x y+ = .

First of all, we will learn how to plot the graph of this. Once, we know how to plot, we can find out

the area and perimeter orally without plotting the graph.

If there is a single modulus involved in expression, we get two equations. But if two mods are

involved, we will get four equations under different conditions. In this case also, we will get 4

equations.

1st

equation: x + y = 5 if x 0 and y 0.

2nd

equation: x y = 5 if x 0 and y < 0.

3rd

equation: x + y = 5 if x < 0 and y 0.

4th

equation: x y = 5 if x < 0 and y < 0.

So, we get four different equations with different conditions. So we need to plot these lines and use

the conditions in which quadrant x and y is positive or negative.



Let us plot the graphs of all the four lines and see what we get.



If we merge all four graphs, we will get a square or a rhombus or a diamond. If we did this much of

hard work in plotting, we can easily plot any graph of this type orally now if point changes from 5 to

any other point.



One thing we should understand in case of ABSOLUTE VALUE or MODULUS graphs like 5x y+ = , 5

is the distance from the origin. In this case x becomes zero at 0 and y also becomes zero at 0.

Thats why the graph is being made at (0, 0) and the distance is 5 units from origin in left-hand

direction, right-hand direction, upwards and downwards. So, we should understand that even if

origin changes or distance changes, logic remains the same.

Now, let us get the answer of question asked.

So, if the same question was asked to find the area of 7x y+ = , we can find out the answer orally. Again, the shape will be of a square and made at point (7, 0), (0, 7), (7, 0) & (0, 7). So, the side of

square will be 72 and area can be found out by squaring the side of square.



E.g. 2: Find the area of the region enclosed by the graph of 1 1 6x y + + = .

If we understood the previous graph thoroughly, we can find solve this orally.

Let us try to understand. In the question 5x y+ = , the origin of the graph is (0, 0) as the ,x ywill become zero at 0. And the distance from the origin is 5. Only difference in the new question is

that origin has shifted to (1, 1) as ,x y will become zero at (1) and (1) respectively. But, the

distance will be 6 from the origin leftwards, rightwards, upwards and downwards.

So, whatever is the area enclosed by 6x y+ = , the same will be the area also for

1 1 6x y + + = .

The points where the square will be made can be found out by adding or subtracting 6 with the

origin depending whether we are moving in left/right or upwards or downwards.

So, the points will be (1, 1) is the origin, (7, 1), (1, 5), (5, 1) and (1, 7).



E.g. 3: How many integral points will satisfy 5x y+ = ?

There are two ways of solving these types of questions.

Algebraic method:

When x = 0, y can take two value which is 5. (2 solutions).

When x = 1, y can take two values which is 4. [Total of 4 solutions i.e. (1, 4), (1, 4), (1, 4) and (1,

4)].

When x = 2, y can take two values which is 3. [Total of 4 solutions].

When x = 3, y can take two values which is 2. [Total 4 solutions].

When x = 4, y can take two values which is 1. [Total 4 solutions].

When x = 5, y can take just one value which is 0. [Total of 2 solutions].

So, when we add up all the solutions, we get a total of 20 solutions.



Graphical Method:

We can also find out number integral points by plotting the graph.

So, we can find out number of integral points by counting all the integral points on four lines of the

square. And answer can be found out orally by multiplying by 4 with whatever is the distance from

the origin.

In this case, question was 5x y+ = , distance from the origin was 5, so we could have directly multiplied distance 5 by 4 to get 20 integral points.

If the question was to find the number of integral points which will satisfy 6x y+ = , we could have directly found the integral points by multiplying distance 6 with 4 giving us 24 points.



E.g. 4: How many integral points will satisfy 5x y+ ?

In this question also, we need to find the integral points, but only difference with previous question

is that in this question we will have to count integral points on the four lines as well as in the region

enclosed by four lines also.

We will learn this question how to count in such cases and after that such cases can be solved

without plotting the graph.



Let us see how to count points.

We already know integral solutions for square made at 5: 20 integral solutions.

Integral solutions for square made at 4: 4 4 = 16 integral solutions.




And dont forget to count the origin (0, 0): 1 integral point.

So, total number of integral solutions to 5x y+ : 20 + 16 + 12 + 8 + 4 + 1 = 61 integral solutions.



E.g. 5: How many integral points will satisfy 7x y+ < ?

In this question, we do not need to plot the graph if we have understood the previous question well.

We need to find integral points for less than 7, means that we have to count integral points for

square made at 6, 5, 4, 3, 2, 1 and the origin.

We can just write the solutions for each of them and add them to get final solution.


Integral solutions for square made at 5: 5 4 = 20 integral solutions





And dont forget to count the origin (0, 0): 1 integral point.

On adding, we get total number of solutions as 105 integral solutions.

E.g. 6: How many integral points will satisfy 1 1 5x y + + ?

We have already seen the shape of graph of 1 1 5x y + + will be same as that of 5x y+ = .

Only difference will be that origin has shifted from (0, 0) to (1, 1). But the number of integral points

remains the same as the distance from the origin is still 5.

So, the answer for 1 1 5x y + + will remain same as that for 5x y+ .

We had calculated answer for this question previously as 61 integral solutions.

E.g. 7: Plot the graph of 4x y = .

Let us learn how to plot this graph also. If we are doing it for the first time, we need to use the

definition of modulus and frame four equations under four different conditions.

1st

equation: x y = 4; if x 0 & y 0.

2nd

equation: x + y = 4; if x 0 & y < 0.

3rd

equation: x y = 4; if x < 0 & y 0.

4th

equation: x + y = 4; if x < 0 & y < 0.

Let us plot all of these four lines.



On merging all four graphs, we will get this final figure.

E.g. 8: Plot the graph of 4y x = .

In this question also, we will get four different lines with four different conditions.

1st

equation: y x = 4; if x 0 & y 0.

2nd

equation: x + y = 4; if x < 0 & y 0.

3rd

equation: x y = 4; if x 0 & y < 0.

4th

equation: y + x = 4; if x < 0 & y < 0.

Let us plot all of these four lines.

If we understood how to plot the graph of 4x y = , then we can plot the graph of 4y x = in the same way, and all of u should have guessed also the graph.

The graph will look like this.



E.g. 9: Find the area of the region enclosed by the graphs of 4y x = and 7y = .

We know the graph of 4y x = . And the graph of 7y = will be line made at (0, 7) and (0, 7). In

other words, we can also say that lines of 7y = will be parallel to X-axis.

By now, most of you would have guessed the graph of 7x = will be lines made at (7, 0) and (7, 0).

So, the lines of 7x = will be parallel to Y-axis.

Let us plot both the graphs and find the area of the enclosed region.



ARITHMETIC PROGRESSIONS 93

55.1 Introduction

You must have observed that in nature, many things follow a certain pattern, such asthe petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, thespirals on a pineapple and on a pine cone etc.

We now look for some patterns which occur in our day-to-day life. Some suchexamples are :

(i) Reena applied for a job and got selected. Shehas been offered a job with a starting monthlysalary of Rs 8000, with an annual increment ofRs 500 in her salary. Her salary (in Rs) for the1st, 2nd, 3rd, . . . years will be, respectively

8000, 8500, 9000, . . . .(ii) The lengths of the rungs of a ladder decrease

uniformly by 2 cm from bottom to top(see Fig. 5.1). The bottom rung is 45 cm inlength. The lengths (in cm) of the 1st, 2nd,3rd, . . ., 8th rung from the bottom to the topare, respectively

45, 43, 41, 39, 37, 35, 33, 31

(iii) In a savings scheme, the amount becomes 54

times of itself after every 3 years.

The maturity amount (in Rs) of an investment of Rs 8000 after 3, 6, 9 and 12years will be, respectively :

10000, 12500, 15625, 19531.25

ARITHMETIC PROGRESSIONS

Fig. 5.1



94 MATHEMATICS

(iv) The number of unit squares in squares with side 1, 2, 3, . . . units (see Fig. 5.2)are, respectively

12, 22, 32, . . . .

Fig. 5.2

(v) Shakila put Rs 100 into her daughters money box when she was one year oldand increased the amount by Rs 50 every year. The amounts of money (in Rs) inthe box on the 1st, 2nd, 3rd, 4th, . . . birthday were

100, 150, 200, 250, . . ., respectively.(vi) A pair of rabbits are too young to produce in their first month. In the second, and

every subsequent month, they produce a new pair. Each new pair of rabbitsproduce a new pair in their second month and in every subsequent month (seeFig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start ofthe 1st, 2nd, 3rd, . . ., 6th month, respectively are :

1, 1, 2, 3, 5, 8

Fig. 5.3




In the examples above, we observe some patterns. In some, we find that thesucceeding terms are obtained by adding a fixed number, in other by multiplyingwith a fixed number, in another we find that they are squares of consecutivenumbers, and so on.

In this chapter, we shall discuss one of these patterns in which succeeding termsare obtained by adding a fixed number to the preceding terms. We shall also see howto find their nth terms and the sum of n consecutive terms, and use this knowledge insolving some daily life problems.

5.2 Arithmetic ProgressionsConsider the following lists of numbers :

(i) 1, 2, 3, 4, . . .(ii) 100, 70, 40, 10, . . .(iii) 3, 2, 1, 0, . . .(iv) 3, 3, 3, 3, . . .(v) 1.0, 1.5, 2.0, 2.5, . . .

Each of the numbers in the list is called a term.Given a term, can you write the next term in each of the lists above? If so, how

will you write it? Perhaps by following a pattern or rule. Let us observe and write therule.

In (i), each term is 1 more than the term preceding it.In (ii), each term is 30 less than the term preceding it.In (iii), each term is obtained by adding 1 to the term preceding it.In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding

(or subtracting) 0 to the term preceding it.In (v), each term is obtained by adding 0.5 to (i.e., subtracting 0.5 from) the

term preceding it.In all the lists above, we see that successive terms are obtained by adding a fixed

number to the preceding terms. Such list of numbers is said to form an ArithmeticProgression ( AP ).

So, an arithmetic progression is a list of numbers in which each term isobtained by adding a fixed number to the preceding term except the firstterm.

This fixed number is called the common difference of the AP. Remember thatit can be positive, negative or zero.



96 MATHEMATICS

Let us denote the first term of an AP by a1, second term by a2, . . ., nth term byan and the common difference by d. Then the AP becomes a1, a2, a3, . . ., an.So, a2 a1 = a3 a2 = . . . = an an 1 = d.Some more examples of AP are:

(a) The heights ( in cm ) of some students of a school standing in a queue in themorning assembly are 147 , 148, 149, . . ., 157.

(b) The minimum temperatures ( in degree celsius ) recorded for a week in themonth of January in a city, arranged in ascending order are

3.1, 3.0, 2.9, 2.8, 2.7, 2.6, 2.5(c) The balance money ( in Rs ) after paying 5 % of the total loan of Rs 1000 every

month is 950, 900, 850, 800, . . ., 50.(d) The cash prizes ( in Rs ) given by a school to the toppers of Classes I to XII are,

respectively, 200, 250, 300, 350, . . ., 750.(e) The total savings (in Rs) after every month for 10 months when Rs 50 are saved

each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.It is left as an exercise for you to explain why each of the lists above is an AP.You can see that

a, a + d, a + 2d, a + 3d, . . .represents an arithmetic progression where a is the first term and d the commondifference. This is called the general form of an AP.

Note that in examples (a) to (e) above, there are only a finite number of terms.Such an AP is called a finite AP. Also note that each of these Arithmetic Progressions(APs) has a last term. The APs in examples (i) to (v) in this section, are not finite APsand so they are called infinite Arithmetic Progressions. Such APs do not have alast term.

Now, to know about an AP, what is the minimum information that you need? Is itenough to know the first term? Or, is it enough to know only the common difference?You will find that you will need to know both the first term a and the commondifference d.

For instance if the first term a is 6 and the common difference d is 3, thenthe AP is

6, 9,12, 15, . . .and if a is 6 and d is 3, then the AP is

6, 3, 0, 3, . . .



sarithaTypewritten Text


Similarly, whena = 7, d = 2, the AP is 7, 9, 11, 13, . . .a = 1.0, d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3, . . .

a = 0, d = 112

, the AP is 0, 112

, 3, 412

, 6, . . .

a = 2, d = 0, the AP is 2, 2, 2, 2, . . .So, if you know what a and d are, you can list the AP. What about the other way

round? That is, if you are given a list of numbers can you say that it is an AP and thenfind a and d? Since a is the first term, it can easily be written. We know that in an AP,every succeeding term is obtained by adding d to the preceding term. So, d found bysubtracting any term from its succeeding term, i.e., the term which immediately followsit should be same for an AP.

For example, for the list of numbers :6, 9, 12, 15, . . . ,

We have a2 a1 = 9 6 = 3,a3 a2 = 12 9 = 3,a4 a3 = 15 12 = 3

Here the difference of any two consecutive terms in each case is 3. So, thegiven list is an AP whose first term a is 6 and common difference d is 3.

For the list of numbers : 6, 3, 0, 3, . . .,a2 a1 = 3 6 = 3a3 a2 = 0 3 = 3a4 a3 = 3 0 = 3

Similarly this is also an AP whose first term is 6 and the common differenceis 3.

In general, for an AP a1, a2, . . ., an, we haved = ak + 1 ak

where ak + 1 and ak are the ( k + 1)th and the kth terms respectively.To obtain d in a given AP, we need not find all of a2 a1, a3 a2, a4 a3, . . . .

It is enough to find only one of them.Consider the list of numbers 1, 1, 2, 3, 5, . . . . By looking at it, you can tell that the

difference between any two consecutive terms is not the same. So, this is not an AP.



98 MATHEMATICS

Note that to find d in the AP : 6, 3, 0, 3, . . ., we have subtracted 6 from 3and not 3 from 6, i.e., we should subtract the k th term from the (k + 1) th termeven if the (k + 1) th term is smaller.

Let us make the concept more clear through some examples.

Example 1 : For the AP : 32

, 12

, 12

, 32

, . . ., write the first term a and the

common difference d.

Solution : Here, a =32 , d =

12

32 = 1.

Remember that we can find d using any two consecutive terms, once we know thatthe numbers are in AP.

Example 2 : Which of the following list of numbers form an AP? If they form an AP,write the next two terms :

(i) 4, 10, 16, 22, . . . (ii) 1, 1, 3, 5, . . .(iii) 2, 2, 2, 2, 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .

Solution : (i) We have a2 a1 = 10 4 = 6a3 a2 = 16 10 = 6a4 a3 = 22 16 = 6

i.e., ak + 1 ak is the same every time.So, the given list of numbers forms an AP with the common difference d = 6.The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.

(ii) a2 a 1 = 1 1 = 2a3 a 2 = 3 ( 1 ) = 3 + 1 = 2

a 4 a3 = 5 ( 3 ) = 5 + 3 = 2i.e., ak + 1 ak is the same every time.So, the given list of numbers forms an AP with the common difference d = 2.The next two terms are:

5 + ( 2 ) = 7 and 7 + ( 2 ) = 9(iii) a2 a 1 = 2 ( 2) = 2 + 2 = 4

a3 a 2 = 2 2 = 4As a2 a1 a3 a 2 , the given list of numbers does not form an AP.




(iv) a2 a1 = 1 1 = 0a3 a2 = 1 1 = 0a4 a3 = 2 1 = 1

Here, a2 a1 = a3 a2 a4 a3.So, the given list of numbers does not form an AP.

EXERCISE 5.11. In which of the following situations, does the list of numbers involved make an arithmetic

progression, and why?(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each

additional km.(ii) The amount of air present in a cylinder when a vacuum pump removes 1

4 of the

air remaining in the cylinder at a time.(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for

the first metre and rises by Rs 50 for each subsequent metre.(iv) The amount of money in the account every year, when Rs 10000 is deposited at

compound interest at 8 % per annum.2. Write first four terms of the AP, when the first term a and the common difference d are

given as follows:

(i) a = 10, d = 10 (ii) a = 2, d = 0

(iii) a = 4, d = 3 (iv) a = 1, d = 12

(v) a = 1.25, d = 0.25

3. For the following APs, write the first term and the common difference:(i) 3, 1, 1, 3, . . . (ii) 5, 1, 3, 7, . . .

(iii)1 5 9 13, , , ,3 3 3 3 . . . (iv) 0.6, 1.7, 2.8, 3.9, . . .

4. Which of the following are APs ? If they form an AP, find the common difference d andwrite three more terms.

(i) 2, 4, 8, 16, . . . (ii) 5 72, , 3, ,2 2

. . .

(iii) 1.2, 3.2, 5.2, 7.2, . . . (iv) 10, 6, 2, 2, . . .

(v) 3, 3 2+ , 3 2 2+ , 3 3 2,+ . . . (vi) 0.2, 0.22, 0.222, 0.2222, . . .(vii) 0, 4, 8, 12, . . . (viii)

12

, 12

, 12

, 12

, . . .



100 MATHEMATICS

(ix) 1, 3, 9, 27, . . . (x) a, 2a,

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