Vedanta Opt. Maths Book 10 Final (2078)

vedanta Excel In Opt. Mathematics - Book 10

( )= a b x
x2 + a +c

{ ( ) ( ) }= a b
x2 + 2.x. 2a + b 2- b2 +c
2a 2a

{( ) }= ax+ b 2 - b2 +c
2a 4a

{( ) }= ax + b 2 + 4ac-b2
2a 4a

Which is in the from of y = a(x - h)2 + k

Where, h = - b , k = 4ac-b2
2a 4a

(h, k) is called vertex of the parabola

( )?
(h, k) = – b , 4ac-b2
2a 4a
The line of symmetry of the parabola is x = - 2ba.

Worked out Examples

Example 1. Draw the graph of y = x2 - 2x - 3.
Solution:
Given equation is y = x2 - 2x - 3
Example 2.
Solution: This equation is quadratic in x and represents a parabola. Comparing the
given equation with y = ax2 + bx + c, we get,

a = 1, b = -2 and c = -3

( )Vertex (h, k)
= - b , 4ac-b2
2a 4a
Y
( )= 6
-2 , 4.1.-3-(-2)2 = (1, -4) 5
2.1 4a 4
3
To draw the curve of parabola, we make a table, 2
1
x -2 -1 0 1 2 3 4 X' -5 -4 -3 -2 -1-1O 1 2 3 4 5 X
y 5 0 -3 -4 -3 0 5 -2
-3
Plotting these points on graph paper, we get the -4
curve of parabola as shown in the figure. -5

Draw the graph of y = x2 + 5x + 6 Y'

Here, y = x2 + 5x + 6

This is quadratic function of the form

y = ax2 + bx + c. It represents a parabola.

Here, a = 1, b = 5, c = 6 to find the vertex.

x - coordinate of vertex = (h) = - b = - 5 = - 2.5
2a 2

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y - Coordinate of vertex = (k) = 4ac - b2
4a
4.1.6 - 52
= 4.1

= 1 = 0.25 Y
4 12
Vertex = (h,k) = (-2.5, –0.25) 11
10
To draw the curve, we make table. 9
8
x -6 -5 -4 -3 -2.5 -1 0 7
6
y 12 6 2 0 -0.25 2 6 5
4
These points are plotted on a graph, we get a X' 3 X
parabola as shown in the figure. 2
1O
-5 -4 -3 -2 -1 1 2 3 4 5

Y'

Example 3. Draw graph y = x2 - 6x + 8. Also find the line of symmetry of the curve.
Solution:
Given equation is y = x2 - 6x + 8

comparing it to y = ax2 + bx + c, we get,

a = 1, b = - 6 and c = 8

To find the vertex of parabola.

h =x - coordinate of vertex = - b = - (-6) = 3
2a 2.1

k = y - coordinate of vertex

= 32 - 6.3 + 8 = 9-18 + 8 =-1 [ y = x2 - 6x + 8]

vertex = (3, -1)

To draw the curve, we make a table for points.

x -1 0 1 2 3 4 5 6 7
y 15 8 3 0 -1 0 3 8 15

Plotting the points and joining them, we get a curve of parabola.

From the curve, we observe that a line passing Y
through the vertex (3, -1), parallel to
Y-axis divides the curve into two equal 16
halves. Let the equation of line of 14
symmetry be y = b. 12
10
It passes through the vertex (3, -1) 8
? b=3 6
Hence, the line of symmetry of the parabola is 4
2O
y = 3. X' -10 -8 -6 -4 -2-2 2 4 6 8 10 X

-4
-6

To find the equation of the parabola. Y'

From the given conditions and graph of parabola, the following conditions can be used to
find the equation of parabola.

(i) Equation of the parabola is y = ax2, when the vertex is at the origin and a point in the
curve is given.

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(ii) Equation of the parabola is y = ax2 + c, when the vertex is on Y - axis and a point on
the curve is given.

(iii) Equation of parabola is y = a(x - h)2, when the vertex is on the X - axis and a point on
the curve is given.

(iv) Equation of parabola is y = ax2 + bx + c, when at least three points on the curve are
given.

Example 4. From the given curve of parabola, find the equation of it.
Solution:
The given curve of parabola has its vertex ar the origin O. A point (2,4) is on
Example 5.
Solution: the curve. Y X
Let its equation be y = ax2. 10
Example 6. It passes through the point (2,4). 9
Solution: 8
4 = a.22 7
or, a = 1 6
Hence, the required equation of parabola is 5
4 (2, 4)
y =x2. 3
2
1
X' -5 -4 -3 -2 -1O 1 2 3 4 5

Y'

Find the equation of given parabola passing through points (0, 1) and (2, 5).

The parabola has its vertex at y-axis.

Let equation of parabola be y = ax2 + k.

The parabola passes through the points (0,1) and (2,5).

1 = a.02 + k Y

k=1 9
8
7
and 5 = a.22 + 1 6
5
or, 5 -1 = 4a (-2, 5) 4 (2, 5)
3
2
or, 4a = 4 1 (0, 1) X
? a=1 X' -5 -4 -3 -2 -1-1 O1 2 3 4 5

Hence, the equation of parabola is y = x2 + 1. -2

Y'
Find the equation of parabola passing through the point (1, 0), (2, 0) and (4, 6).

Let the required equation of parabola be y = ax2 + bx + c ......... (i)

Putting the point (1, 0) in equation (i), we get, 0 = a.l2 + b.l + c

or, a + b + c = 0 ..... (ii)

Putting the point (2,0) in equation (i), we get,

4a + 2b + c = 0 ........ (iii)

Subtracting (ii) from (iii), we get 3a + b = 0 ..... (iv).

Again, putting the point (4,6) in equation (i), we get,

16a + 4b + c = 6 ...............(v)

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Subtracting (ii) from (v), we get,

15a + 3b = 6 ...............(vi)

From equation (iv), b = - 3a

Putting the value of b in equation (vi), we get,

15a - 9b = 6

or, 6a = 6 ? a=1

Putting the value of a in equation (iv)

b = -3.1 ? b=-3

Putting the values of a and b in equation (i),

we get c = 2

Putting the values of a, b and c in equation (i),

We get, y = x2 - 3x + 2

5.5 Graph of a cubic Function

A cubic function is defined as y = f(x) = ax3 + bx2 + cx + d, a ≠ 0, Y

Where a, b, c, and d are constants. 8

Let y = x3 be a cubic function 7
To draw curve of y = x3 6
5
4
3
2
x -2 -1 0 1 2 -6 -5 -4 -3 -2 -11
y -8 -1 0 1 8 X' O-11 2 3 4 5 6 X

-2
-3
-4
-5
-6
-7
-8

Y'

Again, let y = -x3 be a cubic function. Y
To draw graph of y = -x3
8
x -2 -1 0 1 2 7
y 8 1 0 -1 -8 6
5
4
3
2
1
X' -6 -5 -4 -3 -2 -1-O1 1 2 3 4 5 6 X
-2
-3
-4
-5
-6
-7
-8

Y'

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Example 8. Draw graph of the functions :
Solutions:
(i) y = - 2x3

(ii) y = (x + 1)3 + 3 Y

(i) Given function is y = -2x3 16
To draw graph of the function. 15
14
x -2 -1 0 12 13
y 16 2 0 -2 -16 12
11
Plotting the points on the graph 10
and joining them, we get a curve as 9
shown in the graph. 8
7
6
5
4
3
2
1

(ii) Given function is y = (x + 1)3 + 3 X' -6 -5 -4 -3 -2 -1-1O 1 2 3 4 5 6 X
To draw graph of given function -2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12
-13
-14
-15 y = –2x2
-16

x -3 -2 -1 0 1 Y'

y -5 2 3 4 11

The points are plotted on the graph and joined them, we get a curve as
shown in the graph.

Y

11 X
10

9
8
7
6
5
4
3
2
1
X' -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-O1 1 2 3 4 5 6 7 8 9 10
-2
-3
-4
-5

Y'

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Exercise 5.1

Very Short Questions

1. (a) Define a quadratic function and a cubic function with their general equations.

(b) Define the following terms of parabola:

(i) Vertex (ii) Line of symmetry

(c) Write the vertex of parabola if equation is given by y=ax2 + bx + c. Also write
equation of line of symmetry of parabola for this equation.

(d) From the equation y = x2 + 5x + 6, answer the following questions.

(i) Find the points on x-axis of the curve.

(ii) Find the vertex of the parabola. Y
(iii) Find the equation of line symmetry. 10
2. Observe the given curve and answer the following 9
questions: 8
(a) Name the curve. 7
(b) Find the vertex of the curve. 6
(c) Write the coordinates of the points M and N. 5
(d) Name the line PQ which is parallel to y-axis. 4
Short Questions 3 N
3. Draw the graphs for the following linear equations:
2
1A M
X' -6 -5 -4 -3 -2 -1-1O 1 2 3 4 5 6 X

-2
-3
-4 Q P

Y'

(a) x = y (b) y + x = 0,

(c) 4x + 3y = 12 (d) y = 3x

4. Draw the graphs for the quadratic following equations:

(a) y = x2 (b) y = -x2 (c) y = x2 + 4

(d) y = x2 - 4 (e) y = (x + 1)2 (f) y = 1 x2
2
5. Draw the graphs for the following equations:

(a) y = x3 (b) y = -x3 (c) y = (x+1)3 (d) y = -(x+1)3

Long Questions

6. Draw the graphs for the following quadratic functions:

(a) y = 4x2 - 8x + 3 (b) y = x2 + 2x - 8 (c) y = - x2 - 2x + 5

(d) y = 4x2 + 8x + 5 (e) y = x2 + 2x - 3 (f) y = (x - 5)2

(g) y = -(x + 3)2

7. Find the equation of the parabola under the given conditions:

(a) Vertex at the origin and passing through the point (2, 3)

(b) Vertex at the origin and passing through the point (1, -4)

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vedanta Excel In Opt. Mathematics - Book 10

8. Find the equation of the parabola under the following conditions.
(a) Vertex = (2, -1), passing through (1, 0)
(b) Vertex = (-1, -4), passing through (2,5) [Hints use y = a(x - h)2 + k]

9. Find the equations of parabola under the following stated conditions.
(a) passing through the points (1,0), (0, -3), and (-3,0)
(b) passing through the points (0,-3), (-1, -1), and (3,3)
(c) passing through the points (2,4), (-1, -2,) and (3, 10)
(d) passing through the points (1,4), (2,1), and (4,1)

10. From the given graphs of parabola. Find their equations:

(a) Y (b) Y
10 10
9 9
8 8
7 7
6 6
5 5
4 4
3 3
2 2
1 1
X X
X' -6 -5 -4 -3 -2 -1O -11 2 3 4 5 6 X' -6 -5 -4 -3 -2 -1O -11 2 3 4 5 6
-2 -2
-3 -3
-4 -4

Y' Y'

(c) Y (d) Y X

10 10
9 9
8 8
7 7
6 6
5 5
4 4
3 3
2 2
1 1

X' -6 -5 -4 -3 -2 -1O -11 2 3 4 5 6 X X' -5 -4 -3 -2 -1O -11 2 3 4 5 6 7
-2 -2
-3 -3
-4
Y' Y'

1. (c) - b , 4ac - b2 , x = - b (d) (i) (-2, 0), (-3, 0) (ii) - 5 , 1 (iii) 2x + 5 = 0
2a 4a 2a 2 4

2. (a) Parabola (b) (1, 0) (c) M(2, 1), N(3, 4)

7. (a) y = 3 x2 (b) y = -4x2 8.(a) y = x2 - 4x + 3 (b) y = x2 + 2x - 3
4 (c) y = x2 + x - 2 (d) y = x2 - 6x + 2

9. (a) y = x2 + 2x - 3 (b) y = x2 - x - 3

10. (a) y = x2 (b) y = x2 + 2x - 3 (c) y = x2 - 2x (d) y = x2 - 7x + 10

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5.6 Solution of Linear and Quadratic Equations

Review:

Let us consider two linear equations:
2x + y = 5 ...........(i)
x + y = 3 ............ (ii)

Then answer the following questions:
(a) In how many ways above two equations can be solved ?
(b) Solve above equations using any one methods you preferred.
(c) What is meaning of simultaneous equations ?
(d) Can we say that equations (i) and (ii) are simultaneous linear equations ?

Discuss the answer of above questions.

Introduction :
In general, we see that a straight line may cut the curve at two points. These points are called
solution of the curve and the straight line.

The point of intersection of a curve of quadratic equation and a straight line with linear
equation is called the solution of linear and quadratic equation.

Here, our study is focused on substitution and graphical methods of solution of linear and
quadratic equations.

Quadratic Equation :

An equation of the from ax2 + bx + c = 0 is called quadratic equation. In general a quadratic

equation has two solutions or roots. Y

Solution of Quadratic Equation Graphically.

If we draw a curve of quadratic equation

ax2 + bx + c = 0, the curve cuts two points on x-axis. These

two points are solution of quadratic equation. O
Let ax2 + bx + c = 0 be a quadratic equation in x. X' B(a2, 0) X
A(a1, 0)

We know that the equation represents a parabola. If it cuts

X-axis at two points A(a1, 0), B(a2, 0), the values of a1, and a2 Y'
be the solution of the quadratic equation i.e. x = a1 , a2 .

Worked Out Examples

Example 1. Solve x2 - 4x - 5 = 0 graphically
Solution: Let y = x2 - 4x - 5 = 0
Then, y = x2 - 4x - 5 ............ (ii)
108 Here, equation (i) represents a parabola and y = 0 represents equation of X -

axis. The point of intersection of x-axis and parabola gives the solution.

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vedanta Excel In Opt. Mathematics - Book 10

From equation (i), y = x2 - 4x - 5

Comparing it with y = ax2 + bx + c, a = 1, b = -4, and c = - 5

x - coordinate of vertex = - b = - (-4) =2
2a 2.1

y - coordinate of vertex, y = 22 - 4.2 - 5 = 4 - 13 = - 9

Vertex of parabola = (2, -9)

To draw the curve of y = x2 - 4x - 5

Y

X' (–1, 0) O (5, 0) X

Y'

x -2 -1 0 1 2 3 4 5 6
y 7 0 -5 -8 -9 -8 -5 0 7

Plotting these points on the graph and joining them, we get a parabola.
The parabola cuts X-axis at points A(5,0) and B(-1,0). The x-coordinates of

these points, x = 5 and x = 1 are the solutions of given equations.
x = -1, 5
Note :
Do not write x = (-1, 5)

Alternative Method

Here, x2 - 4x - 5 = 0
or, x2 = 4x + 5
Let y = x2 = 4x + 5
Then y = x2 ........(i) and y = 4x + 5 ..... (ii)
Equation (i) represents a parabola with vertex at the origin and equation (ii) represents
a straight line.

From equation (i) y = x2

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x -3 -2 1 0 1 3 3 Y
y 9 4 10149 25 (5, 25)
From equation (ii) y = 4x + 5

x 1 0 -1 20
y951

Plotting the points of parabola and straight 15
line, joining them. We get the point of
intersection of the parabola and the straight
line (-1,1) and (5,25).

Hence, the required solutions are x = -1,5. 10

Note: In factorization method,

x2 - 4x -5 = 0 5
or, x2 - 5x + x - 5 = 0

or, x(x-5) + 1(x-5) = 0 (–1, 1)

or, (x-5) (x + 1) = 0 X' O X

Either x - 5 = 0 o x = 5

or, x + 1 = 0 o x = -1

? x = -1, 5 Y'
Example 2. Solve graphically x2 - 6x + 9 = 0

Solution: Here, x2 - 6x + 9 = 0

Let y = x2 - 6x + 9 = 0

Then, y = x2 - 6x + 9

= (x - 3)2 ................(i)

and y = 0 ............(ii)

From equation (i), y = (x-3)2

which is in the form of y = a(x - h)2 + k

a = 1, h = 3, k = 0 Y X
Vertex (h,k) = (3,0) 10
To plot graph, we have y = (x-3)2 9
8
x3214506 7
y0141499 6
Plotting these points on the graph and joining 5
them, we get the curve of parabola. The X' 4
parabola cuts X - axis at point (3,0). 3
Hence solution is x = 3. The parabola touches 2
X-axis at only one point. 1
-6 -5 -4 -3 -2 -1O -11 2 3 4 5 6

-2
-3
-4

Y'

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Example 3. Solve the given equations:
Solution:
(a) By substitution Method and (b) Graphical Method

y = x2 - 2x and y = x

Given equations are :

y = x2 - 2x ......... (i)

y = x .......... (ii)

Solution by substitution method.

Putting the value of y is equation (i), we get

x = x2 - 2x

or, x2 - 3x = 0

or, x(x-3) = 0

Either x = 0 or, x - 3 = 0 or, x = 3

From equation ... (ii)

When x = 3, then y = 3 Also for x = 0, y = 0

Hence, the required solutions are (0,0) and (3,3)

Solution by graphical method

From equation (i), y = x2 - 2x

Comparing with y = ax2 + bx + c, we get

a = 1, b = 2, c = 0 to find vertex of parabola.

x-coordinate of vertex = - b = -(-2) = 1.
2a 2.1

When x = 1, y = 12 - 2.1 = - 1

Vertex = (1, -1)

To draw graph of parabola for y = x2 - 2x

x -2 -1 0 1 2 3 4 Y X
y 8 3 0 -1 0 3 8 10
9
From equation (ii), y = x 8
7
x124 6
y124 5
4
Plotting the above points and joining them we get curve of 3
parabola and straight line. 2
From the graph we observed that the points of intersection 1
of the parabola and straight line are (0, 0) and (3, 3). X' -6 -5 -4 -3 -2 -1O -11 2 3 4 5 6
Hence, the required solutions are (0, 0) and (3, 3)
-2
-3
-4

Y'

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Exercise 5.2

Short Questions :

1. Solve the following equation of parabola and straight line by substitution method.

(a) y = x2, y = x (b) y = x2 - 3x, y = x

(c) y = x2 + 2x - 8, y = -5 (d) y = 4x2 + 8x + 5, x + y = 3

(e) y = x2 + 7x + 12, x - y + 7 = 0

Long Questions :

2. Solve the following equation graphically:

(a) x2 + 6x + 8 = 0 (b) x2 - 2x - 15 = 0 (c) x2 + 2x - 8 = 0

(d) x2 - x - 2 = 0 (e) x2 + 7x + 12 = 0 (f) x2 - 4x + 3 = 0

3. Solve the following equation graphically:

(a) y = x2, y = 3 - 2x (b) y = x2, y = 2x -1

(c) y = x2 - 5, y = 2x + 3 (d) y = x2 + 2x - 8, y = - 5

(e) y = x2 + 7x + 12, x - y + 7 = 0 (f) y = 6x2 - 2x - 15, y = 4x - 3

(g) y = 4x2 + 8x + 5, x + y = 3

Project Work

4. Draw the parabola from the equation y = ax2, take

(a) a = - 2, - 1, - 1 , 0 and (b) a = 2, 1, 1 , 0
2 2

Draw graphs of parabola in both cases on the same graph, compare them, and write
the conclusions. Use different colors for different values of a.

1. (a) (0, 0), (1, 1) (b) (0, 0), (4, 4) (c) (1, -5), (-3, -5) (d) (-2, 5), - 1 , 13
(e) (-1, 6), (-5, 2) (b) -5, -3 4 4
(d) 2, -1 (f) 1, 3
(b) (1, 1) 2.(a) -4, -2 (d) (1, -5), (-3, -5) (c) 2, -4

(f) (2, 5), (-1, -7) (c) -4, -3 3.(a) (1, 1), (-3, 9)

(c) (-2, -1), (4, 11) (e) (-1, 6), (-5, 2)

(g) (-2, 5), - 1 , 13
4 4

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vedanta Excel In Opt. Mathematics - Book 10 6

Continuity

6.0 Review

Let us discuss the answer of the following questions:
(i) What are counting numbers?
(ii) What are rational and irrational numbers?
(iii) Define integers.
(iv) In daily life, give three examples in which we use limit or continuity.
(v) What did we study in class 9 about limit?

6.1 Continuity in the set of numbers.

Let us study the following figures.

(a) 1 2 3 4 5 6 7 8
In this figure, every point in the number line represents a natural number. There
is continuity of natural numbers in the number line.

(b) 1 2 3 4 5 6 7 8

In this figure, natural number 4 is excluded in the number line. It does not
represent continuity of natural numbers.

(c) 0 1 2 3 4 5 6 7 8

In this figure, every point in the number line represents a whole number. There is
continuity of whole numbers in the number line.

(d) -4 -3 -2 -1 0 1 2 3 4

In this figure, every point in the number line represents an integer. There is
continuity of integer numbers in the number line.

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(e) -4 -3 -2 -1 0 1 2 3 4

In this figure, whole number 2 is excluded. The number line does not show the
continuity of whole numbers.

(f)

1 2 3 4 5 67 8

In this figure, there are numbers between any two consecutive numbers which
do not have properties of the numbers mentioned in the figure. For example there
are rational and irrational numbers between 4 and 5. There is continuity of real
numbers between 1 and 8.

In a race competition, every competitor touches their feet on the ground for a certain
distance. There is continuity of their steps on the ground.
Mr Sanjit did his job in an office for 20 years continuously. We can say that there is
continuity of his job for 20 years.

Exercise 6.1

Very Short Questions
1. Show the following numbers in number line:

(a) Natural numbers from 1 to 10.
(b) Whole numbers from 4 to 12.
(c) Integers from –2 to + 6.
(d) Integers from –4 to + 7.
2. (a) Write one difference between the given two number lines:

01 2 3 4 5 67 89

1 23 4 5 67 89

(b) What are natural numbers used for?

(c) What are the starting and ending numbers in each of the number lines?

(d) Can you say any two numbers between 4 and 5?

(e) Find any two rational numbers between 4 and 5.

(f) Can you join the following numbers with a straight line?

1 35

114 24

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Short Questions

3. Give any three examples which show “continuity” in our daily life.
4. What is difference between diagrammatic representation of real number and natural numbers?
Long Questions
5. Draw a real number line. Answer the following questions:

(i) Show integers from -5 to +5 in the number line.
(ii) Which numbers can be seen continuously in the number line?
6. A bamboo plant is measured 10 m on Saturday. The plant is supposed to grow 15 cm
everyday continuously. What will be its height on Friday of that week.
7. An employee saves Rs 4000 every month in a saving account of a bank. He continuously
saves Rs. 4000 monthly for a year. Show the saving in a number line. Find the total
saving at the end of the year.

6. 10.50 m 7. Rs. 48000

6.2 Investigation of Continuity in Graph

(a) Let us study the following diagrams of number lines:

(a) , (2, 8)

01 2 34 5 6 7 8 9

(b) , [2, 8]

01 2 34 5 6 7 8 9

(c) , [2, 8)

01 2 34 5 6 7 8 9

(d) , (2, 8]

01 2 34 5 6 7 8 9

Let us define about above number lines.
(a) In this figure, both of ending points 2 and 8 are excluded. It is denoted by (2, 8). It
is called open interval.

(b) In this figure, both of the ending points 2 and 8 are included. It is denoted by
[2, 8]. It is called closed interval.

(c) In this figure, left end point is included and right end point is excluded. It is
denoted by [2, 8). It is called left closed and right open interval.

(d) In this figure, the left end point is excluded and right end point is included. It is
called left open right closed interval. It is denoted by (2, 8]

(b) -4 -3 -2 -1 0 1 23 4

This number line does not represent continuity of real numbers because real number 1
is excluded. There is breakage at x = 1.

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(c) This represents a continuous curve. From 0 to 10 it increases uniformly and there is
no breakage in it.

Y

7
6
5
4
3
2
1
X' O 1 2 3 4 5 6 7 8 9 10 X

Y'

(d) This curve breaks at x = 2. Hence, it is not a continuous.

Y

7
6
5
4
3
2
1
X' O 1 2 3 4 5 6 7 8 9 10 X

Y'

Definition: The continuity means to have elements from beginning to the end. The continuity
of a simple function can be checked by drawing a curve of it. If there is no breakage at any
point on the curve, then the function is continuous. If there is breakage at a point of the
curve, the function is called discontinuous at that particular point.

Worked Out Examples

Example 1. The given graph is of a rational function. At what points is the curve is
Solution:
discontinuous? Y
116
In the given curve, there are breakages a 10
point 9
8
x=3 7
6
and x = 6 5
4
Hence, the curve is discontinuous at X' 3 X
x=3 2
1
O 1 2 3 4 5 6 7 8 9 10

and x = 6 Y'

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Example 2. In the given curves, comment on discontinuity or continuity of the curve.

(a) Y (b) Y

10 10
9 9
8 8
7 7
6 6
5 5
4 4
3 3
2 2
1 1

X' O 1 2 3 4 5 6 7 8 9 10 11 12 X X' O 1 2 3 4 5 6 7 8 9 10 11 12 X

Solution: Y' Y'

(a) In the given figure, curve is drawn from x = 0 to x = 12. There are gaps
at x = 3 and x = 10. Henc,e the curve is discontinuous at x = 3 and x = 10.

(b) In the given curve, the curve is drawn from x = 0 to x = 10. It has
breakage at x = 8. Hence, the curve is discontinuous at x = 8. The curve is
continuous except at x = 8.

Exercise 6.2

Very Short Questions :
1. In the following given curves:

(i) Find the initial and the terminating points of the curve.
(ii) State the continuity or discontinuity of the curve.

(a) Y (b) Y

7 7
6 6
5 5
4 4
3 3
2 2
1 1
X' O 1 2 3 4 5 6 7 8 9 10 11 12 13 14 X X' O 1 2 3 4 5 6 7 8 9 10 X

Y' Y'

(c) Y (d) Y

X' 7 1234567 X X' 7 1234567 X
6 6
5 5
4 4
3 3
2 2
1 1
-4 -3 -2 -1-O1 -4 -3 -2 -1-O1
-2 -2

Y' Y'

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(e) Y (f) Y

7 7
6 6
5 5
4 4
3 3
2 2
-4 -3 -2 -11 1
X' X X' -4 -3 -2 -1-O1 1 2 3 4 5 6 7 X
O 1234567 -2
-1
-2 Y'

Y'

Short Questions

2. Discuss the continuity and discontinuity of the following curves from points x = –6 to
x = 6 (stating the intervals for continuity and points of discontinuity for discontinuity).

(a) Y X (b) Y X

6 6
5 5
4 4
3 3
2 2
X' O1 1
-6 -5 -4 -3 -2 -1-1 1 2 3 4 5 6 X' -6 -5 -4 -3 -2 -1-O1 1 2 3 4 5 6
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6
Y'
Y'

(c) Y (d) Y

6 6
5 5
4 4
3 3
2 2
-6 -5 -4 -3 -2 -11 -6 -5 -4 -3 -2 -11
X' X X' X
O 123456 O 123456
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6

Y' Y'

Project Work

3. Ask the marks obtained in Optional Maths in class 9 final exam of your friends in the
last year. Fill in the table.

Marks obtained 40-50 50-60 60-70 70-80 80-90 90-100

No, of students

From the above data draw "less than" and "more than" ogive curves. Discuss the
continuity or discontinuity of the curve in your class.

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1. (a) (i) From x = 1 to x = 14 (ii) discontinuous at x = 10

(b) (i) From x = 0 to x = 6 (ii) discontinuous at x = 2 and x = 4

(c) (i) From x = 0 to x = 7 (ii) continuous

(d) (i) From x = 0 to x = 4 (ii) discontinuous at x = 2 and x = 3

(e) (i) From x = -3 to x = 3 (ii) discontinuous at x = 2

(f) (i) From x = -4 to x = 3 (ii) discontinuous at x = 0

2. (a) Continuous in [-6, -1) and discontinuous at x = -1, continuous in (-1, 6]

(b) Continuous in [-6, 3), discontinuous at x = 3, continuous in (3, 6]

(c) Continuous in [-6, -3), discontinuous at x = -3 and x = 2,

continuous is (-3, 2), continuous in (0, 6]

(d) Continuous in [-6, -2), discontinuous at x = -2 and x = 5,

continuous in (-2, 5), continuous is (5, 6]

6.3 Notational Representation of Continuity

(a) Let us consider a real valued function

f : R o R defined by f(x) = 3x – 1.

Discus the following questions:

(i) What are the domain and range of the function?

(ii) What are values of the function at x = –2, x = 4, x = 6, i.e. f(–2), f(4) and f(6)?

(iii) Can we show the points –2, 4 and 6 and the functional values f(–1), f(3) and f(6) in
graph paper?

(b) If fi(nx)gr=apx1h,pwahpaetr.is the value of f(0)? In this case f(0) is not defined and we cannot show
it

(c) Let us suppose a function f(x) = x + 5 whose domain and range both are real numbers.
Let us observe the table given below:

x 1.9 1.99 1.999 1.9999 ........ 2

f(x) 6.9 6.99 6.999 6.9999 ........ 7

Find difference between f(1.9) and f(1.9999). Also, find difference between f(1.9999)
and f(2).

From the above table, we observe that as the value of x approaches to 2, the value of f(x)
approaches to 7.

For x = 1.9, 1.99, 1.999, 1.9999, we write x o 2 – 0

or, x o 2–. As x takes its value from left of 2, we say that it is left hand limit of f(x) at x

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= 2.

We write lximo 2– f(x) = 7 or, xlimo 2 – 0 f(x) = 7

When x approaches 'a' from left hand side of a, the function f(x) tends to a'l1t' a definite
number this definite number l1 is called left hand limit of the function f x = a.

We write,

lximo a– f(x) = l1 or, lximo a – 0 f(x) = l1

Again, observe the table given below for f(x) = x + 5

x 2.1 2.01 2.001 2.0001 ......... 2

f(x) 7.1 7.01 7.001 7.0001 ......... 7

For x = 2.1, 2.01, 2.001, 2.0001, .... we write x o 2 + 0 or x o 2+ which means x takes
its value from right of 2, then value of f(x) is called right hand limit of (x) at x = 2.

We write,

lximo 2+ f(x) = 7 or, lximo 2 + 0 f(x) = 7

When x approaches 'a' from right hand side of a, the function f(x) taetnxds=toa,l2w, aedwefriinteite
is called right hand limit of f(x) f(x) = l2
number. This definite number l2
lximo a+ f(x) = l2 or, xlimo a + 0

If the left hand limit and right hand limit of a function f(x) at x = a are equal, we write

lim f(x) = lim f(x)
x o a– x o a+
Then we say that the limit of the function f(x) exists at x = a. We write xlimo a f(x) is
finite. Limit of a function exists means xlimo a f(x) is finite.
In the above example, 7 is called limit of the function f(x) at x = 2.

If the functional value and limit of a function defined at a point x = a are equal, then
we say that the function is continuous at x = a.

OR

A function f(x) is said to be continuous at x = a if the following conditions are
satisfied.

1. f(a) exists 2. lximo a f(x) exists 3. xlimo a f(x) = f(a)

Exercise 6.3

Very Short Questions :

1. Write a sentence for each of the following notations.

(a) xlimo a– f(x) or lximo a – 0 f(x)
(b) lim f(x) or xlimo a + 0 f(x)
x o a+

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(c) lximo a f(x)
(d) lximo a+ f(x) = xlimo a– f(x)
(e) lximo a f(x) = f(a)

(f) Write conditions for continuity of function f(x) at a point x = a, using notations.

Long Questions

2. Let f : R o R be a real valued function defined by f(x) = x + 4

(a) for x = 3.9, 3.99, 3.999, 3.9999, find the value of f(x)

(b) for x = 4.1, 4.01, 4.001, 4.0001, find the values of f(x).

(c) Find the value of f(x) at x = 4.

(d) Find the values of xlimo 4– f(x) and lximo 4+ f(x).
(e) Does limit of the function f(x) exist at x = 4 ?

(f) Write the notation to show above function is continuous at x = 4.

3. Let f : R o R be a real valued function defined by f(x) = x + 3, 1 d x < 2

(a) Find lim f(x) (b) Find lim 4x – 3, x t 2
x o 2– x o 2+ f(x)

(c) Is lximo 2– f(x) = lximo 2+ f(x) ? (d) Find f(2).

(e) Draw your conclusion.

4. Discuss the continuity of the function f(x) at x = 2,

2x – 1 when x < 2
f(x) 3
when x = 2 at x = 2

x + 1 when x > 2

Project Work

5. In daily life "continuity" is used frequently. For example, continuity of study is one. List
any four other examples of continuity used in daily life with reasons.

1. (a) Left hand limit of f(x) at x = a (b) Right hand limit of f(x) at x = a

(c) Limit of function f(x) at x = a (d) Limit of the function f(x) exists at x = a

(e) Limit of f(x) at x = a is equal to functional value f(a)

(f) f(a) is finite, lximo a f(x) exists and lximo a f(x) = f(a)
2. (a) 7.9, 7.99, 7.999, 7.9999 (b) 8.1, 8.01, 8.001, 8.0001

(c) 8 (d) 8, 8 (e) Yes (f) xlimo 4 f(x) = f(4)
3. (a) 5 (b) 5 (c) Yes (d) 5

(e) f(x) is continuous at x = 2 4. Continuous at x = 2

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6.4 Investigation of Continuity

Intervals

Let a and b be to real numbers in a number line. Then, set of points in the real line between
a and b is called an interval. The points a and b are called end points of the interval.
An interval not containing the end points a and b is called open interval. It is denoted by
(a, b)

a (a, b) b 4 (4, 8) 8

An interval containing the end points a and b is called closed interval. It is denoted by [a, b].

a [a, b] b –2 [–2, 4] 4

Definition of Continuity

Let y = f(x) be a function defined in an open interval (m, n). Let x = a be a point in the
interval (m, n). Then, the function f(x) is said to be continuous at a point x = a, if

(i) f(a) is finite

(ii) xlimo a f(x) exists i.e. lim f(x) = lim f(x)
x o a+ x o a–
(iii) lximo a f(x) = f(a)
In other words, a function f(x) is said to be continuous at a point x = a if f(a) = lximo a f(x).
If the function is not continuous at a point, then, it is said to be discontinuous at that point.

Worked out Examples

Example 1: Examine the continuity or discontinuity of the following functions at the
points mentioned.

(i) f(x) = x + 6 at x = 3
at x = 4
(ii) f(x) = x 1 4 at x = 3
– at x = 3
x+1
(iii) f(x) = 4x + 4

Solutions: (a) Here, f(x) = x + 6

Functional value at x = 3

f(2) = 3 + 6 = 9, finite.

limit at x = 3, xlimo 3 f(x) = xlimo (x + 6) = 3 + 6 = 9
? f(3) = xlimo 3 f(x)

Hence, f(x) is continuous at x = 3.

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(b) Here, f(x) = x 1 4
–

When, x = 4

f(4) = 4 1 4 = 1 = ∞
– 0

which is not finite.

Hence, f(x) is discontinuous at x = 4.

(c) Here, f(x) = x+1 at x = 3
4x + 4
Functional value at x = 4

f(3) = 3+1 = 4 = 4 = 1
4.3 + 4 12 + 4 16 4
For limit at x = 3

xlimo 3 f(x) = lximo 3 x+1 = 3+1 = 4 = 4 = 1
4x + 4 4.3 + 4 12 + 4 16 4

? f(3) = lximo 3 f(x).
Example 2: Hence, f(x) is continuous at x = 3.

Solutions: Are the following functions continuous at the points mentioned?

(a) f(x) = x2 – 5x at x z 1
x–5

1 at x = 1
x2 – 7x
(b) f(x) = x–7 at x z 7

5 at x = 7

(a) Here, (i) f(x) = x2 – 5x
x–5

1

Functional value at x = 1, f(1) = 1, finite.

Limit at x = 1, lximo 1 = xlimo 1 12 – 5.1 = 1––45= –4 = 1.
1–5 –4
which is finite.

? f(1) = xlimo 1 f(x)
Hence, f(x) is continuous at x = 1.

(b) f(x) = x2 – 9x at x z 9 at x = 9
x–9 at x = 9
8

Functional value of x = 9.

f(9) = 8 finite.

limit at x = 9,

xlimo 9 f(x) = xlimo 7 x2 – 9x ( 0 form)
x–9 0
x(x – 9)
= lximo 9 x–9

= xlimo 9 x = 9

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? f(9) z lximo 9 f(x)
Hence, the function f(x) is discontinuous at x = 9.

Example 3: Discuss the continuity of the function f(x) at x = 3.
Solutions:
f(x) = 2x – 1 when x < 3 at x = 3
Example 4: 5 when x = 3
Solutions:
x + 2 when x > 3
Functional value at x = 3,

f(3) = 5

Right hand limit

lximo 3+ f(x) = xlimo 3+ (x + 2) = 3 + 2 = 5
Left hand limit,

xlimo 3– f(x) = xlimo 3– (2x – 1) = 2 × 3 – 1 = 5

? lim f(x) = lim f(x) = 5
x o 3+ x o 3–

Hence, the limit exists at x = 3.

? f(3) = xlimo 3 f(x)
Hence, f(x) is continuous at x = 3.

A function f(x) is defined by:

f(x) = 2x2 + 1 when x < 2 at x = 2
8 when x = 2

4x + 1 when x > 2
Discuss the continuity of the above function at the point mentioned. If f(x) is
not continuous, how can it be made continuous?

Functional value at x = 2 is,

f(2) = 8.

For the right hand limit we have,

lim f(x) = lim (4x + 1) = 4 × 2 + 1 = 9
x o 2+ x o 2+
For the left hand limit we have,

lim f(x) = lim (2x2 + 1) = 2.22 + 1 = 9
x o 2– x o 2–
Since left hand limit = right hand limit

The limit exists

lximo 2 f(x) = 9
But, xlimo 2 f(x) z f(2)
Hence, the given function f(x) is discontinuous at x = 2.

To make above function continuous, f(2) must be 9.

i.e. f(2) = 9

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Example 5: Find the value of p if the given function is continuous at x = 2.
Solutions:
f(x) = 3x2 – 3x at x z 1 at x = 1
x–1 at x = 1
p

For x = 1, the functional value is given

f(1) = p

To find limit at x = 1, we have

xlimo 1 f(x) = xlimo 1 3x2 – 3x ( 0 form)
x–1 0
= xlimo 1 3x (x – 1)
x–1
= xlimo 1 3x
= 3.1

=3

For the function f(x) to be continuous, we have

f(1) = xlimo 1 f(x) ? p=3
i.e. p = 3

Exercise 6.4

Short Questions

1. Examine the continuity or discontinuity of the following functions at the points
mentioned:

(a) f(x) = 4x + 1 at x = 3

(b) f(x) = 3x2 + 2x at x = 2
at x = 5
(c) f(x) = 1 at x = 4
x–5 at x = 8
2 at x = 10
(d) f(x) = 3x – 5

(e) f(x) = x2 – 64
x – 8

(f) f(x) = x2 – 100
x – 10

Long Questions

2. Examine the continuity or discontinuity of the following functions at the points
mentioned:

(a) f(x) = x2 – 7x when x z 7 at x = 7
x–7 when x = 7
3

(b) f(x) = x2 – 9 when x z 2 at x = 2
x–3 when x = 2
6

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(c) f(x) = x2 – 2x when x z 2 at x = 2
x–2 when x = 2
2

(d) f(x) = x2 – 5x when x z 5 at x = 5
x–5 when x = 5
5

(e) f(x) = 4x2 – 16x when x z 2 at x = 2
x–4 when x = 2

8

(f) f(x) = x2 – x – 6 when x z 3 at x = 3
x–3 when x = 3

5

(g) f(x) = x2 – 5x + 6 when x z 2 at x = 2
x–2 when x = 2

–1

(h) f(x) = x2 – 3x + 2 when x z 2 at x = 2
x2 + x – 6 when x = 2
1
5

3. Examine the continuity or discontinuity of the following functions at the points
mentioned by calculating left hand limit, right hand limit and functional values:

(a) f(x) = 3–x for x ≤ 0
3 for x > 0 at x = 0

(b) f(x) = 4x – 1 for x < 1
3x for x ≥ 1 at x = 1

(c) f(x) = 5x + 2 for x ≤ 2
6x for x > 2 at x = 2

(d) f(x) = 2x + 1 for x < 1
2x for x = 1 at x = 1

(e) f(x) = 3x for x > 1
2x2 + 1 for x < 2

9 for x = 2 at x = 2

4x + 1 for x > 2
4. Show the following functions are continuous at the points mentioned:

(a) f(x) = x3 – 8 for x z 2 at x = 2
x–2 for x = 2
12

(b) f(x) = x3 – 125 when x z 5 at x = 5
x–5 when x = 5
75

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5. A function is defined as follows:

(a) f(x) = 3 + 2x for –3/2 ≤ x < 0
3 – 2x for 0 ≤ x < 3/2
x ≥ 3/2
– 3 – 2x for

Show that f(x) is continuous at x = 0 and discontinuous at x = 3/2.

6. (a) Find the value of m if f(x) is continuous at x = 5.

f(x) = x2 – 2x for x z 5
x–2 for x = 5
m

(b) Find the value of n, if f(x) is continuous at x = 2.

f(x) = x2 – 4 for x z 2
x–2 for x = 2
n

7. Define continuity of a function at a point x = 2.

Discuss the continuity of given function at x = 2

f(x) = x2 – 4 at x ≠ 2
x–2 at x = 2
5

If f(x) is not continuous, how can you make it continuous at the point x = 2.

1. (a) Continuous (b) Continuous (c) Discontinuous (d) Continuous

(e) Discontinuous (f) Discontinuous (d) Continuous
(h) Continuous
2. (a) Discontinuous (b) Discontinuous (c) Continuous (d) Discontinuous
(b) n = 4
(e) Continuous (f) Continuous (g) Continuous

3. (a) Continuous (b) Continuous (c) Continuous

(e) Continuous x2 – 4 6.(a) m = 5
7. Discontinuous, f(x) = x–2 at x ≠ 2
4 at x = 2

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Matrices

7.0 Review

Let us consider the following matrices:

P= 1 26 ,Q= 2 03 , R= 1 2 , S= 2 -2 4
3 21 4 12 3 5 32 1
0 32 2 1

Then answer the following questions.

(a) Write the orders of each of above matrices.

(b) Which of above matrices can be added ?

(c) Are PQ and QP defined ?

(d) Write QT and ST.

(e) Is Q2 = Q.Q ?

(f) Find 2P - 3S.
Discuss the answers of the above questions and conclude the conclusions.

7.1 Determinant of a Matrix

Let A = a b be a square matrix.
c d

Then write |A| = a b . It represents a number associated with given matrix A. It is
c d
called determinant of A. We also write det. A to represent |A|.

It is defined by |A| = a b =ad - bc.
c d

Example : Let A = 2 3
4 5

Then |A| = 2 3 = 2 × 5 - 4 × 3 = 10 - 12 = - 2
4 5

Definition: A determinant is a number associated with a square matrix. When the elements
of a square matrix are enclosed within vertical lines || without changing their positions, it
is called determinant of the given matrix. The determinant of a matrix A is denoted by |A|
or determinant of A or det. (A).

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Let A = a b then,
c d

|A| = a b = ad - bc
c d

Example : Let A = 1 2 , its determinant is given by ,
3 4

|A| = 1 2 = 1 × 4 - 3 × 2 = 4 - 6 = -2
3 4

Determinant of a square matrix of order 1.

Let A = [7], it is a square matrix of order 1.

Then |A| = |7| = 7

If B = [-7] then |B| = |-7| = - 7

Note : Absolute value of -7 = |-7| = 7.
det. [-7] = |-7| = - 7

Determinant of square matrix of order 2

Let A = 1 4 . It is a square matrix of order 2 × 2.
6 8
Then, its determinant is given by,

|A| = 1 4 = 1 × 8 - 6 × 4 = 8 - 24 = - 16

68

Example : Evaluate 2 5
7 3

Here, 2 5 = 2 × 3 - 7 × 5 = 6 - 35 = - 29
7 3

Worked out Examples

Example 1. Evaluate the following :

(a) |-10| (b) 2 3 (c) 1 -cos2x
4 5 1 sin2x
Solution: (a) |-10| = - 10, (determinant of (-10) = -10)

(b) 2 3 = 2 × 5 - 4 × 3 = 10 - 12 = -2
4 5

(c) 1 -cos2x = sin2x + cos2x = 1
1 sin2x

Example 2. Evaluate the determinant of the following matrices.

(a) [-20] (b) 4 5 (c) 2 4
6 7 -3 3

Solution: (a) Determinant of [-20] = |-20| = -20

(b) Determinant of 4 5 = 4 5 = 4 × 7 - 6 × 5 = 28 - 30 = -2
6 7 6 7

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(c) Determinant of 2 4 = 2 × 3 - (-3 × 4) = 6 + 12 = 18.
-3 3

Example 3. Find the value of x under the given conditions:

(a) 4 5 =x (b) x 3+2x =0
3 6 1 x

Solution: (a) Here, 4 5 =x
3 6
or, 4 × 6 - 3 × 5 = x

or, 24 - 15 = x

? x=9

(b) Here, x 3+2x =0
1 x
or, x2 - 3 - 2x = 0

or, x2 - 2x - 3 = 0

or, x2 - 3x + x - 3 = 0

or, x(x - 3) + 1 (x - 3) = 0

or, (x - 3) (x + 1) = 0

Either, x - 3 = 0 ? x=3

or, x + 1 = 0 ? x=-1

? x = -1,3

Example 4. If P = 2 3 and Q = 5 2 then find the determinant of,
Solution: 4 5 3 1
(a) 2P - 3Q (b) P2 - Q2

(a) Here, P = 2 3 ,Q= 5 2
4 5 3 1

Now, 2P - 3Q =2 2 3 -3 5 2
4 5 3 1

= 4 6 - 15 6
8 10 9 3

= 4-15 6-6 = -11 0
8-9 10-3 -1 7

Determinant of (2P - 3Q) = |2P - 3Q|

= -11 0 = -77 - 0 = - 77
-1 7
(b) For determinant of P2 - Q2

Here, P2 = P.P = 2 3 23
4 5 45

= 4+12 6 + 15 = 16 21
8+20 12+25 28 37

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Q2 = Q.Q = 5 2 5 2 = 25+6 10+2 = 31 12
3 1 3 1 15+3 6+1 18 7

Now, P2 - Q2 = 16 21 - 31 12
28 37 18 7

= 16-31 21-12 = -15 9
28-18 37-7 10 30

Determinant of (P2 - Q2 ) = |P2 - Q2| = -15 9
10 30

= –450 – 90 = –540

Example 5. If P = 2 1 , Q= 4 6 ,I= 1 0 , find the determinant of
Solution: -1 3 2 4 0 1
1 5
5P - 2 Q + 3I 15

Here, 5P = 5 2 1 = 10
-1 3 -5

1 Q = 1 4 6 = 2 3
2 2 2 4 1 2

3I =3 1 0 = 3 0
0 1 0 3

Now, 5P - 1 Q +3I = 10 5 - 2 3 +3 0

2 -5 15 12 03

10-2+3 5-3+0 = 11 2
= -5 -1 +0 15-2+3 -6 16

Determinant of 5P - 1 Q + 3I = 11 2
2 -6 16

= 176 + 12 = 188

Example 6. If A = 1 2 and B = 1 2
Solution: 3 4 4 5

Show that : |AB| = |A| |B|

Here, |A| = 1 2 = 4 - 6 = -2
3 4

|B| = 1 2 = 5 - 8 = -3
4 5

|A| |B| = (-2) × (-3) = 6

Also, AB = 1 2 12 = 1+8 2+10 = 9 12
3 4 45 3+16 6+20 19 26

|AB| = 9 12 = 234 - 228 =6
19 26

Therefore, |AB| = |A| |B| Proved.

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Exercise 7.1

Very Short Questions

1. (a) Write the condition to have determinant of a matrix.

(b) Evaluate (i) a b (ii) secɵ tanɵ
c d tanɵ secɵ

(c) Write the determinant of [-8]

(d) If A = [4] is a square matrix, find its determinant

(e) |-8| = -8 and |-8| = 8, write a difference between them.

2. Evaluate : (a) 1 0 (b) 0 1
3. Evaluate the following : 0 1 1 0

(a) 4 5 (b) 4 8 (c) x+y x-y
3 2 3 6 x-y x+y

4. Evaluate the determinant of the following matrices:

(a) [-8] (b) 23 (c) a b
Short Questions 4 -5 -b a

5. (a) If P = 1 0 ,Q= 3 1 , find the determinant of (i) P+Q+I (ii) P+Q - I
3 1 5 3

(b) If M = 2 4 ,N= 2 1 , find the determinant
5 2 3 4

(i) of 2M + 3N (ii) 3M - 2N (iii) M2 - N2

6. Evaluate (a) a+b a+b (b) a2+ab + b2 b2 + bc + c2
a+b a–b b–c a-b

7. Find the value of x in the following cases.

(a) 5 -4 =8 (b) x 4 = 0 (c) x x =9
x 4 x 2x 3x 4x

8. In each of the following cases, show that each |AB| = |A| |B|

(a) A= 2 3 , B= 0 -1
4 -1 5 2

(b) A= 2 3 , B = -3 4
4 5 2 1

Long Questions

9. (a) If P = 2 3 ,Q= 1 2 , find the determinants of PQ and QP.
4 5 4 5

(b) If A = 1 4 ,B= 2 3 , find the determinants of AB and BA.
4 5 3 4

10. If M = 1 2 and N = 2 3 , find the determinants of
4 5 3 5

(a) MT + NT (b) (M+N)T (c) (MN)T

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11. (a) If P = 0 -2 , find the determinant of 2P2 - 5P + 4I, where I is a unit matrix of
3 4

order 2 × 2.

(b) If P = 1 -2 and Q = -3 0 , find the determinant of 5P - Q2 + 5I, where I
3 4 1 -2
is a unit matrix of order 2 × 2.

12. If P = 1 2 and Q = 2 3 is |(P + Q)2| = |P2 + 2PQ + Q2|
4 5 4 5

1. (a) ad - bc (b) 1 (c) -8 (d) 4

(e) determinant and absolute value 2.(a) 1 (b) -1

3. (a) -7 (b) 0 (c) 4xy 4.(a) -8

(b) -22 (c) a2 + b2 5.(a) (i) 17 (ii) 1

(b) (i) -49 (ii) -94 (iii) 65

6. (a) -2ab - b2 (b) a3 - 2b3 + c3 7.(a) -3 (b) 0, 2

(c) ± 3 9.(a) 6, 6, |PQ| = |QP| (b) 11, 11, |AB| = |BA|

10. (a) -5 (b) -5 (c) -3

11. (a) 22 (b) 221 12. No

7.2 Inverse Matrix

Let us consider two matrices A = 2 1 and B = 3 -1 .
5 3 -5 2

Then, let us find AB and BA.

AB = 2 1 3 -1
and BA 5 3 -5 2

= 6-5 -2 + 2 = 1 0 =I
15 - 15 -5 + 6 0 1

= 3 -1 2 1
-5 2 5 3

= 6-5 3-3 = 1 0 =I
-10+10 -5 + 6 0 1

In both cases AB = I and BA = I where I is a unit matrix of order 2 × 2. Then B is called the
inverse matrix of A or vice versa. The inverse matrix of A is denoted by A-1 .

Before defining inverse matrix, let us define singular and non-singular matrix.

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Singular and Non-singular Matrices

Let A be a square matrix. Then, the matrix A is said to be singular if |A| = 0. Then the
matrix A is said to be non-singular if |A| ≠ 0.

Example 1. Let A = 4 5
8 10

Then |A| = 4 5 = 4 × 10 - 8 × 5 = 40 - 40 = 0 .
8 10
Since |A| = 0. the matrix A is a singular matrix.

Example 2. Let B = 5 2
3 6

Then |B| = 5 2 = 5 × 6 - 3 × 2 = 30 - 6 = 24 ≠ 0
3 6
Since |B| ≠ 0, B is not a singular matrix.

Definition of inverse Matrix

Let A be a square matrix whose determinant is not zero. If there exists a matrix B such that
AB = BA = I, where I is a unit matrix of order A or B, then the matrix B is called inverse of
matrix A. The inverse of matrix A is denoted by A-1.

Then, AA-1 = A-1A = I

Warning A-1 ≠ 1 , where A-1 is the inverse matrix of A.
A
Only non- singular matrices have their inverses.

Adjoint of a 2 × 2 Matrix

Let A = a b be a square matrix. then interchange the elements in the major diagonal
c d
(or principal diagonal) and change the signs of the element in the minor diagonal.

Then, the new matrix obtained is called adjoint of matrix A, i.e., d -b is called the
adjoint matrix of A. It is denoted by Adj. A -c a

Adj. A = d -b
-c a
2 3
Example : Find the adjoint of matrix A = 6 5 .

Here, adjoint of matrix A = Adj. A = 5 -3
-6 2
To find the inverse matrix of 2 × 2 order.

Let A = a b be a non-singular square matrix of order 2 × 2.
c d

Then |A| = a b = ad - bc.
c d
w x
Let the inverse of matrix of A be A-1 = y z .

Then by definition of inverse matrix, we write AA-1 = A-1 A = I, where I is a unit matrix of

order 2 × 2.

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Now, ab w x = 1 0
cd y z 0 1

or, aw + by ax + bz = 1 0
cw + dy cx + dz 0 1
Equating the corresponding elements of equal matrices:

aw + by = 1 .......... (i)

ax + bz = 0 .............. (ii)

cw + dy = 0 ............. (iii)

cx + dz = 1 ............... (iv)

Multiplying equation (i) by c and (iii) by a and subtracting (i) from (iii)
acw + ady = 0

acw + bcy = c

-- -

y(ad - bc) = - c

? y = -c
ad-bc
Putting the value of y in equation (iii)

cw + d × -c =0
ad-bc

or, cw = –d (– c )
? ad-bc
d
w = ad-bc

Again, multiplying equation (ii) by c and (iv) by a and subtracting (ii) from (iv), we get

acx + adz =a

acx + bcz = 0

-- -

z(ad - bc) = a

? z = a
ad-bc
Putting the value of z in equation (ii), we get,

a
ad-bc
( )ax = - b
b
? x = - ad-bc

Putting the values of w, x, y and z in A-1, we get,

A-1 = 1 d -b
ad-bc -c a

= 1 Adj. A, |A| ≠ 0.
|A|

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Working Rule to find inverse of a matrix.

Let A = a b be a non-singular square matrix.
c d
Its inverse can be found by using the following steps.

(i) Find |A| = ad - bc. If |A| = 0, the inverse of matrix A i.e. A-1 does not exists.

(ii) Find the adjoint matrix of A.

i.e. Adj. A = d -b
-c a

(iii) Find A-1 by using the formula,

A-1 = 1 Adj.A = 1 d -b
|A| (ad-bc) -c a

Note :
The following conditions are necessary for the existence of A-1.

(i) The matrix A must be square.

(ii) The matrix A must be non-singular. i.e. |A| ≠ 0.

(iii) AA-1 = A-1A = I must be satisfied.

Properties of Inverse of Matrices

Let A and B be two square matrices of same order. Then, the following properties are satisfied.

(i) The inverse of the product of two non-square matrices is equal to the product of inverses
taken in the reverse order.

i.e., (AB)-1 = B-1 A-1

(ii) For a non- singular matrix A. (A-1)-1 = A

(iii) For a non- singular matrix A (AT)-1 = (A-1)T

To prove above properties, here, we take 2 × 2 square matrices,

Let A= 4 5 ,B= 2 1
1 2 5 3

|A| = 4 5 =8-5=3
1 2

|B| = 2 1 =6-5=1
5 3

(i) To prove : (AB)-1 = B-1A-1

Here, AB = 4 5 . 2 1
1 2 5 3

= 8 + 25 4 + 15 = 33 19
2 + 10 1+6 12 7

|AB| = 33 19 = 231 - 228 = 3
12 7
1
LHS = (AB)-1 = |A| Adj. (AB)

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= 1 7 -19 7 - 19
3 -12 33 =3 3

-4 11 2 5
3
1 1 2 -5 3 -
|A| 3 -1 4
Also, A-1 = Adj. A = = 1 4
3 3
-

B-1 = 1 Adj.B = 1 3 1 = 3 1
|B| 1 -5 2 -5 2

2 - 5
3 -1 3 3
RHS = B-1A-1 = -5 2 ×
= 1 4
- 3 3
–139
2 + 1 –5 – 4 7
3 3 3
=
–130 – 2 25 + 8
3 3 3 –4 11

(ii) To show : (A–1)–1= A

2 - 5
3 3
|A-1| = = 8 - 5 = 3 = 1
- 1 4 9 9 9 3
3 3
1
Now, (A-1)-1 = |A-1| Adj. A–1.

4 5
3 3 4 5
= 1 1 2 = 1 2 =A
3 3
3
? (A-1)-1 = A

(iii) To show: (A-1)T = (AT)-1

From (i), we have,

2 - 5
3 3

A-1 = - 1 4
3 3

Again, AT = 4 1
5 2

|AT| = 4 1 =8-5=3
5 2
2 -13
LHS = (A–1)T = 3 4
-53 3
1 1 2 -1 2 -31
Now, RHS =(AT)-1 = |AT| Adj. AT = 3 -5 4 = 3 4
-35 3

? (A-1)T = (AT)-1 Proved.

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Worked out Examples

Example 1. Find the inverse matrices of the following matrices:
Solution :
(a) A= 2 3 (b) B = 3 6
4 5 2 4

(a) Here, A = 2 3
4 5

|A| = 2 3 = 10 - 12 = -2 ≠ 0
4 5
Since |A| ≠ 0, A-1 exists.

Adj. A = 5 -3
-4 2

Now, the inverse matrix of A is given by

A-1 = 1 Adj. A = 1 5 -3 = - 5 3
|A| -2 -4 2 2 2
2 -1
3 6
(b) Here, B = 2 4

|B| = 3 6 = 12 - 12 = 0
2 4

Since |B| = 0, B–1 does not have its inverse, i.e., B-1 does not exists.

Example 2. If P = 3 5 and Q = -3 5 , then, show that they are inverse to each
Solution: 2 3 2 -3
other.
Example 3.
Solution: Here, PQ = 3 5 -3 5
2 3 2 -3

= -9 + 10 15 - 15 = 1 0
-6 + 6 10 - 9 0 1

and QP = -3 5 35
2 -3 23

= -9 + 10 -15+15 = 1 0
6-6 10 - 9 0 1

Since PQ = QP = I, P, and Q are inverse matrix to each other.

If y 1 and 2 -1 are inverse matrices to each other,
5 x -5 3

find the values of x and y.

Since y 1 and 2 -1 are inverse to each other, we can write,
5 x -5 3

y 1 2 -1 = 1 0
x -5 3 0 1
5

or, 2y - 5 - y+3 = 1 0
10 - 5x -5 + 3x 0 1

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Equating the corresponding elements, we get,

2y - 5 = 1 or, 2y = 6 or, y = 3

-y + 3 = 0 or, y = 3

10 - 5x = 0 x=2 -5 + 3x = 1

or, 3x = 6 ? x = 2 ? x = 2, y = 3

Example 4. For what value of x the matrix P = x-3 0 does not have its
Solution: inverse? 0 3x+1

Here, P = x-3 0
0 3x+1
If P is a singular matrix, it does not have its inverse.

x-3 0
|P| = 0 3x+1 = 0

or, (x – 3). (3x + 1) = 0

Either x – 3 = 0, or, x = 3

or, 3x + 1 = 0 or, x = –1/3

? x = 3, – 1
3

Exercise 7.2