Prime Optional Mathematics 10

Solution of quadratic equation and straight Line

The quadratic equation ax2 + bx + c = 0 and a straight line with equation ax + by + c = 0 can
be solved by using substitution method as well as graphical method. Here we discuss both
substitution method as well as graphical method.

1. Substitution method:
The value of the variable ‘x' have to be found out from the equation of straight
line ax + by + c = 0 as
–by – c
x = a ...............................(i)

It has to be substituted in the quadratic equation ax2 + bc + c = 0 as,
–by – c 2 –by – c
a a a + b a a k + c = 0
k

By solving the above form of equation two values of ‘y' can be obtained and then value of
‘x' can be obtained from equation (i)

2. Graphical method:
The parabolic curve given by ax2 + bx + c = 0 and the straight line given by ax + by + c
= 0 intersect at two points A(x1, y1) and B(x2, y2). The solution is given by x = x1
y = y1 and y2. and x2,

Worked out Examples

1. Solve the quadratic equation = x2 – 2x – 3 and y = – 3 by substitution method.
Solution:
The given equation are,

y = x2 – 2x – 3 and
y = – 3
Here, y = x2 – 2x – 3 ...........................(i)
y = – 3 .........................(ii)

Substituting the value of ‘y' in equation (i).
y = x2 – 2x – 3
or, – 3 = x2 – 3
or, x2 – 2x = 0
or, x(x – 2) = 0

Either, x = 0
or, x – 2 = 0 ⇒ x = 2
\ The intersecting points of given equations are A(0, –3) & B(2, – 3).

96 PRIME Opt. Maths Book - X

2. Solve the quadratic equation y = x2 – 5x + 4 and 4x + y – 6 = 0 by substitution method.

Solution:

The given equations are:
y = x2 – 5x + 4 ...........................(i) and
4x + y – 6 = 0 ⇒ y = 6 – 4x ...........................(ii)
Substituting the ‘y' in equation (i)
or, 6 – 4x = x2 – 5x + 4

or, x2 – x –2 = 0
or, x2 – (2 – 1) x – 2 = 0
or, x2 – 2x + x – 2 = 0
or, x(x – 2) + (x – 2) = 0
or, (x – 2) (x + 1) = 0

Either x –2 = 0 OR
x + 1 = 0
\ x = 2 x = –1

Again,

Substituting the value of ‘x' in equation (ii),
y = 6 – 4x y = 6 – 4x
or, y = 6 – 4 × 2 y = 6 – 4(– 1)
\ y = – 2 y = 10
\ The intersecting point of the given equations are (2, – 2) and (– 1, 10).

3. Solve the quadratic equations = x2 – 2x – 3 and straight line y = – 3 graphically.

Solution:

The given equations are:
y = x2 – 2x – 3 ..............................(i)
y = – 3 ...............................(ii)
From equation (i),
y = x2 – 2x – 3.
b2
vertex of the parabola = a– b 4ac – k
2a , 4a y = x2 – 2x – 3

=

= (1, –4)

x 0 1 – 1 2 – 2 3 – 3 4 5

y – 3 – 4 0 – 3 5 0 12 5 12

From equation (ii), B A y = –3
y = – 3(It is parallel to x - axis)

From graph we get,
The straight line cuts the parabola at points A(2, – 3)

and B(0, – 3).
\ x = 2 & 0
y = – 3

PRIME Opt. Maths Book - X 97

4. Solve the quadratic equation y = x2 – 5x + 4 and a straight line 4x + y – 6 = 0 graphically.

Solution:

The given equations are:
y = x2 – 5x + 4 ......................................................(i)
4x + y – 6 = 0 ⇒ y = 6 – 4x ............................ (ii)
From equation (i)
y = x2 – 5x + 4
b2
vertex of the parabola = a– b 4ac – k
2a , 4a

=
= (2.5, – 2.25)

Points satisfying (i) are
x 0 1 – 1 2 – 2 3 4 5 6 7
y 4 0 10 – 2 18 – 2 0 4 10 18

From equation (ii), 3
y = 6 – 4x – 6

x01
y62

From graph, we get
The straight line cuts the parabola at points A (2, – 2)

and B(– 1, 10).
\ x = 2 & – 1
y = – 2 , 10

Exercise 1.5 'B'

1. Solve the following quadratic equation and straight line by substitution method.
i) y = x2 – 2x – 3 and y = 5 ii) y = x2 and y = x + 2
iii) y = x2 and y = 3 – 2x iv) y = x2 – 6x + 5 and y = –3
v) y = x2 – x – 6 and y = 6

2. Solve the following quadratic equation and equation of straight line by substitution
method.

i) y = x2 + 4x – 5 and 5x – y – 3 = 0 ii) y = x2 – 2x – 8 and 3x + y + 2 = 0
iii) y = x2 + 4x + 3 and 3x – y + 5 = 0 iv) y = x2 – 5x + 4 and x + y – 1 = 0
v) y = x2 – x + 10 and x – y + 13 = 0

98 PRIME Opt. Maths Book - X

3. Solve the following quadratic equation and equation of straight line graphically.
i) y = x2 and y = – 2x + 3 ii) y = x2 and y = x + 2
iii) y = x2 – 6x + 5 and y = – 3 iv) y = x2 – x – 6 and y = 6
v) y = x2 – 2x – 3 and y = 5

4. Solve the given equations graphically.
i) y = x2 – 5x + 4 and x + y – 1 = 0
ii) y = x2 – 3x + 10 and x – y + 13 = 0
iii) y = x2 – 2x – 8 and 3x + y + 2 = 0
iv) y = x2 + 4x – 5 and 5x – y – 3 = 0
v) y = x2 + 4x + 3 and 3x – y + 5 = 0

5. PRIME more creative questions:
i) Prove that the straight line x – y – 3 = 0 touches the parabola having equation x2 – x

– 2 = 0 at a single point.
ii) Prove that the straight line 2x – y – 12 = 0 is tangent to the parabola of equation x2

– 2x – 8 = 0.
iii) From the quadratic equation x2 – 2x – 15 = 0, find the vertex, points where x - axis

cuts the parabola, equation of axis of parabola and graph of parabola.
iv) Find vertex, equation of axis of parabola, roots of the variable and draw the graph of

x2 – 4x + 3 = 0.
v. At what points the straight line x + y + 6 = 0 cuts the parabola y = 6 – x2 ? Find by

using graphical method.

6. Project work
Prepare a report of solving quadratic equation and linear programming in a chart paper

and present the report into your classroom.

Answer

1. i) (–2, 5), (4, 5) ii) (–1, 1), (2, 4) iii) (1, 1), (–3, 9)
iv) (2, –3), (4, – 3) v) (–3, 6) iii) (1, 8), (–2, –1)
iii) (2, –3), (4, –3)
2. i) (–1, –8), (2, 7) ii) (2, –8), (–3, 7) iii) (2, –8), (–3, 7)
iv) (1, 0), (3, –2) v) (–1, 12), (3, 16)

3. i) (1, 1), (–3, 9) ii) (–1, 1), (2, 4)
iv) (–3, 6), (4, 6) v) (–2, 5), (4, 5)

4. i) (1, 0), (3, –2) ii) (–1, 12), (3, 16)
iv) (–1, –8), (2, 7) v) (1, 8), (–2, –1)

5. iii) (1, –16), (–3, 0), (5, 0), x – 1 = 0, graph.
iv) (1, –9), x – 2 = 0, (1, 3), graph
v) (–3, 6) and (4, 6)

PRIME Opt. Maths Book - X 99

Quadratic equation and liner programming

Unit Test Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. Write down the formula to find the vertex of the parabola.

2. a) Solve the equations x – y + 6 = 0 and y = x2.
b) Find the vertex of the parabola having equation x2 – 4x – 12 = 0.
c) Draw the graph of the function y = x3.

3. a) Solve graphically that x2 – 2x – 8 = 0.

b) Maximize the function F = 3x + 4y under the constraints x + y # 8, 2x – 3y # 6, x $
0 and y $ 0.

4. Find the inequalities which represents the given feasible region and maximize the
function z = 2x + 5y.
Y
C(0, 6)

B(4, 3)

X' O A(2, 0) X
Y'

100 PRIME Opt. Maths Book - X

Unit 2 Limit and Continuity

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions 1 – 1 – 2 5 10
Weight 1 – 4 –

K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total
Questions, TM = Total Marks

Objectives : At the end of the lesson
• Students are able to know the number system and their relation.
• Students are able to know the limit value of a function.
• Students are able to simplify the algebraic function using limit.
• Students can find the situation of continuity and discontinuity.
• Students can find the conditions of continuity and discontinuity.

Materials
• Chart of number of system.
• Chart of showing concept of limit.
• Application chart of limit.
• Chart of different form of limit.
• Geo-board.
• Graph paper and Graph board.

PRIME Opt. Maths Book - X 101

Enjoy the recall
i. Natural numbers is the set of counting number.

N = {1, 2, 3, 4, 5, ....................}.

ii. Whole number is the set of counting number including zero.

W = {0, 1, 2, 3, 4, 5, ....................}

iii. Rational number is the set of numbers in the form of p where q ≠ 0.
q

Z = {.................... , – 7 , –3, – 5 , –2, – 1 , – 1 , 0, 1 , 1, 3 , 2, ............. }
2 2 2 4 8 4

iv. Integers is the set of negative as well as positive numbers including zero.

I = {............., –4, –3, –2, –1, 0 1, 2, 3, 4, 5, ...................}

v. Irrational numbers is the set of numbers which can not be expressed in p form and they
are taken in root is called irrational numbers. q

Q = {– 3, – 2, ................., 5, 7, 8................ }

vi. Real numbers is the set of rational as well as irrational numbers 'R'.
vii. Real number is denoted by a real line in number line system.

-4 -3 -2 -1 0 1 2 3 4 5

viii. 0.9, 0.99, 0.999, 0.9999, ....................... can be taken a sequence whose limit value will be '1'
but not exactly equal to one.

A function f(x) is said to be a limit ‘l' when x → a if the numerical
difference between the value of f(x) and ‘l' can be made vary small
as we please by making x sufficiently close to 'a'. We write

lim
x " a f (x) = l.

Properties of limit

i. lim f(x) = finite number (limit exists)
x"a

ii. lim k = k (where k is any constant)
x"a

iii. lim [f(x) + g(x)] = lim f(x) + lim g(x)
x"a x"a x"a

iv. lim [f(x) – g(x)] = lim f(x) – lim g(x)
x"a x"a x"a

v. lim [f(x) . g(x)] = lim f(x) . lim g(x)
x"a x"a x"a

102 PRIME Opt. Maths Book - X

lim f (x) = lim f (x) lim g(x) ≠ 0.
x"a g (x) x"a x"a
vi. Provided
lim
x"a g (x)

vii. Indeterminate form like : 0 , 3 , 3 + 3 , 3 – 3 cannot be defined for the limit.
0 3

viii. Left hand limit = value of the variable approaches to the value from left side
Eg. x → 2– means value of x approaches to 2 from the left side.
ix. Right hand limit = value of the variable approaches to the value from right side
Eg. x → 2+ means value of x approaches to 2 from right side.

Continuity :

Continuity in the set of number system
i. Set of natural numbers :
1, 2, 3, 4, 5, 6, 7, 8, ...................

Here,
The numbers between any two natural numbers are not defined natural due to which the

set of natural numbers does not show continuity.
1, ..... 2, ..... 3, ..... 4, ..... 5, ..... 6, ..... 7, ..... 8,

ii. Set of integers:
–4, –3, –2, –1, 0, 1, 2, 3, 4, .......................

Here,
The numbers between any two integers are not defined integers due to which the set of

integers does not show continuity.
–4, ..... –3, ..... –2, ..... –1, ..... 0, ..... 1, ..... 2, ..... 3, ..... 4

iii. Rational numbers : 1
3 1 2 3
–3, –2, – 2 , –1, – 2 , 0, , 1, 2 , 2, ..................

Here,

The numbers between any two rational numbers are also rational due to which the set of

rational numbers shows continuity.

–3, –2, – 3 , –1, – 1 , 0, 1 , 1, 3 , 2
2 2 2 2

In mathematics, a continuous function is a function for which sufficiently small changes in the
input result in arbitrarily small changes in output. Otherwise it is said to be is discontinuous.

For an example, consider the function h(t) which describes the height of a growing plant at
time 't'. This is continuous. But another function M(t) denotes the amount of the money in a
bank account at the 't', then the function jumps at each point in time by deposited and with
drawn, so the function M(t) is discontinuous.

PRIME Opt. Maths Book - X 103

Let us consider the following curves. Y

Y f

f X' O X
g
X' O X h
g
h

Y' Y'

Here, The curves f, g & h are unbroken Here, The curves f, g, h are broken at a
and they are continuous point and they are discontinuous.

Note : Any function is not continuous and discontinuous generally in any interval but it is
continuous or discontinuous at a point or an interval.

Discontinuity:

Concept of interval

i. -4 -3 -2 -1 0 1 2 3 4 5
Here,

1 and 5 are not included in the given set of rational numbers to 5 which is written as
(1, 5). It is called open interval.

ii. -4 -3 -2 -1 0 1 2 3 4 5
Here,

1 is not included but 5 is included in the given set of rational numbers 1 to 5 which is
written as (1, 5].

iii. -4 -3 -2 -1 0 1 2 3 4 5
Here,

Both number 1 and 5 are also included between the set of rational numbers 1 to 5 which
is written as [1, 5]. It is called closed interval.

Here, In the above example in the interval (1, 5) both are not included and in the interval
[1, 5), 1 is included but not 5 and so on.

104 PRIME Opt. Maths Book - X

The curve of a function y = f(x) is broken at any points between the
interval [a, b] as the given diagram, the function y = f(x) is called the
discontinuous.

Y

X' a X
O b

Y'

Discontinuity of the function can be observed by discussing the following activities.
Activity : 1

Let us consider a function f(x) = 1 x<0
Here, ( x>0
2

Left hand limit is for x < 0
Where value of x approaches to '1' from left side

i.e.
Y

1

X' O X

Y' Y X
Right hand limit is for x > 0, 2
Where value of x approaches to '2' from right hand side. O

i.e.

X'

PRIME Opt. Maths Book - X Y'

105

Let observe the graph of function at x = 0 and its near by area.
Y

2
1

X' O X

Y'

Here,
The two curves are not meet at a point and we can say the curve is broken and is not

continuous at x = 0 which is called discontinuous function at x = 0.

Activity : 2

Let us consider a function f(x) = 1 x<0
( x>0
1

Here,
Left hand limit is for x < 0
Where value of f(x) approaches to 1 from left side

i.e.
Y

1

X' O X

Y'
Right hand limit is for x > 0,
Where value of f(x) approaches to 1 from right hand side.

i.e.
Y

1 X Y
X' O

Y' 1
OX
Let observes the graph of the function at x = 0 and X'
nearer region.

Y'

106 PRIME Opt. Maths Book - X

As x approaches to O, f(x) approaches to 1 from both sides, but at x = 0, f(x). doesn't exist
as function is not defined for x = 0. So, the curve is discontinuous at x = 0, so the function
is discontinuous at x = 0.

Hence, left hand limit = right hand limit but not equal to functional value 1.
So, It is discontinuous.

Activity : 3

Let us consider a function f(x) = –x2 1 x<0
)x2 + x>0

Here,

Left hand limit = x lim f(x) lim (–x2)= – 02 = 0
" 0– x"0

Right hand limit = x lim f(x) lim (x2 + 1)= 02 + 1 = 1
" 0+ x"0

Let observe the graph of the function :

Y

(2, 5)

(1, 2)

(0, 1)
X' O X

(–1, –1)

(–2, –5)

Y'
The curves are not meeting any point. Hence it is discontinuous function.

Conditions of discontinuity:
i. The curve of the function is broken at any point.

ii. If lim f(x) does not exist.
x"a

i.e. x lim f(x) ≠ x lim f(x).
" a– " a+

iii. If lim f(x) ! f(a).
x"a

It is removable discontinuity.

iv. If lim f(x) " + 3 or –3 .
x"a

It is infinite discontinuity.

PRIME Opt. Maths Book - X 107

Exercise 2.1

1. Answer the following questions.
i) What do you mean by continuity of a function?
ii) What do you mean by discontinuity of a function?
iii) Write down open interval from 1 to 7 in symbol. Is 1 included in the function?
iv) Write down closed interval of –2 to 3. Is 3 included in the function?
v) Write down any 5 rational numbers between 0 and 1 with equal interval.

2. Write down the following in number line with features.
i) 1 to 8 natural numbers where 1 and 8 are not included.
ii) –5 to 6 integers where –5 is included but not 6.
iii) –3 to 5 rational numbers where both of them are included.
iv) An open interval of (2, 7).
v) A closed interval of [–4, 5]

3. Answer the following questions.
i) Discuss the continuity or discontinuity between the integers and rational numbers

by using number line.
ii) Write down the difference between the following interval shown in number line of

rational numbers.

-4 -3 -2 -1 0 1 2 3 4 5

-4 -3 -2 -1 0 1 2 3 4 5
iii) Discuss the running of a tortoise and a here from a point A to B deeding upon the

concept of continuity and discontinuity.
iv) Discuss the increasing height of a plant and deposited amount in a bank on

depending upon the concept of continuity and discontinuity.
v) There are 20 students in a class of a school on Sunday during new admission time

and 4 students will be added continuously every day in that week. How many
students will be there on Friday? Show in a sequence.

4. Which of the following curves of the function show the continuity at the given point? Give

reason also. ii) Y
i) Y

O a x = 2 b X O a x = a b X
PRIME Opt. Maths Book - X
108

iii) Y iv) Y

O x = 3 X a x = 4 b X
v) Y O Y

vi)

X' x = –3 O x = 4 X X' x = –2 O x = 1 x = 3 X

Y' Y'

5. a. If f(x) = x2 – 1 is a function. Draw the graph of it from the following and discuss LHL and
x –1

RHL.

x 0.5 0.9 0.99 0.999 0.9999 0.99999

f(x) = x2 – 1 ................ ................ ................ ................ ................ ................
x –1

x 1.5 1.1 1.01 1.001 1.0001 1.00001

f(x) = x2 – 1 ................ ................ ................ ................ ................ ................
x –1

b. Collects the mark obtained by students of your class in optional mathematics in first
terminal examination and tabulate in frequency distribution table by taking an class
interval 10. Also draw the frequency ogive curve of both nature (is less then and is more
than) and discuss continuity and discontinuity by taking a point in such curves.

Answer

Show to your teacher.

PRIME Opt. Maths Book - X 109

Continuity and its symbolic representation

Let us consider a curve of a function which is drawing without Y

lifting a pencil in a sheet of paper in between the given interval
[a, b] as,
It is called the continuous function. Here, the function f(x) is defined
at x → x0 which exists f(x) = f(x0).
Hence, it is the continuous function at x' → x0 X
O a x0 b
lim
i.e. x " x0 f(x) = f(x0).

or, lim f(x) = f(a).
x"a

For continuity of a function at x = a,

i. x lim f(x) = x lim f(x) = lim f(x) (i.e. limit exists at x = a)
" a+ " a– x"a

ii. lim f(x) = f(a)
x"a

A function f(x) is said to be continuous at x = a, if lim f(x) = f(a) and
x"a

is said to be continuous on the interval [a, b] if it is continuous at each

point of the interval.

i.e. x lim f(x) = x lim f(x) = lim f(x) = f(a)
" a– " a+ x"a

Continuity of function can be observed by discussing the following activities.
Activity : 4
x#0
Let us consider a function f(x) = 1 x$0
(
1

Here, Y
Left hand limit is as x ≤ 0
Where value of f(x) approaches to 1 from left side
and 1 is also included.

i.e.

1

X' O X

Y'

110 PRIME Opt. Maths Book - X

Right hand limit is as x ≥ 0,
Where value of f(x) approaches to '1' from right hand side and also equal to 1.

i.e.
Y

1 X
X' O

Y'
If we observes the graph of both limits,

Y

1

X' O X

Y'
Here, as x approaches to 'O' from left, f(x) approaches to '1'. Similarly as x approaches to

'O' from right, f(x) approaches to '1' and at x = 0, f(x) = 1. So, the graph is continuous near
by area of x = 0.
So, It is Continuous function.

Activity : 5

Let us consider a function f(x) = –x2 x<0 Y
)2x2 x$0 (2, 8)

Here, (1, 2)

Left hand limit = x lim f(x) lim = – 02 = 0
" 0– x"0

Right hand limit = x lim f(x) lim = 2 × 02 = 0
" 0+ x"0

Functional value of f(x) = f(0) = 2 × 02 = 0

X' O X
(–1, –1)

(–2, –4)
Y'

PRIME Opt. Maths Book - X 111

Let us taking graph of the function :

The portions of graph meet at a point O at x = 0 from both left hand as well as right hand

limit.

i.e.

x lim f(x) = x lim f(x) = f(0) = 0
" 0– " 0+

Hence it is continuous function.

Worked out Examples

1. Examine the function f(x) = 2x2 – 3x + 10 at x = 1 for continuity.

Solution:
f(x) = 2x2 – 3x + 10
Taking left hand limit at x = 1–,
f(x) = 2 × 12 – 3 × 1 + 10 = 9
Right hand limit at x = 1+,
f(x) = 2 × 12 – 3 × 1 + 10 = 9
Also, functional value of f(x) at x = 1,
f(x) = 2 × 12 – 3 × 10 = 9

\ x lim f(x) = lim f(x) = lim f(x) = 9
" 1– 1+ x "1

Here
f(x) = 2x2 – 3x + 10

Now lim f(x)
x "1

lim (2x2 – 3x + 10)
x "1

= 2 × 12 – 3 × 1 + 10
= 2 – 3 + 10
= 9

Again f(1) = 2 × 12 – 3 × 1 + 10
= 9

Since lim f(x) = f(1) = 9
x "1

So it is continuous at x = 1.

2. Determine wheather the function f(x) = 1 at x = 1 is continuous or not.
x–1
Solution:

Here, f(x) = 1
x –1

Now, f(x) is not defined at x – 1 = 0

So, at x = 1, it is discontinuous.

112 PRIME Opt. Maths Book - X

3. A function f(x) is defined as,
3x + 2 for x < 1

f(x) = 4 for x = 1

7x – 2 for x > 1

Is the function f(x) continuous at x = 1? If not, how do you make it continuous?

Solution:
left hand limit at x = 1

x lim f(x) = x lim (3x + 2) = 3 × 1 + 2 = 5
" 1– " 1–

Again,
Right hand limit at x = 1

x lim f(x) = x lim (7x – 2) = 7 × 1 – 2 = 5
" 1+ " 1+

Again, f(1) = 4

\ lim f(x) = 5
x "1

So,

Since lim f(x) ≠ f(1)
x "1

So, it is not continuous function.
It can be made continuous by redefining the function as the following.
f(x) = 3x + 2 for x < 1
5 for x = 1
7x – 2 for x > 1

4. A function f(x) is continuous at x = 2 which is defined as.

f(x) = 3x2 – 2 for x < 2
kx for x ≥ 2

Find the value of ‘k'.
Solution:

Taking left hand limit and right hand limit.

x lim f(x) = lim (3x2 – 2) = 3 × 22 – 2 = 10
" 2– x"2 = 2k.
= 2k
Again,

x lim f(x) = lim (kx) = k × 2
" 2+ x"2

Then, for the function to be continuous

x lim f(x) = lim f(x) = f(2)
" 2– x"2

by the question, the function f(x) is continuous,

so, x lim f(x) = lim f(x)
" 2+ x"2

PRIME Opt. Maths Book - X 113

or, 2k = 10
10
or, k = 2

\ k = 5.

Exercise 2.2

1. If f(x) = x + 1; f : R → R is a function. Answer the following questions.
i) Find f(x) where domain = {2.1, 2.01, 2.001, 2.0001, 2.00001}
ii) Find f(x) where domain = {1.9, 1.99, 1.999, 1.9999, 1.99999}
iii) Find the functional value f(2).
iv) Find left hand and right hand limit f(2).
v) Discuss continuity of f(x) for x = 2.

2. Which of the following functions are continuous?
i) f(x) = x2 at x = 3 ii) f(x) = 2 – 3x2 at x = 0
1
iii) f(x) = 3x2 – 2x + 4 at x = 1 iv) f(x) = 2x at x = 0
1
v) f(x) = 1– x at x = 1

3. Discuss the continuity of functions for the followings:
2x + 1 for x < 1

i) f(x) = 2 for x = 1 at x = 1.
x + 2 for x > 1

ii) f(x) = 2 – x2 for x ≤ 2 at x = 2.
x – 4 for x > 2

iii) f(x) = 3x – 1 for x ≤ 2 at x = 2.
2x + 1 for x > 2

iv) f(x) = x2 + 3 for x ≤ 3 at x = 3.
4x for x > 3

x2 + 4 for x < 4
v) f(x) = 20 for x = 4 at x = 4

3x + 8 for x > 4

4. Is the following function continuous? Also redefine the function to make continuous.
2x –1 for x < 2 3x + 1 for x < 1
i) f(x) = 4 for x = 2 ii) f(x) = 8 for x = 1
x + 1 for x > 2 4x for x > 1

x2 – 1 for x < 3 2x2 – 7 for x < 4
iii) f(x) = 5 for x = 3 iv) f(x) = 25 for x = 4

2x + 2 for x > 3 6x + 1 for x > 4

3x2 – 2 for x < 2
v) f(x) = 10 for x = 2

5x for x > 2

114 PRIME Opt. Maths Book - X

5. PRIME more creative questions:
Find the value of ‘k' from the following continuity function at the given point.
kx + 3 for x < 3 kx2 – 1 for x < 2
i) f(x) = 12 for x ≥ 3 ii) f(x) = 7 for x ≥ 2

iii) f(x) = 5x – 1 for x < 1 3x + 2 for x < 4
kx for x ≥ 1 iv) f(x) = 2k for x = 4

x2 + 5 for x < 5 2x + 6 for x > 4
v) f(x) = kx for x = 5

7x – 5 for x > 5

6. Project work
Draw the curve of the functions f(x) = Sinx, f(x) = Cosx and f(x) = Tanx on graph paper. Discuss
the continuity and discontinuity of the functions by taking the domain = {–360 ≤ x ≤ 360°}

Answer

1. Show to your subject teacher. ii) Continuous iii) Continuous
2. i) Continuous iii) Continuous
iv) Discontinuous v) Discontinuous
3. i) Discontinuous ii) Continuous iii) 4
iv) Continuous v) Continuous
2x – 1 for x < 2
4. i) Discontinuous; f(x) = 3 for x = 2
x + 1 for x > 2

3x + 1 for x < 1
ii) Discontinuous; f(x) = 4 for x = 1

4x – 1 for x > 1

x2 – 1 for x < 3
iii) Discontinuous; f(x) = 8 for x = 3

2x + 2 for x > 2

iv) Continuous v) Continuous
5. i) 3 ii) 2
iv) 7 v) 6

PRIME Opt. Maths Book - X 115

Limit and Continuity

Unit Test

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Time : 30 minutes

Attempt all the questions:
1. Is the given curve continuous at x = k for the given interval [a, b]?

Y

O a x = k b X

2. a. Is a function f(x) = 2x2 at x = 2 continuous?
b. Is a set of natural numbers continuous? Give reason for your answer.

c. Discuss continuity for the given number line where f : R → R.

-4 -3 -2 -1 0 1 2 3 4 5

3. a. Draw the graph of function f(x) = Sinx for the interval –2p ≤ x ≤ 2p. Discuss continuity
or not in it with reason.

b. If the function f(x) is continuous, find ‘k'.
kx + 5 at x < 3
f(x) = 14 at x = 3

c. Is the function f(x) continuous? If not how can you make continuous?
4x – 3 at x < 2

f(x) = 4 at x = 2 at x = 2
2x + 1 at x > 2

116 PRIME Opt. Maths Book - X

Unit 3 Matrices

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions 1 2 1 – 4 9 15
Weight 1 4 4 –

K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total
Questions, TM = Total Marks

Objectives : At the end of the lesson
• Students will be able to find the determinant and inverse of a square

matrix.
• Students can understand the Cramer's rule.
• Students can solve the two linear equations using inverse matrix and

Cramer's rule.
• Students can prove the properties of inverse of matrices.

Materials
• Price list of goods of a market.
• Chart of finding inverse matrix.
• Chart of informations of Cramer's rule.
• Chart of informations of equation solving by matrix method.

PRIME Opt. Maths Book - X 117

3.1 Matrices

Enjoy the recall:
The rectangular array of the numbers in rows and columns enclosed by a square [ ] or round
( ) bracket is called matrix.
• The matrix is denoted by capital letters like A, B, P & Q.
• Elements are denoted by small letters like a, b, c, d, e, ... , etc. (members)
• No. of rows ‘m' and columns ‘n' are written as m x n which is called

order of the matrix.

A = a b c
< e F
d f 2×3

i) Null matrix (Zero matrix) :

The matrix having all the elements zero is called null matrix.
0 0
O = 0 0

ii) Identity Matrix (Unit matrix) :
The square matrix having each element in main diagonal is 1 (one) and remaining all non

diagonal elements zero is called identity matrix.

I = 1 0
<0 1F

• Main diagonal is the diagonal from left top to right bottom.

iii) Equal matrices :

Any two matrices having same order and equal corresponding elements are called equal

matrices. SRTSSSSSSS04+–21 WVWWWWWWW

6

A = 2 3 , B = 2 are the equal matrices.
< F
1 – 4
–1 4

X

iv) Transpose of matrix :

A new matrix which obtains by interchanging the rows and columns of the given matrix

is called the transpose of the matrix. We denote it by A' or AT.

If A = a b
< F , then transpose of A is,
c d

AT = a c
< F
b d

Properties of tranpose
i. (AT)T = A
ii. (A + B)T = AT + BT
iii. (AB)T = BTAT.

v) Addition of matrices:
A new matrix made by adding the corresponding elements of two or more matrices

having same order is called the matrix after addition. We denote the addition of the
matrices A and B by A + B.

118 PRIME Opt. Maths Book - X

If A = a b , B = p q
< F < F
c r s
d

Then A + B = =a + p b + qG
c + r d+s

vi) Multiplication of the matrices:
The matrix AB of order m × n is obtained after the multiplication of any two matrices A

and B of order m × p and p × n respectively is called the matrix multiplication.
If any two matrices A and B of order m × p and p × n respectively, then AB is defined as,

A B AB

order: m × p p ×n m×n

Conditions for matrix multiplication :
• No. of columns of first matrix and no. of rows of second matrix should be equal.
• Elements of rows of first matrix should be multiplied to the corresponding elements of

columns of second matrix should and to be added to each other.

Like [a b] × c = [a × c + b × d] = [ac + bd]
<F
d

Example:

A = 2 3 and B = 3 –2
< F < F
–2 4
1 1

Then,

AB = 2 3 3 –2
< F < F
–2 4
1 1

=

= 18 –1
< F
–2
5

3.1.1 Determinant of the matrix:

Let, A = [5] is a square matrix of order 1 × 1 where 5 is a number associated with A.

Let, B = 2 1 is a square matrix of order 2 × 2 where 2 × 4 – 3 × 1 = 5 is a number associated
< F
3
4

with the matrix B. In both examples a single real number 5 is associated with the given square

matrices. Hence, 5 is called the determinant of the matrices A and B.

A single real number which is associated to a square
matrix is called the determinant.
Here, |A| is the determinant of A.

PRIME Opt. Maths Book - X 119

• Determinant of A is denoted by |A| or D or det(A).

• If A = a b , then determinant of A is,
< F
c
d

|A| = a b
c d

= a × d – c × b
= ad – bc.

3.1.2 Singular matrix and non-singular matrix:

Let, A = 2 3 the value of its determinant is,
<–2 –3F

|A| = 2 3 = 2 × (–3) – (–2) × 3
– 2 –3 = –6 + 6

= 0

Here,
Matrix A is singular matrix

The square matrix which has the value of determinant
zero is called singular matrix. It has no inverse.
Here, A is singular, if |A| = 0.

Let, A = 1 3
< F , the value of its determinant is,
4 2

|A| = 1 3 = 1 × 2 – 4 × 3
4 2 = 2 – 12

= – 10 ≠ 0

Here,
Matrix A is non-singular matrix .

The square matrix which has the value of determinant
non-zero is called non-singular matrix.
A is non-singular, if |A| ≠ 0.

120 PRIME Opt. Maths Book - X

3.1.3 Inverse of matrix:

If A = 2 5 and B = –7 5
< F < F are any two square matrices where,
3 –2
7 3

AB = 2 5 –7 5
< F < F
3
7 3 –2

=

= 1 0
< F
0
1

BA = –7 5 2 5
< F < F
3
3 –2 7

=

= 1 0
< F
0
1

Here, we get AB = BA = I.
As a result,

A is taken as inverse of B and B is taken as inverse of A.

If any two square matrices of same order A and B are in
such a way that AB = BA = I where |A| ≠ 0 and |B| ≠ 0.
A and B are called inverse matrices to each other.

Here, Inverse matrix of A is written as A– I

Then, AA– I = A– IA = I = 1 0
< F
0
1

Properties of inverse matrix:
i) ^A–1h–1 = A where A is non - singular square matrix.
ii) ^ABh–1 = B–1A– I where A and B are square matrices of same order and A , B ] 0.

3.1.4 Adjoint of a matrix:

If A = <a11 a12F is a square matrix.
a21 a22

where, (sign is positive according to 1 + 1 = 2 even)
CCCCooooffffaaaaccccttttoooorrrr ooooffff aaaa21122112 ==== aa––21aa1221 21 (sign is negative according to 1 + 2 = 3 odd)
(sign is negative according to 2 + 1 = 3 odd)
(sign is positive according to 2 + 2 = 4 even)


PRIME Opt. Maths Book - X 121

\ Matrix of cofactors of A = <a22 –a21F • For 2 × 2 matrix, the sign of
–a12 a11
+–
cofactor is = +G .
Transpose of cofactors of A = <a22 –a12F –
–a21 a11
• The sign of cofactor is (–1)i + j
Which is the adjoint matrix of A. Where i = row
j = column
\ Adjoint of A = <a22 –a12F
–a21 a11

The transpose matrix of the matrix formed by taking the co - factor
elements of the elements of matrix A is called the adjoint matrix of A.

• Inter change the position of elements of main diagonals change
the sign of elements of secondary diagram in a matrix to obtain its
adjoin for 2 × 2 matrix.

If A = a b
< F
c
d

\ adjoint of A = d –b
< F
–c
a

3.1.5 Formula to find inverse of a matrix:

Let, A = a b = a x d – c x b = ad – bc ≠ 0
< F is a square matrix where A
c d

(say) (i.e. A– I exists)

Let, A– I = k l
<m nF

Then, Taking the condition of inverse matrix, AA– I = I

a b k l = 1 0
< F <m nF < F
c 0
d 1

= 1 0
< F
0
1

Equating the corresponding elements,
ak + bm = 1 .......... (i)
ck + dm = 0 .......... (ii)
al + bn = 0 .......... (iii)
cl + dn = 1 .......... (iv)

Solving equation (i) and (ii), k = d bc and m = bc b ad = – ad b bc
ad – – –

Solving equation (iii) and (iv) l = bc c ad = ad c bc and n = ad a bc
– – –

122 PRIME Opt. Maths Book - X

Now,

A– I = k l = = 1 d –b
< F ad – bc < F
m n –c
a

\ A– I = 1 (adjoint of A)
A

It is the calculating formula of A– I.

Worked out Examples

1. If A = <–41 –32F is a square matrix, find its determinant.

Solution:

A = 4 –2
< F
–1
3

Determinant of A is,

|A| = <4 –2F
–1 3

= 4 × 3 – (–1) (–2)
= 12 – 2
= 10

2. Evaluate: a+b –b
Solution: b a–b

a + b –b

b a–b

= (a + b) (a – b) – b(–b)
= a2 – b2 + b2
= a2

3. If determinate of A = <1x 3xF is 13, find the value of ‘x'.

Solution: A = x 3
< F
1 x

We have,

|A| = x 3
1 x

or, 13 = x × x – 1 × 3
or, 13 + 3 = x2
or, x = 16
\ x = +4.

PRIME Opt. Maths Book - X 123

4. If A = <–13 –22F , find the determinant of A2 – 2A + 3I.

Solution: A = 3 2
< F
Then, –1
–2

A2 – 2A + 3I = 3 2 3 2 – 2 3 2 + 3 1 0
< F < F < F < F
–1 –1 –1 0
–2 –2 –2 1

= – 6 4 + 3 0
< F < F
–2 0
–4 3

=

= 4 –2
< F
1
9

Again,
Determinant of A2 – 2a + 3I is,

4 –2 = 4 × 9 – 1 (–2)
1 9 = 36 + 2
= 38.

5. If A = <–34 12F , find A– I.

32
Solution: A < F
Here, –4 1

|A| = 3 × 1 – (–4) × 2
= 3 + 8
= 11 ≠ 0 (inverse exists)

Then,

A– I = 1 (adjoint of A)
A

== 1SSSTRSSSSS11111411<14 –2 F

3
WWVWWWXWWW
–2
11
3
11

124 PRIME Opt. Maths Book - X

6. If A– I = <63 ––52F , find the matrix A. Also find A– IA.

Solution: A– I = 3 –2
< F
6 –5
Here,

A–1 = 3 × (–5) –6(–2)
= –15 + 12
= –3 ≠ 0 (inverse exists)
\ A = (A– I)–1
1
= A–1 (Adjoint of A– I)

= – 1 –5 2
3 < F
–6
3

5 – 2 H
3
= >3
2 –1

Again,

A– IA = 3 –2 × – 1 –5 2
< F 3 < F
6 –6
–5 3

= – 1
3

= – 1 –3 0
3 < F

0 –3

= 1 0
< F
0
1

= I
\ A– IA = I

7. Prove that the matrices <32 75F and <–52 73F are inverse to each other.
Solution:

A = 3 7 , B = 5 7
< F < F
2 –2
5 3

Now,

AB = 3 7 5 7
< F < F
2 –2
5 3

=

= 1 0 = I
< F
0
1

BA = 5 7 3 7
< F < F
–2 2
3 5

=

= 1 0 = I
< F
0
1

PRIME Opt. Maths Book - X 125

Here, AB = BA = I,
Hence, they are inverse to each other.

8. If the matrices <5x 74F and <–34 –y5F are inverse to each other, find the value of ‘x' and ‘y'.
Solution

A = 5 7 and B = –4 y
< F < F
x
4 3 –5

Since, A and B are inverse to each other,
AB = BA = I

or, 5 7 –4 y = 1 0
<x 4F <3 –5F <0 1F

or, = 1 0
< F
0
1

or, = 1 0
< F
0
1

By equating the corresponding elements,
–4x + 12 = 0 and 5y – 35 = 0
or –4x = – 12 and 5y = 35
or, x = 3 and y = 7
\ x = 3
y = 7

9. If A = <32x 6xF is a singular matrix. Find the value of ‘x'

Solution:

A = 2x 6 is a singular i.e. |A| = 0
3 x

or, 2x 6 = 0
3 x

or, 2x × x – 3 × 6 = 0
or, 2x2 = 18

or, x2 = 9
\ x = + 3.

10. If A = <12 –02F , B = <01 –21F and C = <–04 00F , prove that AB – C is a singular matrix.

Solution:

A = 2 0
< F
1
–2

B = 1 –1
< F
0
2

C = 0 0
< F
–4
0

126 PRIME Opt. Maths Book - X

Then,

AB – C = 2 0 1 –1 – 0 0
< F < F < F
1 0 –4
–2 2 0

= – 0 0
< F
–4
0

= 2 –2 – 0 0
< F < F
1 –4
–5 0

= 2 –2
< F
5
–5

Here, Determinant,

AB – C = 2 × (–5) –5(–2)
= – 10 + 10
= 0 Hence, it is singular matrix.

Exercise 3.1

1. Answer the following questions.
i) What do you mean by inverse matrix?
ii) What do you mean by singular matrix?

iii) Is A = –2 3 a singular?
< F
–4
6

iv) What do you mean by adjoint of a matrix.

v) Write down the adjoint matrix of A = 3 –2
< 4 F.
–1

2. Find the determinant of the following matrices.

i) A = 3 4 ii) B = –3 –2
< F < F
7
9 5 7

iii) A = 2 × 2 matrix having element aij = 3i – 2j.

iv) P = 2 × 2 matrix having element aij = i + 2j.

v) Determinant of 2A where A = 1 –2
< F
–2 3 3
< F 2
–4
vi) A2 – 2A + 3I where A = 5

3. Evaluate the followings:

i) –7 6 ii) a + b b –a
5 –9 ab

iii) m2 –n2 m + n iv)
m–n 1

v) AB where A = 2 –1 and B = 3 2
3 –1 < F
1
–1

PRIME Opt. Maths Book - X 127

4. i) If determinant of A = 2 x is 7, find the value of x.
< F
–3
–4

ii) If M = 2m –5 M = –4, find the value of m.
< F and
–3
4

iii) If B = 7 5 = 3, find the value of ‘a'.
< F and B
a 4

iv) If A 1 2 , B = 4 1 and AB = 6, find the value of ‘P'
< F < F
3 P
4 3

v) If A = 2 3 , B = 3 5 and AB = 0, find the value of ‘m'.
< F < F
m 1
6 7

5. i) Prove that A = 6 3 is a singular matrix.
< F
4
2

ii) Prove that AB is a singular matrix where A = 2 3 and B = 3 5
< F < F .
4 1 7
6

iii) If P = 1 2 and Q = 2 –1
< F < 3 F , prove that PQ has no inverse.
3 0
6

iv) Show that A = 3 4
< F has inverse.
–2 –5

v) If A = 1 0 and B = –1 2 B = AB
< F < F , show that
–5 –3 4 A
2

6. Find the inverse of the following matrices.

i) A = –3 –2 ii) B = 8 –5
< F < F
3
7 5 –2

iii) A = 2 × 2 matrix having elements aij = 3i – j.

iv) Inverse of 3A where A = 2 –1
< F .
1 –2

v) Inverse of (A + B) where A = 4 –1 and B = 3 4
< F < F .
2 3 1
1

7. i) If A = 3 2 , find AA– I.
< F
1 –2

ii) Prove that P = –3 2 and Q = –5 2 are inverse to each other.
< F < F
–7 –7
5 3

iii) If A– I = –7 –5 , find the matrix ‘A'.
< F

4 3

iv) If P–1 = 3 2 , find P–1P.
< F
25
17

v) If determinant of A = m –1 is 6, find A– I.
< F
2
1

128 PRIME Opt. Maths Book - X

8. i) If a matrix n –5 8 –5 find the value of ‘m' and ‘n'.
< F , is the inverse of matrix < F
3 –8 m
–2

ii) If A = a 3 and B = >1 –3
< F
2 –1 2 H are inverse to each other, find ‘a' & ‘b'.
2
b

iii) If the matrices <p –7F and <3 –7F are inverse to each other, find the value of p and
q. 2 –3 q –5

iv) If matrices < –1 2F has its inverse <n 2F , find the value of ‘m' and ‘n'.
2m –7 4 1

v) If >–1 – 5 H is the inverse of matrix –6 a , find the value of ‘a' and ‘b'.
2 < F
–1
–3 b –2

9. PRIME more creative questions:

i) If A = –1 2 , find the determinant of A2 – 3A + 2I.
< F
–2
3

ii) If A = 3 2 , B = 2 –1 and AB does not have inverse, find the value of x.
< F < F
x 3
4 2

iii) If A = 3 2 and B = –6 5 , find (AB)–1 Also find (B–1A– I) & write down the
< F < F
7 5 –2 2

conclusion.

iv) If P = 3 5 and Q = 5 4 , prove that (PQ)–l = Q–lP–l .
< F < F
4 3
7 2

v. If A– I = 5 7 and B–l = 2 5 , prove that (AB)T = BTAT.
< F < F
3 3
4 7

Answer

1. Show to your teacher.

2. i) –1 ii) –11 iii) 6 iv) –2 v) 32
iii) 0
3. i) 33, ii) a2 + b2 iii) 5 iv) 1 v) –5
iii) SSSSSSTSSR–4533 –2133WWWWWWVXWW
4. i) 5 ii) 4 iv) 15 v) 4
2TSSSSSSSSSRSSRSSSSST–21, 17199 63 3699WVXWWWWWWWWWWWWWWWXV
6. i) –5 –2 ii) 2 –5 iv) –1 v) –2 3
< F < F –2 v) < F
3 v)
7 3 –8 iv) 1 5 –7
2 5, 2
7. i) 1 0 iii) –3 –5 iv) 1 0
< F < F < F
0 0
1 4 7 1

8. i) 3, 2 ii) 4, 2 iii) 5, 2
–45 19
9. i) 0 ii) 6 iv) (AB)–l = B–lA–l
iii) > 2 2H

–26 11

PRIME Opt. Maths Book - X 129

3.2.1 Solution of simultaneous equations using matrix method

Two linear equation ax + by + c = 0 and px + qy + r = 0 are called the simultaneous equations

which can be solved by using matrix method by converting the linear equations in matrix

form as follows.
ax + by = – c
px + qy = – r

a b x –c
The equations in matrix form, < F< F =< F
p qy –r

Let, A = a b , X = x and B = –c
< F <F <F
p y –r
q

The equations becomes, AX = B
Multiplying both sides by A– I,
or, A– IAx = A– IB
or, I X = A– IB ( a A– IA = I)
or, X = A– IB ( a I X = X)
It is possible only for A ! 0 to calculate A– I, and we get a unique of the system.

Example: Solve the linear equations x + 3y = 7 and 5x – y – 3 = 0 by matrix method.
Solution :
The given equation are

x + 3y = 7 ....................................(i)
5x – y = 3 ................................(ii)

The equations in matrix form:

13 x 7
< F< F =< F
5 –1 y 3

Let, A = 1 3 , X = x and B = 7
< F <F <F
5 y 3
–1

The equations becomes,
A X = B
or, A– IA X = A– IB
or, I X = A– I B [ a A-1A = I]
\ X = A– IB .................................(iii)
[ a IA = A]

Here, |A| = 1 3
5 –1

= 1 × (–1) – 5 × 3
= –1 – 5
= – 16 ≠ 0 (inverse exists)

Then, From equation (iii),
X = A– IB
1
= A (adjoint of A). B

= 1 –1 –3 7
–16 < F< F
–5
13

130 PRIME Opt. Maths Book - X

= 1
–16

= – 1 – 16
16 <F
– 32

x = 1
<F <F
y 2

\ x = 1
y = 2

3.2.2 Solution of simultaneous equations using Cramer's rule

Cramer's rule:
System of linear equation of two variables can be solved by using different rules like
substitution method, elimination method, cross - multiplication method, graphical method,
etc. where above methods were already discussed in previous classes in compulsory
mathematics and matrix method is discussed already in this chapter.

Here, we introduce a new method of solving the system of linear equations known as Cramer's
rule.

It is the method of solving the system of linear equation
of two variables using determinant of a square matrix of
2 × 2 order.

Here, Taking two simultaneous equations,
aa21xx ++ bb21yy == cc21 ................................................................................ ((iii))

Multiplying equation (i) by b2 and equation (ii) by b1 and subtracted
(–)aa12bb21xx( ++– )bb11bb22yy ==(– bb)21cc12

x(a1b2 – a2b1) = (b2c1 – b1c2)

c1 b1

\ x = b2 c1 – b1 c2 = c2 b2 = Dx
a1 b2 – a2 b1 a1 b1 D

a2 b2

PRIME Opt. Maths Book - X 131

Similarly,
M(–u)laat11iaap22lxxy (i++n– g)aa 12ebbq12uyya ==(t– iaao)21ncc 21(i) by a2 and equation (ii) by a1 and subtracted,

y(a2b1 – a1b2) = (a2c1 – a1c2)

a1 c1

\ y = a2 c1 – a1 c2 = a2 c2 = Dy
a2 b1 – a1 b2 a1 b1 D

a2 b2

\ X = Dx and y = Dy where D ≠ 0.
D D

According to Cramer's rule.

Coefficient of x Coefficient of y Constant

a1 b1 c1

a2 b2 c2

Where,
D = Determinant of coefficients of two equations.

= a1 b1
a2 b2

Dx = a1 and a2 of D are replaced by constants c1 and c2.
c1 b1
= c2 b2 . It is also written as D1

Dy = b1 and b2 of D are replaced by constants c1 and c2.
a1 c1
= a2 c2 . It is also written as D2

Note :

• Two simultaneous equations should be written in order of x and y and constants should be

taken in right hand side.

a1x + b1y = c1
a2x + b2y = c2
• Finding D, Dx and Dy as above,

• Finding x = Dx and y = Dy
D D

132 PRIME Opt. Maths Book - X

Example: Solve 3x – 2y = 5 and 5x + y = 17 by using Cramer's rule.
Solution: The given equations are:

3x – 2y = 5 and
5x + y = 17

Coefficient of x Coefficient of y Constant
3 –2 5

5 1 17

Here,

D = 3 –2 = 3 × 1 – 5(–2) = 13
5 1

Dx = 5 –2 = 5 × 1 – 17(–2) = 39
17 1

Dy = 3 5 = 3 × 17 – 5 × 5 = 26
5 17

Then, values of the variables of the given equations are:

x = Dx = 39 =3
D 13

y = Dy = 26 =2
D 13

\ x = 3
y = 2

Worked out Examples

1. If <12 –53F <xyF = <–92F , find ‘x' and ‘y' using inverse of the 2 x 2 matrix.

2 –3 x 9 Steps of solving simultaneous
Solution: The given matrix form is, < F< F =< F • Writing the equations in
1 5 y –2

Let, A = 2 –3 , X = x and B = 9 the form of :
< F <F <F
1 y –2 aa21xx ++ bb12yy == cc12
5 Write the equations

We get, A X = B • in
or, A– I A X = A– I B
matrix from
or, I X = A– I B • Finding X = A– I B
or, X = A– I B ..........(i)
• Multiplying A and B for x
<F
y

Here,

A = 2 –3 = 2 x 5 –(–3) = 13 ≠ 0 (inverse exists)
1 5

Then From equation (i),
X = A– I B
1
or, X = A (adjoint of A) . B

PRIME Opt. Maths Book - X 133

= 1 5 39
13 < F< F
–1
2 –2

= 1
13

= 1 39
13 <F
–13

x = 3
<F <F
y –1

\ x = 3
y = –1

2. Solve by matrix method of 3x + 4 = 7 and 5xy – 8 = 8y.
y
Solution: The given equations are:

3x + 4 = 7 and 5xy – 8 = 8y
y

or, 3x + 4 = 7 and y(5x – 8) = 8
y

or, 3x + 4 = 7 and 5x – 8 = 8.
y y

Now, The equations in matrix form are,

3 4 x H = 7
< F <F
5 >1 8
–8
y

Let, A = 3 4 , X = x H , B = 7
< F <F
5 >1 8
–8
y

Then, we get,
A X = B

or, A– I A X = A– I B
or, I X = A– I B
or, X = A– I B ..........(i)

Here, |A| = 3 4
5 –8

= 3(–8) – 5(4)
= – 44 ≠ 0 (Inverse exists)

From equation (i),
X = A– IB
1
or, X = A (adjoint of A) . B

= 1 <–8 –4F <7F
–44 –5 38

= – 1
44

134 PRIME Opt. Maths Book - X

= – 1 –88
44 <F
–11

x2

or, >1 H= >1 H

y4

\ x = 2, y = 4

3. Solve the equation x + 5y = – 3 and x – 3y = 10 by using Cramer's rule.
Solution: 2 3 4

The given equations are:

x 5y = –3 and x – 3y = 10
2 +3 4

or, 3x + 10y = – 18 and x – 12y = 40

The equations in tabular form are: Constant
Coefficient of x Coefficient of y –18
3 10
40
1 –12

Here, D = 3 10 = 3 × (–12) – 1 x 10 = – 46
1 –12

Dx = –18 10 = (– 18) × (–12) – 40 × 10 = 184
40 –12

Dy = 3 –18 = 3 × 40 – 1(–18) = 138
< F
1
40

Then, The value of the variables,

x = Dx = –184 = 4
D –46

y = Dy = 138 = – 3
D –46

\ x = 4
y = – 3

4. Solve the equations 6 – 12 = – 1 and 8x – 9y = – xy by using cramer's rule.
x y

Solution: The given equations are,

6 – 12 = – 1 and 8x – 9y = – xy
x y

or, 6 – 12 = – 1 and 8x 9y = xy
x y xy – xy – xy

or, 6 – 12 = – 1 and 8 – 9 = – 1
x y y x

or, 6 – 12 = – 1 and 9 – 8 = 1
x y x y

PRIME Opt. Maths Book - X 135

The equations in tabular form are, Constant
–1
Coefficient of x Coefficient of y
6 –12

9 –8 1

Here,

D = 6 –12 = 6(–8) – (9)(–12) = 60
9 –8

Dx = –1 –12 = (–1)(–8) – 1(–12) = 20
1 –8

Dy = 6 –1 = (6) × 1 – (9)(–1) = 15
< F
9
1

Then, The value of the variables are: 1 = Dx = 20 = 1 & x =3
x D 60 3

1 = Dy = 15 = 1 & y =4
y D 60 4

\ x =3, y = 4
5. The cost of 4 chairs and 3 tables is Rs.7700 and cost of 5 chairs and 2 tables is Rs.7000, find

the cost of a chair and a table by using Cramer's rule.
Solution:
Let, cost of a chair be x and cost of a table be y
Then,
4x + 3y = 7700 and
5x + 2y = 7000
The equations in tabular form are:

Coefficient of x Coefficient of y Constant

4 3 7700

5 2 7000

Here,

D = 4 3 = 4 × 2 – 5 × 3 = – 7
5 2

Dx = 7700 3 = 7700 × 2 – 7000 × 3 = – 5600
7000 2

Dy = 4 7700 = 4 × 7000 – 5 × 7700 = – 10500
5 7000

Then, value of the variables,

x = Dx = –5600 = Rs. 800
D –7

y = Dy = –10500 = Rs. 1500
D –7

\ Cost of a chair = Rs.800
Cost of a table = Rs.1500

136 PRIME Opt. Maths Book - X

Exercise 3.2

1. Find the value of ‘x' and ‘y' from the followings by matrix method.

i) 3 1 x = 10 ii) 7 –3 x = 17
< F <F <F < F <F <F
4 y 14 8 y 21
2 –5 SSRSSSSSS11yx WWWWWWWWV

iii) 9 –7 1 H = 10 iv) 8 3 = 1
< F < F <F
5 >x >2 H 4 3
1 –6
y 3

v) A x = B where A = 2 7 23 TX
<F < F , B =< F
y 8 –1
5

2. Find the value of x and y from the following by using Cramer's rule.
i) Write down the formula of finding x and y variables according to Cramer's rule.
ii) IFfi nDdx =D 2x f0r,o Dmy =th 2e4 e aqnuda tDio =n s4 2, fxin –d 3 xy a=n 3d ayn uds i3nxg + C yra =m 2e1r.'s rule.
iii)

23 7 2 23 1 3 81

iv) x = 5 –1 and y = 8 5 v) 1 = 3 –6 and 1 = 4 3
2 7 2 7 x 8 3 y 8 3

8 –1 8 –1 4 –6 4 –6

vi) 7 –3 x = 17 vii) 9 –7 1 10
< F <F <F < F
8 y 21 5 >xH=>2 H
–5 1
y3

3. Solve the following linear equations by using matrix method.
a. i) 5x – 2y = 9 and 7x – 3y = 13 ii) 3x + 5y = 19 and x – 4y + 5 = 0
x y 5x y
iii) 2x – 3y + 16 = 0 and 5y = 14 – 3x iv) 4 + 6 = 3 and 2 – 12 = 9

v) 3x 5y = 2 and x + 7y = 9
2 –3 4 6

b. i) 6 + 5y = 7 and 12 – 3y = 1 ii) 7x – 4 = 12 and 12 – 5x = – 4
x x y y

iii) 12 + 10 = 7 and 4 – 6 = 1 iv) 5x + 3 = 3 and 3x – 6 = – 1
x y y x 2y y 2

v) 12 – 2y = 1 and 2 5y = 1
5x x –2

4. Solve the following linear equations by using Cramer's rule.
i) x – 4y = – 5 and 3x + 5y = 19 ii) 3x + 5y = 14 and 2x –3y = – 16
8 12
iii) 3x = 36 – 2y and 30x – y = 108 iv) x + 3y = 8 and x – y =1

v) 9x – 4 = 19 and 3x – 8 =8
y y

5. PRIME more creative questions:
i) Solve 3x – 4y = 5x – 7y = 1, by using Cramer's rule.
2x + 5y 3x – 7y
ii) Solve 3 = 19 = 1, by using matrix method.

iii) Solve 4 – 7xy = 12x and 3xy – 4 = – 4x by using Cramer's rule.

PRIME Opt. Maths Book - X 137

iv) Solve 8x + 12y = 5xy and 2 – 2 = 0 by using matrix method.
x y

v) Cost of 10 pencils and 3 pens is Rs. 160 and cost of 6 pencils and 8 pens is
Rs. 220, find the cost of a pencil and a pen by using matrix method.
vi) Cost of 2 books and 5 copies is Rs. 550 and 5 books and a copy is Rs. 800,
find the cost of a book and a copy using Cramer's rule.

6. Project work
Collect the cost of two types of vegetables of your local market on the specific day of two

different weeks. Express the cost of them of 2kg and 3kg in first week and 5 kg and 8 kg
in the second week in simultaneous equations. Find the rate of the their cost by solving
using matrix method and using Cramer's rule separately.

Answer

1. i) 3, 1 ii) 2, –1 iii) 3, –1 iv) –3, 4 v) 1, 3
2. ii) 5, 6 iv) 1, 3 v) –3, 4
3.a. i) 1, –2 ii) 3, 2 iii) –2, 4 vi) 2, –1 vii) 3, –1

b. i) 3, 1 ii) 2, 2 iv) 4, 12 v) 8, 6
4. i) 3, 2 ii) –2, 4
5. i) 3, 2 ii) 4, –1 iii) 6, 2 iv) 1 , 3 v) 4, – 1
2 5

iii) 4, 12 iv) 4, 2 v) 2, –4

iii) –2, –2 iv) 4, 4 v) 10, 20 vi) 150, 50

Matrices

Unit Test Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. Find the determinant of A = 2 –1
< F
3 2

2. a) Which matrix pre-multiplies to 30 results [7 –1]?
< F
2 1

b) If A = 21 , find the determinant of A2 – 2A + 3I.
< F
3 1

x 5 2 –5
c) If A = < F and B = < y F are inverse to each other, find the value of ‘x' and ‘y'.
1 2 –1

3. a) Solve the equations 3x – 2y – 5 = 0 and 5x + 3y – 21 = 0 using Cramer's rules.
b) Solve the equations x + 5y – 12 = 0 and 3x – y = 4 using matrix method.

4. If A = 3 2 3 –4 , prove that (AB)–1 = B–1A– I.
< F and B = < F
5 4 3 –5

138 PRIME Opt. Maths Book - X

Unit 4 Co-ordinate Geometry

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions 2 2 1 1 6 15 30
Weight 2 4 4 5

K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total
Questions, TM = Total Marks

Objectives : At the end of the lesson
• Students are able to find angle between any straight lines & Condition of

perpendicular and parallel.
• Students are able to find homogeneous equation, separate equations and

angle between the line pairs.
• Students are able to find equation of circle in different forms.
• Students are able to understand the conic section of cone and plane figure.

Materials
• Chart of formula used in co-ordinate.
• Chart of the forms of Conic section.
• Scissors, tape, chart paper etc.
• Chart of trigonometric ratios for standard angles.
• Graph paper.
• Geo-board.

PRIME Opt. Maths Book - X 139

4.1 Co-ordinate

Enjoy the recall:
• Co-ordinate of a point is the pair of abscissa and ordinate of the point.
for the point (x, y).
x = abscissa y = ordinate

• Co-ordinate of the different points in standard positions.
Origin → 0(0, 0)
Point on x-axis → A(x, 0)
Point on y-axis → B(0, y)

• Sign of components in quadrant:
Y

2nd quadrent –– –– –– –– –– –– ++ ++ ++ ++ ++ 1st quadrent
(–, +) (+, +)

X' –– –– –– –– –– –– –– –– + + + + + + + + X
+ + + + + + +

3rd quadrent 4th quadrent
(–, –) (+, –)

Y'
• Inclination of a straight line :

It is the angle made by a straight line with x-axis in positive sense (anticlock wise direction).

• Slope of straight line :
It is the trigonometric tangent of angle of inclination of straight line.
If q be the inclination, slope = Tanq = m.

Also, it is called the gradient of the straight line.

• Formula of slope of a straight line:
Slope (m) = Tanq (for angle with x-axis)

Slope (m) = – Coeffient of x (for equation)
Coefficient y

Slope (m) = y2 – y1 [For two points]
x2 – x1

• Equation of straight lines: x y
a b
In double intercepts form: + = 1.

In slope – intercept form: y = mx + c
In slope – point form: y – y1 = m(x – x1)
y2 – y1
In double points form: y – y1 = x2 – x1 (x – x1)

In perpendicular form: xCosa + ySina = P
First degree equation: ax + by + c = 0

140 PRIME Opt. Maths Book - X

• Distance formula:
d =
for any two points

P = ax1 + by1 + c for a point and a straight line.
a2 + b2

• Section formula:
Internally: (x, y) =

Externally: (x, y) =

Mid- Point: (x, y) =

Centroid of a triangle: (x, y) =

• Area of closed shape
x1 x2 x3 x1
1
Triangle: A = 2 y1 y2 y3 y1

A = 1 × base × height
2

1 x1 x2 x3 x4 x1
2
Quadrilateral: A = y1 y2 y3 y4 y1

Note:
• Above informations can be applied when necessary.
• Such formula can be taken in different necessary situations according to the form

of equations.
• But syllabus for grade X are as follows.
i. Angle between any two straight lines.
ii. Line pairs of second degree homogeneous equation.
iii. Conic Section
iv. Equation of circle.

These are discussing as the following in different topics.

4.1.2 Angle between any two straight lines

Let ‘q' be the angle between any two straight lines Y
AB and CD having equations, O
YY == mm21 xx ++ CC12 -------------------- ((iii))
A C(180° – q)

Where q1 and q2 are the inclinations made by q
them with x-axis respectively.
\ SSllooppee ooff ACDB ((mm21)) == TTaannq q21
X' q2 q1 X
D B

Y'

PRIME Opt. Maths Book - X 141

Geometrically,

we have,

[Exteriqo1r =\ q= + s qu2m o f t w o interior opposite \ s in a triangle]
Tqa =n qq1 = – Tqa2n(q1 – q2) [ a Taking Tan on both sides ]
or,
or,

or, Tanq =

or, Tanq = m1 – m2 .......................... (i)
1 + m1 m2

Also, m1 – m2
1 + m1 m2
Tan (180° – q) = –

Here, we have:

Value of Tanq is positive for acute angle ‘q' and negative for obtuse angle ‘q'.
m1 – m2
\ Tanq = ! 1 + m1 m2

` q = Tan–1 8! m1 – m2 B
1 + m1 m2

It is the required angle between the

straight lines

i) Condition of the two lines being parallel.

For the parallel straight lines,
q = 0°
or, Tanq = Tan0°
m1 – m2 Sin 0°
or, 1 + m1 m2 = Cos0°

or, m1 – m2 = 0
1 + m1 m2 1

or, m1 – m2 = 0
\ m1 = m2

ii) Condition for the two lines being perpendicular,

For the lines perpendicular,
q = 90°

or, Tanq = Tan90°

or, m1 – m2 = Sin90°
1 + m1 m2 Cos90°

or, m1 – m2 = 1
1 + m1 m2 0

or, 1 + m1 m2 = 0
\ m1 m2 = – 1

142 PRIME Opt. Maths Book - X

Angle between the straight lines having equations a1x + b1y + c1 = 0 and a2x + b2y
T+hce2 g=iv0e.n equation of straight lines AB and CD are respectively,
aa21xx ++ bb21yy ++ cc21 == 00 ......................((ii)i )
YA D

Here, a2x + b2y + c2 = 0 q

Slope of AB (m1) = – Coefficient of x a1x + b1y + c1 = 0
Coefficient of y

= – a1 q1
b1 B
q2
Slope of CD (m2) = – a2 X' O X
b2

For the angle q between the st lines 'q' C
m1 – m2 Y'
Tanq = ± 1 + m1 m2

or, Tanq = ±

–a1 b2 + a2 b1

or, Tanq = ± b1 b2
b1 b2 + a1 a2

b1 b2

or, Tanq = ± a2 b1 – a1 b2
a1 a2 + b1 b2

\ q = Tan–1 :! a2 b1 – a1 b2 D
a1 a2 + b1 b2

It is the angle between the given lines.

i) Conditions of the lines being parallel,
m1 = m2
a1 a2
or – b1 = – b2

or, a1 bb22 –= aa22 bb11 = 0
\ a1

ii) Condition of the lines being perpendicular,
m1 m2 = – 1

or, `– a1 j `– a2 j = – 1
b1 b2

or, a1a2 = – b1b2
\ a1a2 + b1b2 = 0

PRIME Opt. Maths Book - X 143

Worked out Examples

1. Find the angle between the straight lines having equations x – 3y + 2 = 0 and 3x – y + 3

= 0

Solution:

The given equation of straight lines are:

x – 3y + 2 = 0 ........................................ (i)
3x – y + 3 = 0 ........................................ (ii)
Coefficient of x
Slope of line(i)(m1) = – Coefficient of y

= – 1
–3

= 1
3

Slope of line (ii) (m 2) = – 3
–1

= 3

Now,
For the angle between the lines ‘q'.

Tanq = ± m1 – m2
1 + m1 m2

1– 3

= ± 1+ 3 # 3
1

3

1–3

= ! 3
2

= ! 1
3

or, q = Tan–1 c! 1 m
3

= 30°, (180° – 30°)
= 30°, 150°

2. If the straight lines having equations (p + 1)x + 3y – 7 = 0 and 3x – 7y + 5 = 0 are perpendicular

to each other, find the value of ‘p'.

Solution:

The equation of straight lines are,
(p + 1)x + 3y – 7 = 0 ........................................ (i)
3x – 7y + 5 = 0 ........................................ (ii)
p+1
Here, slope of line (i) (m1) = – Coefficient of x = – 3
Coefficient of y

Slope of line (ii) (m2) = – 3
–7

= 3
7

144 PRIME Opt. Maths Book - X

By the question,

The lines are perpendicular,

So, m1m2 = –1
+
or, a– p 3 1 k. 3 = –1
7

or, –3p – 3 = – 21
or, 18 = 3p

or, p = 18
3

\ p = 6

3. Prove that the line 8x – 6y + 5 = 0 and line joining the points (3, –2) and (12, 10) are parallel

to each other.

Solution:

The given equation of straight line is,
8x – 6y + 5 = 0 ........................................ (i)
Line joining the points A(3, –2) and B(12, 10) is line ........................................ (ii)

Now, = – Coefficient of x
Slope of line (i) (m1) Coefficient of y

= – 8
–6

= 4
3

Slope of line (ii) (m2) = y2 – y1
x2 – x1

= 10 + 2
12 – 3

= 12
9

= 4 4
3 3

iH.ee.rme,1 S =lo mpe2 (m1) = Slope (m2) = ,

Hence, the given lines are parallel to each other. A (1, 3)

4. Find the slope of diagonal BD of a rhombus ABCD D
C(7, –1)
where two opposite vertices are A(1, 3) and C(7, –1).

Solution:
The two opposite vertices of a rhombus ABCD are
A(1, 3) and C(7, –1).
= y2 – y1
Slope of diagonal AC (m1) x2 – x1 B

= –1 –3
7–1

= –4
6

= – 2
3

PRIME Opt. Maths Book - X 145