11th Business and Statistics Maths Textbook Volume 1

Example 4.21
Find sin105o + cos105o

Solution

sin105o + cos105o

= sin(90o + 15o)+cos105o

= cos15o + cos105o

= cos105o + cos15o = 2 cosb 105c + 15c l cos b 105c - 15c l
2 2
= 2 cosb 1220c lcosb 920c l

= 2cos60o cos45o = 1
2

Example 4.22

Prove that ^cos a + cos bh2 + ^sin a + sin bh2 = 4 cos2 b a - b l
2

Solution

cos a + cos b = 2 cos b a + b l $ cos b a - b l ……….(1)
2 2 ……….(2)

sin a + sin b = 2 sin b a + b l $ cos b a - b l
2 2

Squaring and adding (1) and (2)

^cos a + cos bh2 + ^sin a + sin bh2

= 4 cos2 b a + b l cos2 b a - b l + 4 sin2 b a + b l cos2 b a - b l
2 2 2 2

= 4 cos2 b a - b l;cos2 b a + b l+ sin2 b a + b lE
2 2 2

= 4 cos2 b a - b l
2

Exercise 4.3

1. Express each of the following as the sum or difference of sine or cosine:

(i) sin A8 sin 38A (ii) cos]60c + Agsin]120c + Ag
(iii) cos 73A sin 53A (iv) cos 7i sin 3i

T ri g onom et ry 143

2. Express each of the following as the product of sine and cosine

(i) sinA + sin2A (ii) cos2A + cos4A

(iii) sin 6i - sin 2i (iv) cos 2i - cos i

3. Prove that (ii) tan 20c tan 40c tan 80c = 3
(i) cos 20c cos 40c cos 80c = 18

4. Prove that

(i) ^cos a - cos bh2 + ^sin a - sin bh2 = 4 sin2 b a - b l
2
(ii) sin A sin (60c + A) sin (60c - A) = 14 sin 3A

5. Prove that

(i) sin]A - Bgsin C + sin]B - Cgsin A + sin]C - Agsin B = 0

(ii) 2 cos r cos 91r3 + cos 31r3 + cos 51r3 = 0
13

6. Prove that

(i) csoins 22AA - scions33AA = tan A2
-

(ii) cos 77AA + scions55AA = cot A
sin -

7. Prove that cos 20c cos 40c cos 60c cos 80c = 116

8. Evaluate

(i) cos 20° + cos 100° + cos 140°

(ii) sin 50° - sin 70c + sin 10°

9. If cos A + cos B = 12 and sin A + sin B = 14 , prove that tan b A + B l = 12
2

10. If sin^y + z - xh,sin^z + x - yh, sin^x + y - zh are in A.P, then prove that
tan x , tan y and tan z are in A.P

11. If cosecA + sec A = cosecB + sec B prove that cot b A + B l = tan A tan B
2

144 11th Std. Business Mathematics

4.4 Inverse Trigonometric Functions

4.4.1 Inverse Trigonometric Functions

The quantities such as sin-1 x, cos-1 x, tan-1 x etc., are known as inverse trigonometric
functions.

i.e., if sin i = x , then i = sin-1 x . The domains and ranges (Principal value branches)
of trigonometric functions are given below

Function Domain Range
sin-1 x x
y
[–1, 1]
: -2r , r D
2

cos-1 x [–1, 1] [0, r ]
tan-1 x R = ^-3, 3h
cosec-1 x b -2r , r l
sec-1 x R–(–1, 1) 2
cot-1 x R–(–1, 1)
: -2r , r D, y ! 0
R 2

: -r , r D, y ! r
2 2 2

(0, r )

4.4.2 Properties of Inverse Trigonometric Functions

Property (1) (ii) cos-1 ]cos xg = x
(i) sin-1 ]sin xg = x (iv) cot-1 ]cot xg = x
(iii) tan-1 ]tan xg = x (vi) cosec-1 ]cosec xg = x
(v) sec-1 ]sec xg = x

Property (2)

(i) sin-1 b 1 l = cosec-1 (x) (ii) cos-1 b 1 l = sec-1 (x)
x x
1 1
(iii) tan-1 b x l = cot-1 (x) (iv) cosec-1 b x l = sin-1 (x)

(v) sec-1 b 1 l = cos-1 (x) (vi) cot-1 b 1 l = tan-1 (x)
x x

T ri g onom et ry 145

Property (3) (ii) cos-1 (- x) = r - cos-1 (x)
(i) sin-1 (- x) =- sin-1 (x) (iv) cot-1 (- x) =- cot-1 (x)
(iii) tan-1 (- x) = - tan-1 (x) (vi) cosec-1 (- x) =- cosec-1 (x)
(v) sec-1 (- x) = r - sec-1 (x)

Property (4)

(i) sin-1 (x) + cos-1 (x) = r (ii) tan-1 (x) + cot-1 (x) = r
2 2

(iii) sec-1 (x) + cosec-1 (x) = r
2

Property (5)

If xy<1, then

(i) tan-1 (x) + tan-1 (y) = tan-1 d x+y n
1 - xy

(ii) tan-1 (x) - tan-1 (y) = tan-1 d x-y n
1 + xy

Property (6)
sin-1 (x) + sin-1 ^yh = sin-1 _x 1 - y2 + y 1 - x2 i

Example 4.23
Find the principal value of (i) sin-1 _1 2i (ii) tan-1 ^- 3 h

Solution

(i) Let sin-1 _1 2i = y where - r #y# r
2 2

` sin y = _1 2i = sin b r l
6

y = r
6

The principal value of sin-1 _1 2i is r
6

(ii) Let tan-1 ^- 3 h = y where - r #y# r
2 2

` tan y = – 3 = tan b - r l
3

y = - r
3

The principal value of tan-1 ^- 3h is - r
3

146 11th Std. Business Mathematics

Example 4.24
Evaluate the following (i) cosbsin-1 153 l (ii) tanbcos-1 187 l

Solution ... (1)

(i) Let bsin-1 153 l = i
sin i = 153

cos i = 1 - sin2 i

= 1 - 12659
= 1132
... (2)

From (1) and (2), we get

Now cosbsin-1 153 l = cos i = 1132
(ii)Let bcos-1 187 l = i

cos i = 187

sin i = 1 - cos2 i

= 1 - 26849
= 1175
... (2)

From (1) and (2), we get

Now tanbcos-1 187 l = tan i = sin i
cos i
15 = 185
= 17
8
17

Example 4.25 (i) tan-1 b 17 l + tan-1 b 113 l = tan-1 b 92 l
Prove that (ii) cos-1 b 54 l + tan-1 b 53 l = tan-1 b 1217 l

Solution

(i) tan-1 b 17 l + tan-1 b 113 l = tan-1 >1 1 + 1 1 H
7 1 13 13
7
- _ i_ i

T ri g onom et ry 147

= tan-1 b 9200 l
= tan-1 b 92 l
(ii) Let cos-1 b 54 l = i . Then cos i = 54 & sin i = 53

` tan i = 34 & i = tan-1 b 34 l

` cos-1 b 54 l = tan-1 b 34 l

Now cos-1 b 54 l + tan-1 b 53 l
= tan-1 b 34 l + tan-1 b 53 l

3 + 3
= tan-1 4 5
e 1 3 3 o
- 4 # 5

= tan-1 b 1217 l

Example 4.26

Find the value of tan ; r - tan-1 b 18 lE
4

Solution

Let tan-1 b 18 l =i

Then tan i = 18

` tan ; r - tan-1 b 18 lE = tan b r - il
4 4

= tan r - tan i
1 4 r tan
+ tan 4 i

= 1 - 1 = 79
8
1 1
+ 8

Example 4.27

Solve tan-1 b x - 12 l+ tan-1 b x + 12 l = r
x - x + 4

Solution

tan-1 b x - 21 l + tan-1 b x + 12 l = tan-1 >1 x - 1 + x + 1 1 H = tan-1 b 2x-2 -3 4 l
x - x + x - 2 1 x + 2 2
- 2 +
- _ x - i_ x + i
x x

Given that

tan-1 b x - 1 l+ tan-1 b x + 1 l = r
x - 2 x + 2 4

148 11th Std. Business Mathematics

` tan-1 b 2x-2 -3 4 l = r
4
2x-2 -3 4
= tan r =1
4

2x2 - 4 = –3

& 2x2 –1 = 0

& x =! 1
2

Example 4.28

Simplify: sin-1 ` 13 j + sin-1 ` 32 j

Solution

We know that

sin-1 ]xg + sin-1 (y) = sin-1 7x 1 - y2 + y 1 - x2A

` sin-1 b 13 l + sin-1 b 32 l = sin-1 :13 1 - 94 + 32 1 - 19 D
= sin-1 :13 59 + 32 89 D

= sin-1 c 5 +4 2m
9

Example 4.29
Solve tan-1 ]x + 2g + tan-1 ]2 - xg = tan-1 b 32 l

Solution tan-1 <1^-x + 2h + ^2 - xh F = tan-1 ` 32 j
& ^x + 2h^2 - xh

tan-1 c 1 - 4 x2h m = tan-1 ` 32 j
^4 -

& 2x2 – 6 = 12

& x2 = 9

` x =+3

Example 4.30
If tan(x + y) = 42 and x = tan–1(2), then find y

T ri g onom et ry 149

Solution

tan(x+y) = 42
x + y = tan–1(42)

tan–1(2)+y = tan–1(42)

y = tan–1(42) – tan–1(2)

= tan-1 ;1 42 - 2 2g E
+ ]42 #
= tan-1 b 8450 l

y = tan-1 :187 D

Exercise 4.4

1. Find the pricipal value of the following

(i) sin-1 b- 12 l (ii) tan-1 ]-1g (iii) cosec-1 ^2h (iv) sec-1 ^- 2 h

2. Prove that

(i) 2 tan -1 ]xg = sin-1 c 1 2x m
+ x2

(ii) tan-1 b 34 l + tan-1 b 17 l= r
4

3. Show that tan-1 b 12 l + tan-1 b 121 l = tan-1 b 34 l

4. Solve tan-1 2x + tan-1 3x = r
4

5. Solve tan-1 ]x + 1g + tan-1 ]x - 1g = tan-1 b 74 l

6. Evaluate (i) cos8tan-1 ` 34 jB (ii) sin812 cos-1 ` 54 jB

7. Evaluate: cos`sin-1 ` 54 j + sin-1 `1132 jj

8. Prove that tan-1 b m l - tan-1 b m - n l = r
n m + n 4

9. Show that sin-1 b- 53 l - sin-1 b- 78 l = cos-1 8854

10. Express tan-1 :1 -cossinx x D, - r < x < 32r , in the simplest form.
2

150 11th Std. Business Mathematics

ICT Corner

Expected nal outcomes
Step – 1
Open the Browser and type the URL given (or) Scan
the QR Code.
GeoGebra Work book called “11th BUSINESS
MATHS” will appear. In this several work sheets for
Business Maths are given, Open the worksheet named “Inverse Trigonometric Functions”
Step - 2
Inverse Trigonometric Functions page will open. Click on the check boxes on the Le -hand side
to see the respective Inverse trigonometric functions. You can move and observe the point on
the curve to see the x-value and y-value along the axes. Sine and cosine inverse are up to the max
limit. But Tangent inverse is having customised limit.

St e p 1

St e p 2
St e p 4

B ro w s e in th e lin k
11th Business Maths: https://ggbm.at/qKj9gSTG (or) scan the QR Code

T ri g onom et ry 151

Exercise 4.5

Choose the correct answer

1. The degree measure of r is
8

(a) 20c60l (b) 22c30l (c) 22c60l (d) 20c30l

2. The radian measure of 37c30l is

(a) 52r4 (b) 32r4 (c) 72r4 (d) 92r4

3. If tan i = 1 and i lies in the first quadrant then cos i is
5
1 -1 5 - 5
(a) 6 (b) 6 (c) 6 (d) 6

4. The value of sin 15o is

(a) 3 +1 (b) 3 -1 (c) 3 (d) 3
22 22 2 22

5. The value of sin ]-420cg is

(a) 3 (b) - 3 (c) 12 (d) -21
2 2

6. The value of cos ]-480cg is

(a) 3 (b) - 3 (c) 12 (d) -21
2

7. The value of sin28o cos17o + cos28o sin17o is

(a) 1 (b) 1 (c) -1 (d) 0
2 2 (d) 14

8. The value of sin15º cos15º is

(a) 1 (b) 12 (c) 3
2

9. The value of secA sin(270o + A) is

(a) –1 (b) cos2A (c) sec2A (d) 1

10. If sinA + cosA =1 then sin2A is equal to

(a) 1 (b) 2 (c) 0 (d) 12
152 11th Std. Business Mathematics

11. The value of cos245º – sin245º is

(a) 3 (b) 12 (c) 0 (d) 1
2 (c) 14 2

12. The value of 1–2 sin245º is

(a) 1 (b) 12 (d) 0

13. The value of 4cos340º –3cos40º is

(a) 3 (b) – 12 (c) 12 (d) 1
2 2

14. The value of 2 tan 30c is
1 + tan2 30c
(a) 12 1 3
(b) 3 (c) 2 (d) 3

15. If sin A= 12 then 4 cos3 A - 3 cos A is 3 1
2 2
(a)1 (b) 0 (c) (d)

16. The value of 3 tan 10c - tan3 10c is
1 - 3 tan2 10c
1 (b) 12 3 1
(a) 3 (c) 2 (d) 2

17. The value of cosec-1 d 2 n is
3

(a) r (b) r (c) r (d) r
4 2 3 6

18. sec-1 32 + cosec-1 32 =

(a) -2r (b) r (c) r (d) – r
2

19. If a and b be between 0 and r and if cos^a + bh = 1132 and sin^a - bh = 53
2
then sin 2 a is

(a) 1156 (b) 0 (c) 5665 (d) 6645

20. If tan A = 12 and tan B = 13 then tan^2A + Bh is equal to

(a) 1 (b) 2 (c) 3 (d) 4

21. tan b r - xl is
4
11 ttaann b 11 ttaann
(a) b + x l (b) - x l (c) 1 - tan x (d) 1 + tan x
- x + x T ri g onom et ry 153

22. sinbcos-1 53 l is (b) 53 (c) 54 (d) 54
(a) 53 (c) 2 (d) - 2
(d) sec 50c
23. The value of 1 is (d) 0
cosec ]- 45cg
-1 1
(a) 2 (b) 2

24. If p sec 50c = tan 50c then p is

(a) cos 50c (b) sin 50c (c) tan 50c

25. b ccoosescxx l - 1 - sin2 x 1 - cos2 x is
(a) cos2 x - sin2 x (b) sin2 x - cos2 x (c) 1

Miscellaneous Problems

1. Prove that csoins 44xx + csions 33xx + scions22xx = cot 3x
+ +

2. Prove that 3 cosec 20c - sin 20c = 4

3. Prove that cot 4x]sin 5x + sin 3xg = cot x]sin 5x - sin 3xg .

4. If tan x = 34 , r < x < 32r then find the value of sin 2x and cos 2x

5. Prove that 2 sin2 34r + 2 cos2 r + 2 sec2 34r = 10
4

6. Find the value of (i) sin 75c (ii) tan 15c

7. If sin A = 13 , sin B = 14 then find the value of sin]A + Bg , where A and B are acute angles.

8. Show that cos-1 b 1123 l + sin-1 b 53 l = sin-1 b 5665 l

9. If cos (a + b) = 54 and sin (a - b) = 153 where a+b and a - b are acute, then find
tan 2a

10. Express tan-1 b cos x - sin x l, 0 < x < r in the simplest form.
cos x + sin x

154 11th Std. Business Mathematics

Summary
z If in a circle of radius r, an arc of length l subtends an angle of θ radians,

then l = r θ

z sin2 x + cos2 x = 1

z tan2 x + 1 = sec2 x

z cot2 x + 1 = cosec2 x

z sin(α + β) = sin α cos β + cos α sin β

z sin(α − β) = sin α cos β – cos α sin β

z cos(α + β) = cos α cos β – sin α sin β

z cos^a - bh = cos a $ cos b + sin a sin b

z tan ^a + bh = tan a + tan b
1 - tan a tan b
b2 hsi=n a1tc+aonstaaan-atatannbb
z tan ^a -
z sin 2a =

z cos 2a = cos2 a – sin2 a = 1 - 2 sin2 a = 2 cos2 a - 1

z tan 2a = 1 2 tan a
- tan2 a

z sin 3a = 3 sin a - 4 sin3 a

z cos 3a = 4 cos3 a - 3 cos a

z tan 3a = 3 tan a - tan3 a
1 - 3 tan2 a
x + y x - y
z sin x + sin y = 2 sin 2 cos 2

z sin x - sin y = 2 cos x +y sin x- y
2 2
+ -
z cos x + cos y = 2 cos x 2 y cos x 2 y

z cos x - cos y =- 2 sin x + y sin x - y
2 2
z sin x cos y = 12 6sin^x + yh + sin^x - yh@
z cos x sin y = 12 6sin^x + yh - sin^x - yh@
z cos x cos y = 12 6cos^x + yh + cos^x - yh@
z sin x sin y = 12 6cos^x - yh - cos^x + yh@

T ri g onom et ry 155

Allied angles GLOSSARY
Angle
Componendo and dividendo. துமணக் சகாணஙகள்
Compound angle சகாணம்
Degree measure கூட்டல் ைறறும் கழிததல் விகித ேைம்
Inverse function கூடடுக் சகாணம்
Length of the arc. பாமக அளமவ
Multiple angle சநர்ைாறு ோர்பு
Quadrants வில்லின நீளம்
Radian measure ை்டஙகு சகாணம்
Transformation formulae கால் பகுதிகள்
Trignometric identities சரடியன அளவு / ஆமரயன அளவு
Trignometry Ratios உருைாறறு சூததிரஙகள்
திாிசகாணைிதி முறவறாருமைகள்
திாிசகாணைிதி விகிதஙகள்

156 11th Std. Business Mathematics

Chapter 5 DIFFERENTIAL CALCULUS

Learning Objectives

After studying this chapter, the students will be able to
understand

z the concept of limit and continuity

z formula of limit and definition of continuity

z basic concept of derivative

z how to find derivative of a function by first principle

z derivative of a function by certain direct formula and
its respective application

z how to find higher order derivatives

Introduction

Calculus is a Latin word which means that Pebble or Isaac Newton
a small stone used for calculation. The word calculation is
also derived from the same Latin word. Calculus is primary
mathematical tool for dealing with change. The concept of
derivative is the basic tool in science of calculus. Calculus is
essentially concerned with the rate of change of dependent
variable with respect to an independent variable. Sir Issac
Newton (1642 – 1727 CE) and the German mathematician
G.W. Leibnitz (1646 – 1716 CE) invented and developed the
subject independently and almost simultaneously.

In this chapter we will study about functions and their
graphs, limits, derivatives and differentiation technique.

5.1 Functions and their graphs G.W. Leibnitz
i e entia a cu us 157
Some basic concepts

5.1.1 Quantity

Anything which can be performed on basic mathematical
operations like addition, subtraction, multiplication and
division is called a quantity.

5.1.2 Constant
A quantity which retains the same value throughout a mathematical investigation

is called a constant.

Basically constant quantities are of two types

(i) Absolute constants are those which do not change their values in any
mathematical investigation. In other words, they are fixed for ever.
Examples: 3, 3 , π, ...

(ii) Arbitrary constants are those which retain the same value throughout a
problem, but we may assign different values to get different solutions. The
arbitrary constants are usually denoted by the letters a, b, c, ...

Example: In an equation y=mx+4, m is called arbitrary constant.

5.1.3 Variable

A variable is a quantity which can assume different values in a particular problem.

Variables are generally denoted by the letters x, y, z, ...

Example: In an equation of the straight line x + y = 1,
a b

x and y are variables because they assumes the co-ordinates of a moving point in a
straight line and thus changes its value from point to point. a and b are intercept values
on the axes which are arbitrary.

There are two kinds of variables

(i) A variable is said to be an independent variable when it can have any arbitrary
value.

(ii) A variable is said to be a dependent variable when its value depend on the
value assumed by some other variable.

Example: In the equation y = 5x2 - 2x + 3,
“x” is the independent variable,
“y” is the dependent variable and
“3” is the constant.

5.1.4 Intervals AB ∞

The real numbers can be represented geometrically as ab
points on a number line called real line. The symbol R denotes Fig. 5.1
either the real number system or the real line. A subset of the
real line is an interval. It contains atleast two numbers and all −∞
the real numbers lying between them.

158 11th Std. Business Mathematics

(i) Open interval

The set {x : a < x < b} is an open interval, denoted by (a, b).

In this interval, the boundary points a and b are not AB ∞
included.
()
For example, in an open interval (4, 7), 4 and 7 are not −∞
elements of this interval. ab
Fig. 5.2

But, 4.001 and 6.99 are elements of this interval.

(ii) Closed interval

The set {x : a < x < b} is a closed interval and is denoted by [a, b].

In the interval [a, b], the boundary points a and b are AB ∞
included.
[]
For example, in an interval [4, 7], 4 and 7 are also −∞
elements of this interval. ab
Fig. 5.3

Also we can make a mention about semi closed or semi open intervals.

(a, b] = {x : a < x < b} is called left open interval

and [a, b) = {x : a < x < b} is called right open interval.

In all the above cases, b − a = h is called the length of the interval.

5.1.5 Neighbourhood of a point

Let ‘a’ be any real number. Let ε > 0 be arbitrarily small real number.

Then ( a − ε , a + ε ) is called an ε - neighbourhood of the point ‘a’ and denoted
by Na, ε

For examples,

N5, 1 = b5 - 1 , 5 + 1 l
4 4 4

= & x: 19 < x < 21 0
4 4

N2, 1 = b2 - 1 , 2 + 1 l = &x: 13 < x < 15 0
7 7 7 7 7

i e entia a cu us 159

5.1.6 Function

Let X and Y be two non-empty sets of real

numbers. If there exists a rule f which associates to

every element x ! X, a unique element y ! Y, then e function notation
such a rule f is called a function (or mapping) from the y = f(x)

set X to the set Y. We write f : X $ Y. was rst used by

The set X is called the domain of f, Y is called the Leonhard Euler in 1734 - 1735

co-domain of f and the range of f is defined as f(X) = {f(x) / x ! X}. Clearly f(X) 3 Y.

A function of x is generally denoted by the symbol f(x), and read as “function of x”
or “f of x”.

5.1.7 Classi cation of functions
Functions can be classified into two groups.

(i) Algebraic functions

Algebraic functions are algebraic expressions using a finite number of terms,
involving only the algebraic operations addition, subtraction, multiplication, division and
raising to a fractional power.

Examples : y = 3x4 − x3 + 5x2 − 7, y = 2x3 + 7x - 3 , y= 3x2 + 6x - 1 are algebraic
functions. x3 + x2 + 1

(a) y = 3x4− x3 + 5x2− 7 is a polynomial or rational integral function.

(b) y= 2x3 + 7x − 3 is a rational function.
x3 + x2 +1

(c) y= 3x2 + 6x −1 is an irrational function.

(ii) Transcendental functions
A function which is not algebraic is called transcendental function.
Examples: sin x, sin-1 x, ex, loga x are transcendental functions.
(a) sin x, tan 2x, ... are trigonometric functions.

(b) sin−1 x, cos−1 x, ... are inverse trigonometric functions.
(c) ex, 2x, xx, ... are exponential functions.
(d) loga x, loge (sin x), ... are logarithmic functions.

160 11th Std. Business Mathematics

5.1.8 Even and odd functions
A function f(x) is said to be an even function of x, if f( − x ) = f( x ).

A function f(x) is said to be an odd function of x, if f( − x ) = − f( x ).

Examples: f(x) = x2 and f(x) = cos x NOTE
are even functions.
f(x) = x3 + 5 is neither
f(x) = x3 and f(x) = sin x even nor odd function
are odd functions.

5.1.9 Explicit and implicit functions

A function in which the dependent variable is expressed explicitly in terms of some
independent variables is known as explicit function.

Examples: y = x2 + 3 and y = ex + e−x are explicit functions of x.

If two variables x and y are connected by the relation or function f(x, y) = 0 and
none of the variable is directly expressed in terms of the other, then the function is called
an implicit function.

Example: x3 + y3−xy = 0 is an implicit function.

5.1.10 Constant function Graph of a constant function
y = f(x)
If k is a fixed real number then the function
f(x) given by f(x) = k for all x ! R is called a constant is a straight line parallel to
function. x-axis

Examples: y = 3, f(x) = –5 are constant functions.

5.1.11 Identity function Graph of an identity function
on R is a straight line passing
A function that associates each real number to
itself is called the identity function and is denoted by I. through the origin and
makes an angle of 45c with
i.e. A function defined on R by f(x) = x for all positive direction of x-axis
x ∈ R is an identity function.

Example: The set of ordered pairs {(1, 1), (2, 2),
(3, 3)} defined by f : A " A where A={1, 2, 3} is an
identity function.

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5.1.12 Modulus function

A function f(x) defined by f(x) = |x|, where |x| = )-xx if x$0 is called the modulus
if x10

function. NOTE

It is also called an absolute value function.

Remark Domain is R and range set is [0, ∞) |5| = 5, |–5|
= –(–5) = 5

5.1.13 Signum function

|x | if x ! 0 is called the Signum function.

A function f(x) is defined by f(x) = * x
0 if x = 0

Remark Domain is R and range set is {–1, 0, 1}

5.1.14 Step function

(i) Greatest integer function NOTE

The function whose value at any real number x is the 2.5 = 2, –2.1 = –3,
greatest integer less than or equal to x is called the greatest 0.74 = 0, –0.3 = –1,
integer function. It is denoted by x.
4 = 4.
i.e. f : R $ R defined by f(x) = x is called the
greatest integer function.

(ii) Least integer function NOTE

The function whose value at any real number x is the 4.7 = 5, –7.2 = –7,
smallest integer greater than or equal to x is called the least 5 = 5, 0.75 = 1.
integer function. It is denoted by x.

i.e. f : R $ R defined by f(x) = x is called the least
integer function

Remark Function’s domain is R and range is Z [set of integers]

5.1.15 Rational function

A function f(x) is defined by the rule f (x) = p (x) , q (x) ! 0 is called rational
function. q (x)

Example: f (x) = xx2-+31 , x ! 3 is a rational function.

162 11th Std. Business Mathematics

5.1.16 Polynomial function

For the real numbers a0, a1, a2, ..., a-n1 ; a0 ! 0 and n is a non-negative integer, a
function f(x) given by f]xg = a0 xn + a1 xn + a2 xn - 2 + ... + an is called as a polynomial

function of degree n.

Example: f (x) = 2x3 + 3x2 + 2x - 7 is a polynomial function of degree 3.

5.1.17 Linear function

For the real numbers a and b with a ! 0, a function f(x) = ax + b is called a linear
function.

Example: y = 2x + 3 is a linear function.

5.1.18 Quadratic function

For the real numbers a, b and c with a ! 0, a function f(x) = ax2 + bx + c is called
a quadratic function.

Example: f (x) = 3x2 + 2x - 7 is a quadratic function.

5.1.19 Exponential function
A function f(x) = ax, a ! 1 and a > 0, for all x ∈ R is called an exponential function

Remark Domain is R and range is (0, ∞) and (0, 1) is a point on the graph
Examples: e2x, ex2 + 1 and 2x are exponential functions.

5.1.20 Logarithmic function
For x > 0, a > 0 and a ! 1, a function f(x) defined by f(x) = logax is called the

logarithmic function.

Remark Domain is (0, ∞), range is R and (1, 0) is a point on the graph

Example: f(x) = loge(x+2), f(x) = loge (sinx) are logarithmic functions.

5.1.21 Sum, di erence, product and quotient of two functions
If f(x) and g(x) are two functions having same domain and co-domain, then
(i) (f + g) (x) = f(x) + g(x)
(ii) (fg)(x) = f(x) g(x)

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(iii) c f m(x) = f (x) , g (x) ! 0
g g (x)

(iv) (cf)(x) = cf(x), c is a constant.

5.1.22 Graph of a function NOTE

The graph of a function is the set of points _x, f]xgi It is enough to draw a
where x belongs to the domain of the function and f(x) is the diagram for graph of a
value of the function at x. function in white sheet

To draw the graph of a function, we find a sufficient itself.
number of ordered pairs _x, f]xgi belonging to the function
and join them by a smooth curve.

Example 5.1

Find the domain for which the functions f(x) = 2x2–1 and g(x) = 1–3x are equal.

Solution

Given that f(x) = g(x)
& 2x2 – 1 = 1–3x

2x2 +3x – 2 = 0

(x+2) (2x–1) = 0

` x = –2, x = 1
2

Domain is $- 2, 1 .
2

Example 5.2

f = {(1, 1), (2, 3)} be a function described by the formula f(x) = ax+b. Determine
a and b ?

Solution
Given that f(x) = ax + b , f(1) = 1 and f(2) = 3
` a + b = 1 and 2a + b = 3
& a = 2 and b = –1

164 11th Std. Business Mathematics

Example 5.3

If f(x) =x + 1 , then show that [f(x)]3 = f(x3) + 3 f ` 1 j
x x

Solution f(x) = x+ 1 , f(x3) = x3 + 1 and f` 1 j = f(x)
Now, x x3 x

LHS = [f(x)]3

= 8x + 1 B 3 = x3 + 1 +3`x + 1
x x3 x
j

= f (x3) + 3f (x) = f (x3) + 3f` 1 j
x

= RHS

Example 5.4

Let f be defined by f(x) = x – 5 and g be defined by

g(x) = * x2 - 255 if x ! - 5 Find λ such that f(x) = g(x) for all x ∈ R
x + if x =- 5
m

Solution

Given that f(x) = g(x) for all x ∈ R
` f(–5) = g(–5)

–5 –5 = λ
` λ = –10
Example 5.5
If f(x) = 2x, show that f^xh $ f^yh = f^x + yh

Solution f(x) = 2x (Given)
` f(x + y) = 2x+y

= 2x . 2y

= f^xh $ f^yh

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Example 5.6

If f (x) = x −1, 0,x,1, then show that f [ f (x)] = − 1
x +1 x

Solution

f (x) = x −1 (Given)
x +1

` f 7 f]xgA = f]xg - 1 = x - 1 - 1
f]xg + 1 x + 11 + 1
x - 1
x +

= x −1− x −1 = − 2 = − 1
x −1+ x +1 2x x

Example 5.7

If f (x) = log 1+ x , show that f  2x  = 2 f ( x).
1− x  1+ x2 

Solution

f (x) = log 1+ x (Given)
1− x
2x
` f c 1 +2xx2 m = log 1 + 1 +2xx2
1 - 1 + x2

= log 1 + x2 + 2 x = log (1 + x)2
1 + x2 − 2 x (1 − x)2

= 2 log 1+ x = 2 f (x).
1− x

Example 5.8

If f(x) = x and g(x) = |x|, then find

(i) (f+g)(x) (ii) (f–g)(x) (iii) (fg)(x)

Solution (f + g) (x) = f(x) + g(x) = x + |x|
(i)
= ) x + x if x $0 = )20x if x $ 0
(ii) x - x if x <0 if x <0

(f – g) (x) = f(x) – g(x) = x – |x|

= ) x x -x if x$0 = )20x if x $ 0
- (- x) if x <0 if x <0

166 11th Std. Business Mathematics

(iii) (f g) (x) = f(x) g(x) = x |x|

= )-xx22 if x $ 0
if x < 0

Example 5.9

A group of students wish to charter a bus for an educational tour which can

accommodate atmost 50 students. The bus company requires at least 35 students for that

trip. The bus company charges ` 200 per student up to the strength of 45. For more than

45 students, company cohfasrtguedseenatcshwshtuodpeanrtt`ici2p0a0telsestshe15 times the number more than 45.
Consider the number tour as a function, find the total

cost and its domain.

Solution
Let ‘x’ be the number of students who participate the tour.
Then 35 < x < 50 and x is a positive integer.

Formula : Total cost = (cost per student) × (number of students)
(i) If the number of students are between 35 and 45, then

the cost per student is ` 200

` The total cost is y = 200x.

(ii) If the number of students are between 46 and 50, then

the cost per student is ` $200 - 1 (x - 45). = 209 - x
5 5
x52
` The total cost is y = `209 - x jx = 209x -
5

= 200x ; 35 # x # 45 and
*209x
` y - x2 ; 46 # x # 50
5

the domain is {35, 36, 37, ..., 50}.

Example 5.10
Draw the graph of the function f(x) = |x|

Solution

y = f(x) = |x|

= )-xx if x $ 0
if x <0
i e entia a cu us 167

Choose suitable values for x and determine y.

Thus we get the following table.

x −2 −1 0 1 2
2
y2101

Table : 5.1 y=–x, y > 0
2) 2 (2,
Plot the points (−2, 2), (−1, 1), (0, 0), (−2, x < 0 y=x, x
(1, 1), (2, 2) and draw a smooth curve.
1 (1, 1) 2)

(−1, 1)

The graph is as shown in the figure xl –2 –1 (0, 0) 1 2 x

yl
Fig. 5.4
Example 5.11

Draw the graph of the function f(x) = x2− 5

Solution

Let y = f(x) = x2− 5

Choose suitable values for x and determine y.

Thus we get the following table.

x −3 −2 −1 0 1 2 3

y 4 −1 −4 −5 −4 −1 4

Table : 5.2 y
Plot the points (−3, 4), (−2, −1), (−1, −4), (0, −5),
( 1, −4), (2, −1), (3, 4) and draw a smooth curve. (−3, 4) (3, 4)
The graph is as shown in the figure y=x2–5

Example 5.12 xl (−2,−1) (2, −1) x
Draw the graph of f(x) = ax, a ! 1 and a > 0
(−1, −4) (1, −4)
(0, −5)

yl

Fig 5.5

Solution

We know that, domain set is R, range set is (0, ∞) and the curve passing through
the point (0, 1)

168 11th Std. Business Mathematics

Case (i) w h e n a > 1
< 1 if x < 0

y = f(x) = ax = *= 1 if x = 0

> 1 if x > 0
We noticed that as x increases, y is also increases and y > 0.
` The graph is as shown in the figure.

y y=ax, a>1

(0, l)

xl o x

yl

Fig 5.6

Case (ii) w h e n 0 < a < 1

>1 if x <0

y = f(x) = ax = *= 1 if x = 0

<1 if x >0

We noticed that as x increases, y decreases and y > 0 .

` The graph is as shown in the figure.

y=ax, 0<a<1 y

(0, l) NOTE

xl O (i) The value of e lies between
x 2 and 3. i.e. 2 < e < 3
yl
(ii) The graph of f(x) = ex is identical
Fig 5.7 to that of f(x) = ax if a > 1 and
the graph of f(x) = e–x is identical
to that of f(x) = ax
if 0 < a < 1

Example 5.13
Draw the graph of f(x) = logax ; x > 0, a > 0 and a ! 1.

i e entia a cu us 169

Solution

We know that, domain set is (0, ∞), range set is R and the curve passing the point (1, 0)

Case (i) when a > 1 and x > 0
< 0 if 0 < x < 1

f (x) = loga x = *= 0 if x = 1
> 0 if x > 1
We noticed that as x increases, f(x) is also increases.

` The graph is as shown in the figure.

y
f(x)= logax, a>1

xl O (1, 0) x

yl

Fig 5.8

Case (ii) when 0 < a < 1 and x > 0
> 0 if 0 < x < 1

f (x) = loga x = *= 0 if x = 1
< 0 if x > 1
We noticed that x increases, the value of f(x) decreases.

` The graph is as shown in the figure.

y

f(x)= logax, 0<a<1

xl O (1, 0) x

yl

Fig 5.9

170 11th Std. Business Mathematics

ICT Corner

Expected nal outcomes
Step – 1
Open the Browser and type the URL given (or) Scan
the QR Code.
GeoGebra Work book called “11th BUSINESS
MATHS” will appear. In this several work sheets for
Business Maths are given, Open the worksheet named “Types of Functions”
Step - 2
Types of Functions page will open. Click on the check boxes of the functions on the
Right-hand side to see the respective functions. You can move the slider and change the
function. Observe each function.

St e p 1

St e p 2
St e p 4

B ro w s e in th e lin k
11th Business Maths: https://ggbm.at/qKj9gSTG (or) scan the QR Code

i e entia a cu us 171

Exercise 5.1

1. Determine whether the following functions are odd or even?

(i) f(x) =  ax −1 (ii) f]xg = log_x2 + x2 + 1 i
 
 a x + 1 

(iii) f(x) = sin x + cos x (iv) f(x) = x2− |x| (v) f(x) = x + x2

2. Let f be defined by f(x) = x3− kx2 + 2x, x ∈ R. Find k, if ‘f ’ is an odd function.

3. If f (x) = x3 - 1 , then show that f (x) + f` 1 j = 0
x3 x

4. If f (x) = x+ 1 , then prove that f^ f (x)h = x
x- 1

5. For f (x) = 3xx-+11 , write the expressions of f` 1 j and 1
x f (x)

6. If f (x) = ex and g (x) = loge x , then find

(i) (f+g)(1) (ii) (fg)(1) (iii) (3f)(1) (iv) (5g)(1)

7. Draw the graph of the following functions :

(i) f (x) = 16 - x2 (ii) f(x) = | x - 2 | (iii) f(x) = x | x |
(iv) f (x) = e2x (v) f (x) = e-2x
(vi) f (x) = |x |
x

5.2 Limits and derivatives

In this section, we will discuss about limits, continuity of a function, differentiability
and differentiation from first principle. Limits are used when we have to find the value of
a function near to some value.

Definition 5.1

Let f be a real valued function of x. Let l and a be two fixed numbers. If f(x)
approaches the value l as x approaches to a, then we say that l is the limit of the
function f(x) as x tends to a, and written as

lim f(x) = l

x"a

Notations: (i) L[f(x)]x=a = lim f(x) is known as left hand limit of f(x) at x=a.
(ii)
x " a-

R[f(x)]x=a = lim f(x) is known as right hand limit of f(x) at x=a.

x " a+

172 11th Std. Business Mathematics

5.2.1 Existence of limit

lim f(x) exist if and only if both L[f(x)]x=a and R[f(x)]x=a exist and are equal.

x"a

i.e., lim f(x) exist if and only if L[f(x)]x=a = R[f(x)]x=a .

x"a

NOTE
L[f(x)]x=a and R[f(x)]x=a are shortly written as L[f(a)] and R[f(a)]

5.2.2 Algorithm of le hand limit : L[f(x)]x=a

(i) Write lim f(x).

x " a-

(ii) Put x=a – h, h > 0 and replace x " a– by h" 0.

(iii) Obtain lim f(a–h).

h"0

(iv) The value obtained in step (iii) is called as the value of left hand limit of the

function f(x) at x=a.

5.2.3 Algorithm of right hand limit : R[f(x)]x=a

(i) Write lim f(x).

x " a+

(ii) Put x = a+h, h>0 and replace x " a+ by h" 0.

(iii) Obtain lim f(a+h).

h"0

(iv) The value obtained in step (iii) is called as the value of right hand limit of the

function f(x) at x=a.

Example 5.14

Evaluate the left hand and right hand limits of the function
x- 3
f(x) = * 3 if x ! 3 at x = 3.
x- 0 if x = 3

Solution

L[f(x)]x=3 = lim f(x)

x " 3-

= lim f(3 – h), x = 3 – h
= (3 h) - 3
h"0 (3 - h) - 3
-
lim

h"0

i e entia a cu us 173

= lim -h
-h
h"0

= lim h
-h
h"0

= lim –1

h"0

= –1

R[f(x)]x=3 = lim f(x)

x " 3+

= lim f(3 + h)

h"0

= lim (3 + h) - 3 NOTE
(3 + h) - 3
h"0

= lim h Here,
h
h"0 L[f(3)] ! R[f(3)]

= lim 1 ` lim f(x) does not exist.

h"0 x"3

=1

Example 5.15

Verify the existence of the function f(x) = )54xx3--43 if 0 1x # 1 at x = 1.
if 1 1x 1 2

Solution

L[f(x)]x=1 = lim f(x)

x " 1-

= lim f(1 – h), x = 1 – h

h"0

= lim [5(1–h)–4]

h"0

= lim (1–5h) = 1

h"0

R[f(x)]x=1 = lim f(x)

x " 1+

= lim f(1 + h), x = 1 + h

h"0

= lim [4(1 + h)3 – 3(1 + h)]

h"0

= 4(1)3 – 3(1)

Clearly, L[f(1)] = R[f(1)]

` lim f(x) exists and equal to 1.

x"1

174 11th Std. Business Mathematics

NOTE

Let a be a point and f(x) be a function, then the following stages may happen
(i) lim f(x) exists but f(a) does not exist.

x"a

(ii) The value of f(a) exists but lim f(x) does not exist.

x"a

(iii) Both lim f(x) and f(a) exist but are unequal.

x"a

(iv) Both lim f(x) and f(a) exist and are equal.

x"a

5.2.4 Some results of limits

If lim f(x) and lim g(x) exists, then
x"a x"a

(i) lim (f+g)(x) = lim f(x) + lim g(x)
x"a x"a x"a

(ii) lim (fg)(x) = lim f(x) lim g(x)
x"a x"a
x"a

f lim f (x)
(iii) lim c g m^xh xli"ma g (x) , whenever lim g (x) ! 0
=
x"a x"a x"a

(iv) lim k f(x) = k lim f(x), where k is a constant.

x"a x"a

5.2.5 Indeterminate forms and evaluation of limits
Let f(x) and g(x) are two functions in which the limits exist.

If lim f(x) = lim g(x) = 0, then f^ah takes 0 form which is meaningless but does not
g^ah 0
x"a x"a
f_xi
imply that lim g_xi is meaningless.

x"a

In many cases, this limit exists and have a finite value. Finding the solution of such

limits are called evaluation of the indeterminate form.

The indeterminate forms are 00 , 3 , 0 # 3, 3 - 3, 00, 30 and 13 .
3

Among all these indeterminate forms, 0 is the fundamental one.
0

5.2.6 Methods of evaluation of algebraic limits
(i) Direct substitution
(ii) Factorization

(iii) Rationalisation
(iv) By using some standard limits

i e entia a cu us 175

5.2.7 Some standard limits

(i) lim xn - an = nan - 1 if n ! Q
x - a
x"a
sin i tan i
(ii) lim i = lim i = 1, θ is in radian.

i"0 i"0

(iii) lim ax - 1 = logea if a>0
x
x"0
log (1 + x)
(iv) lim x =1

x"0

(v) lim^1 + 1 =e

x"0 xhx

(vi) lim `1 + 1 jx =e
x
x"3

(vii) lim ex - 1 =1
x
x"0

Example 5.16

Evaluate: xli"m1(3x2 + 4x - 5).
Solution

xli"m1(3x2 + 4x - 5)
= 3(1)2 + 4(1) – 5 = 2

Example 5.17

Evaluate: lim x2 - +4x2+ 6 .
x
x"2

Solution xli"m2xl^i"mx22^-x 4x + 6h
+ 2h
lim x2 - +4x2+ 6 =
x
x"2

= ^2h2 - 4^2h + 6 = 12
2+2

Example 5.18

Evaluate : lim 53 scions22xx - 22csoins 22xx .
r +
x " 4

Solution

lim ^5 sin 2x - 2 cos 2xh
53 scions22xx 22csoins 22xx r
lim - = x " 4
+
x" r lim ^3 cos 2x + 2 sin 2xh
4 r
x " 4

176 11th Std. Business Mathematics

= 5 sin r - 2 cos r
3 cos 2r + 2 sin r2
2 2

= 5^1h - 2^0h = 52 .
3^0h + 2^1h

Example 5.19

Evaluate: lim xx3--11 .

x"1

Solution

lim xx3--11 is of the type 0
xx3--11 0
x"1 - 1h^x2 + x + 1h
= lim ^x ^x - 1h
lim
x"1
x"1

= lim^ x2 + x + 1h

x"1

= (1)2 + 1 + 1 = 3

Example 5.20 2+x - 2.
x
Evaluate: lim

x"0

Solution

lim 2+x - 2 = lim ^ 2 + x - 2 h^ 2 + x + 2h
x x^ 2 + x + 2 h
x"0 x"0

= lim x^ 2+x-2 2h
2+x +
x"0 1
2+x +
= lim 2h
x"0^

= 1 2 = 2 1 .
2+ 2

Example 5.21 x 53 a 53
x 15 a 15
Evaluate: lim - .
-
x"a

Solution 53 53
15
lim x 53 - a 53 = lim x x15 - a
x 15 - a 15 x - a
x"a x"a - a

x-a

= 1553 ^a h- 52 = 3a -52 + 54 = 3a 52 .
^a h- 54

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Example 5.22

Evaluate: lim ssiinn 35xx .

x"0

Solution

lim ssiinn 53xx = xli"m0* 3x # ssiinn53xx53xx 4
5x #
x"0

= 53 lim sin3x3x
sin5x5x
x"0

lim

x"0

= 3 ` 1 j = 3
5 1 5

Example 5.23

Evaluate: lim 6 - 5x2 .
4x + 15x2
x"3

Solution

lim 6 - 5x2 = lim 4xx62+-155
4x + 15x2
x"3 x"3

= 0- 5 =- 1 .
0 + 15 3

Example 5.24

Evaluate: lim x tan ` 1 j.
x
x"3

Solution

Let y= 1 and y " 0 as x " 3
x

lim x tan` 1 j = lim tan y = 1.
x y
x"3 y"0

Example 5.25
/Evaluate: lim
n"3 n2 .
n3

Solution

/lim n2 = lim n^n + 1h^2n + 1h
n3 6n3
n"3 n"3

178 11th Std. Business Mathematics

= lim 16 $` n j` n + 1 j` 2n + 1 j.
n n n
n"3

= 16 lim $`1 + 1 j`2 + 1 j.
n n
n"3

= 1 ^1 h^ 2h
6

= 1 .
3

Example 5.26

Show that lim log^1 + x3h = 1.
sin3 x
x"0

Solution

lim log^1 + x3h = lim ' log^1 + x3h # sinx33 x 1
sin3 x x3
x"0 x"0

= lim log^1 + x3h # 1 j3
x3 sin x
x"0 lim `
x
x"0

= 1 # 1 =1
1

Exercise 5.2

1. Evaluate the following /(iii)

(i) lim xx3++12 (ii) lim 2x + 5 lim n
x2 + 3x + 9 n2
x"2 x"3 x"3

1 + x - 1-x x 5 - 5 sinx223x
5x x -
(iv) lim (v) lim 8 a8 (vi) lim
2
x"0 x"a 2 x"0
3 a3

2. If lim x9 + a9 = xli"m3^x + 6h, find the values of a.
x +a
x"a

3. If lim xn - 22n = 448, then find the least positive integer n.
x -
x"2

4. If f^xh = xx75--13228 , then find lim f^xh

x"2

5. Let f^xh = axx++1b , If lim f^xh = 2 and lim f^xh = 1, then show that f(–2) = 0.

x"0 x"3

i e entia a cu us 179

Derivative
Before going into the topic, let us first discuss about the continuity of a function. A

function f(x) is continues at x = a if its graph has no break at x = a

If there is any break at the point x = a then we say that the function is not continuous
at the point x = a.

If a function is continuous at all the point in an interval, then it is said to be a
continuous in that interval.

5.2.8 Continuous function
A function f(x) is continuous at x = a if

(i) f(a) exists

(ii) lim f(x) exists

x"a

(iii) lim f(x) = f(a)

x"a

Observation
In the above statement, if atleast any one condition is not

satisfied at a point x = a by the function f(x), then it is said to be a
discontinuous function at x = a

5.2.9 Some properties of continuous functions

Let f(x) and g(x) are two real valued continuous functions at x = a , then

(i) f(x) + g(x) is continuous at x = a.

(ii) f(x) g(x) is continuous at x = a.

(iii) kf(x) is continuous at x = a, k be a real number.

(iv) 1 is continuous at x = a, if f(a) ] 0.
f^xh

(v) f^xh is continuous at x = a, if g(a) ] 0.
g^xh

(vi) f^xh is continuous at x = a.

180 11th Std. Business Mathematics

Example 5.27

Show that f(x) = )54xx3--43, x, if 01x#1 is continuous at x = 1
if 11x<2

Solution

L7 f]xgAx = 1 = lim f]xg

x " 1-

= lim f]1 - hg, x =1–h

h"0

= lim 65]1 - hg - 4@

h"0

= 5(1)–4 = 1

R7 f]xgAx = 1 = lim f]xg

x " 1+

= lim f]1 + hg, x = 1+ h

h"0

= lim 74]1 + hg3 - 3]1 - hgA

h"0

= 4(1)3 – 3(1)

= 4–3=1

Now, f(1) = 5(1) – 4 = 5–4 =1

` lim f]xg = lim f]xg = f(1),
x " 1- x " 1+

f(x) is continuous at x = 1.

Example 5.28

Verify the continuity of the function f(x) given by

f]xg = )22 - x if x < 2 at x = 2.
+ x if x $ 2

Solution

L7 f]xgAx = 2 = lim f]xg

x " 2-

= lim f]2 - hg, x=2–h

h"0

= lim "2 - ]2 - hg,
h$0

= lim h = 0

h"0

i e entia a cu us 181

R7 f]xgAx = 2 = lim f]xg

x + 2+

= lim f]2 + hg, x = 2 + h If f(x) = ]]Z\][]]]]]ab (k(xx)) if x<a
if
h"0 if x = a then
x>a
= lim "2 + ]2 + hg,
h"0

= lim]4 + hg = 4 L[f(x)]x=a = lim f (x) = lim a (x)
h"0
x " a- x"a

Now, f(2) = 2+2 = 4. R[f(x)]x=a = lim f (x) = lim b (x)

x " a+ x"a

Here lim f]xg ! lim f]xg It is applicable only when the

x $ 2- x $ 2+ function has di erent de nition on

Hence f(x) is not continuous at x = 2 . both sides of the given point x = a.

Observations
(i) A constant function is everywhere continuous.
(ii) The identity function is everywhere continuous.

(iii) A polynomial function is everywhere continuous.
(iv) The modulus function is everywhere continuous.
(v) The exponential function ax, a>0 is everywhere continuous.
(vi) The logarithmic function is continuous in its domain.
(vii) Every rational function is continuous at every point in its domain.

Exercise 5.3

1. Examine the following functions for continuity at indicated points

(a) f]xg = * xx2-0-,24 , if x!2 at x = 2
if x=2

(b) f]xg = * xx2-6-,39 , if x!3 at x = 3
if x=3

2. Show that f(x) = | x | is continuous at x = 0

5.2.10 Di erentiability at a point
Let f(x) be a real valued function defined on an open interval (a, b) and let

182 11th Std. Business Mathematics

c ! (a, b). f(x) is said to be differentiable or derivable at x = c if and only if lim f]xg - f]cg
x-c
x"c

exists finitely.

This limit is called the derivative (or) differential co-efficient of the function f(x)
d
at x = c and is denoted by (or) D f(c) (or) : dx _ f]xgiD = c .

x

Thus .

5.2.11 Le hand derivative and right hand derivative

(i) lim f]xg - f]cg or lim f]c - hg - f]cg is called the left hand derivative of
x - c -h
x " c- h"0
f(x) at x = c and it denoted by f l^c-h or L[ f l (c)].

(ii) lim f]xg - f]cg or lim f]c + hg - f] c g is called the right hand derivative of
x - c h
x " c+ h"0

f(x) at x = c and it denoted by f l^c+h or R[ f l (c)].

Result f(x) is differentiable at x = c , L7 f l]cgA = R7 f l]cgA

Remarks: A function may be
continuous at a point but
(i) If L7 f l]cgA ! R7 f l]cgA , then we say may not be differentiable at
that f(x) is not differentiable at x=c.
that point.
(ii) f(x) is differentiable at x = c ⇒ f(x)
is continuous at x = c.

Example 5.29
Show that the function f(x) = | x | is not differentiable at x = 0.

Solution

f(x) = | x | = )-xx if x$0
if x<0

L[ f l (0)] = lim f]xg - f]0g ,
x-0
x " 0-

= lim f]0 - hg - f]0g , x = 0–h
0-h-0
h"0

i e entia a cu us 183

= lim f]-hg - f]0g
-h
h"0

= lim -h - 0
-h
h"0

= lim -h
-h
h"0

= lim h
-h
h"0

= lim]-1g

h"0

= –1

R[ f l (0)] = lim f]xg - f]0g
x- 0
x " 0+

= lim f]0 + hg - f]0g , x = 0+h
0+h-0
h"0

= lim f]hg - f]0g
h
h"0

= lim h - 0
h
h"0

= lim h
h
h"0

= lim h
h
h"0

= lim 1

h"0

=1

Here, L[ f l (0)] ! R[ f l (0)]
` f(x) is not differentiable at x=0

Example 5.30
Show that f(x) =x2 differentiable at x=1 and find f l (1).

Solution
f(x) = x2

184 11th Std. Business Mathematics

L[ f l (1)] = lim f]xg - f ]1 g
x- 1
x " 1-

= lim f]1 - hg - f]1g
1-h-1
h"0

= lim ]1 - hg2 - 1
-h
h"0

= lim 1 + h2 - 2h - 1
-h
h"0

= lim]-h + 2g

h"0

=2

R[ f l (1)] = lim f]xg - f ]1 g
x- 1
x " 1+

= lim f]1 + hg - f]1g
1+h-1
h"0

= lim ]1 + hg2 - 1
h
h"0

= lim 1 + h2 + 2h - 1
h
h"0

= lim]h + 2g

h"0

=2

Here, L[ f l (1)] = R[ f l (1)]

` f(x) is differentiable at x = 1

and f l (1) = 2

5.2.12 Di erentiation from rst principle

The process of finding the derivative of a function by using the definition that

f l]xg = lim f]x + hg - f]xg is called the differentiation from first principle. It is
h
h$0

convenient to write f l(x) as dy .
dx

i e entia a cu us 185

5.2.13 Derivatives of some standard functions using rst principle

1. For x ! R, d ^xnh = n xn - 1
dx

Proof:

Let f(x) = xn

then f(x+h) = (x+h)n

d _ f]xgi = lim f]x + hg - f]xg
dx h
h"0

= lim ]x + hgn - ] xgn
h
h"0

= lim ]x + hgn - ]xgn
]x + hg - x
h"0

= lim zn - xn , where z = x + h and z " x as h " 0
z - x
z"x

= nxn–1 <a lim xn - an = nan - 1F
x - a
x"a

i.e., d ^xnh = nxn–1
dx

2. d ^exh = ex
dx

proof:

Let f]xg = ex

then f(x+h) = ex + h

d _ f]xgi = lim f]x + hg - f]xg
dx h
h"0

= lim e x + h- ex
h
h"0

= lim e x eh - e x
h
h"0

= lim e x ; eh - 1 E
h
h"0

186 11th Std. Business Mathematics

= e x lim; eh - 1 E
h
h"0

= ex # 1 <a lim ; eh - 1 E = 1F
h
h"0

= ex

i.e., d ^e xh = ex .
dx

3. d ^loge xh = 1
dx x

Proof:

Let f(x) = logex

Then, f(x+h) = loge(x+h)

d _ f]xgi = lim f]x + hg - f]xg
dx h
h"0

= lim loge ]x + hg - loge x
h
h"0

= lim loge b1 + h l
h x
h"0

= lim loge b1 + h l 1
h x x
h"0 x

= 1 lim loge b1 + h l
x h x
h"0 x

= 1 #1 [a lim loge ]1 + xg = 1]
x x
x"0

= 1
x

Example 5.31

Find d ^ x3h from first principle.
dx

i e entia a cu us 187

Solution
Let f(x) = x3 then
f]x + hg = ]x + hg3

d _ f]xgi = lim f]x + hg - f]xg
dx h
h"0

= lim ]x + hg3 - x3
h
h"0

= lim 3x2 h + 3xh2 + h3
h
h"0

= hli"m0^3x2 + 3xh + h2h

= 3x2 + 3x (0) + (0)2

= 3x2

d ^ x3h = 3x2
dx

Example 5.32

Find d ^e3xh from first principle.
dx

Solution
Let f(x) = e3x

then f(x + h) = e3(x+h)

d _ f]xgi = lim f]x + hg - f]xg
dx h
h"0

= lim e3]x + hg - e3x

h"0 h

= lim e3x $ e3h - e3x
h
h"0

= lim e3x ^e3h - 1h
h
h"0

188 11th Std. Business Mathematics

= 3e3x lim e3h3h- 1

h"0

= 3e3x # 1 [a lim e x - 1 = 1]
x
x"0

= 3e3x

Exercise 5.4

1. Find the derivative of the following functions from first principle.

(i) x2 (ii) e–x (iii) log(x+1)

5.3 Di erentiation techniques

In this section we will discuss about different techniques to obtain the derivatives
of the given functions.

5.3.1 Some standard results [formulae]

1. d ^xnh = nxn - 1
dx

2. d (x) = 1
dx

3. d (k) = 0, k is a constant
dx

4. d (kx) = k, k is a constant
dx

5. d b 1 l = - 1
dx x x2

6. d ^ xh = 1
dx 2x

7. d (ex) = ex
dx

8. d (eax) = aeax
dx

9. d (eax+b) = aeax+b
dx

10. d _e-x2i =- 2x e-x2
dx

11. d ^axh = ax loge a
dx

i e entia a cu us 189

12. d ^loge xh = 1
dx x

13. d loge ]x + ag = 1
dx x+a

14. d _loge ]ax + bgi = a bg
dx ]ax +

15. d ^loga xh = 1
dx x loge a

16. d ]sin xg = cos x
dx

17. d (cosx) = –sinx
dx

18. d (tanx) = sec2x
dx

19. d (cotx) = –cosec2x
dx

20. d (secx) = secx tanx
dx

21. d (cosec x) = –cosecx cotx
dx

5.3.2 General rules for di erentiation

(i) Addition rule

d 7 f]xg + g]xgA = d 7 f]xgA + d 7g]xgA
dx dx dx

(ii) Subtraction rule (or) Difference rule

d 7 f]xg - g]xgA = d 7 f]xgA- d 7g]xgA
dx dx dx

(iii) Product rule

d 7 f]xg $ g]xgA = d 7 f]xgAg]xg + f]xg d 7g]xgA
dx dx dx

(or) If u = f(x) and v = g(x) then, d ]uvg = u d ]v g + v d ]ug
dx dx dx

(iv) Quotient rule

d = f]xg G = g]xg d 7 f]xgA- f]xg d 7g]xgA
dx g]xg dx 7g]xgA2 dx

(or) If u = f(x) and v = g(x) then, d b u l = v d ]ug - u d ]vg
dx v dx v2 dx

190 11th Std. Business Mathematics

(v) Scalar product

d 7cf]xgA = c d 7 f]xgA, where c is a constant.
dx dx

(vi) Chain rule:

d 7 f_ g]xgiA = f l_ g]xgi $ g l]x g
dx
dy dy
(or) If y = f(t) and t = g(x) then, dx = dt $ dt
dx

Here we discuss about the differentiation for the explicit functions using standard
results and general rules for differentiation.

Example 5.33

Differentiate the following functions with respect to x.

(i) x 32 (ii) 7ex (iii) 11 - 33xx
+

(iv) x2 sin x (v) sin3 x (vi) x2 + x + 1

Solution d ` x 32 j = 32 x 32 - 1
(i) dx

= 32 x 12 = 32 x

(ii) d ^7e xh = 7 d ^e xh = 7ex
dx dx

(iii) Differentiating y = 11 - 33xx with respect to x
+

dy = ]1 + 3xg d ]1 - 3xg - ]1 - 3xg d ]1 + 3xg
dx dx ]1 + 3xg2 dx

= ]1 + 3xg]-3g - ]1 - 3xg]3g = -6
]1 + 3xg2 ]1 + 3xg2

(iv) Differentiating y = x2 sin x with respect to x

dy = x2 d ]sin xg + sin x d ^x2h
dx dx dx

= x2 cos x + 2x sin x

= x]x cos x + 2 sin xg

i e entia a cu us 191

(v) y = sin3 x (or) (sin x)3

Let u = sin x then y = u3

Differentiating u= sin x with respect to x

We get, du = cos x
dx

Differentiating y = u3 with respect to u

We get, dy = 3u2 = 3 sin2 x
du
dy dy
dx = du $ du = 3 sin2 x $ cos x
dx

(vi) y = x2 + x + 1

Let u = x2 +x + 1 then y = u

Differentiating u = x2 +x + 1 with respect to x

We get, du = 2x + 1
dx

Differentiating y = u with respect to u.

We get, dy = 1
du 2u

= 2 1
x2 + x + 1

dy = dy $ du = 2 1 $ ]2x + 1g
dx du dx x2 + x + 1

= 2 2x + 1
x2 + x + 1

 d ^sin^log xhh = d xh ^sin ^log xhh d ^log xh
dx d^log dx

 d ^x2 + 3x + 1h5 = d^x2 d 1h ^x2 + 3x + 1h5 d ^x2 + 3x + 1h
dx + 3x + dx

 d ^e^ x3 + 3hh = d 3h ^e^x3 + 3hh d ^ x3 + 3h
dx d^x3 + dx

192 11th Std. Business Mathematics