unit 1_part 1.pdf engg math

8/12/2019 Unit 1_Part 1.PDF Engg Math

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Differential Calculus

UNIT 1 DIFFERENTIAL CALCULUS

Structure

1.1 Introduction

Objectives

1.2 Real Number System

1.3 Functions

1.4 Limits

1.5 Continuity

1.6 Derivative

1.7 Application of Geometry

1.7.1 Geometrical Meaning of the Derivative

1.7.2 Equations of Tangent and Normal at a Point

1.7.3 Angle of Intersection between Two Curves

1.8 Derivative as a Rate Measure

1.8.1 Motion in a Straight Line

1.8.2 Motion under Gravity

1.9 Increasing and Decreasing Functions

1.10 Maxima and Minima

1.11 Mean-value Theorems

1.11.1 Rolles Theorem

1.11.2 Lagranges Mean-value Theorem1.11.3 Cauchys Mean Value Theorem

1.11.4 Taylors Theorem

1.12 Curve Sketching

1.13 Summary

1.14 Answers to SAQs

1.1 INTRODUCTION

Calculus was related to meet the pressing mathematical needs of the 17thcentury scienceand is still the fundamental mathematics for solving problems in Science andTechnology. In this unit, we will introduce the concept of functions and discuss differenttypes of functions, algebra of functions and their limit and continuity. We will alsointroduce the concept of derivative as the instantaneous rate of change, and shall discussmethods of differentiation.

We propose to study the applications of derivative to geometry. We will defineincreasing and decreasing functions and shall discuss how derivatives can be used todetermine the points where a differentiable function has maxima and minima. We shallalso discuss some fundamental theorems of differential calculus.

Objectives

After studying this unit, you should be able to

recall the basic properties of real numbers,

define a function and examine whether a given function is one-one/onto,


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identify whether a given function has an inverse or not,

determine whether a given function is even or odd,

evaluate the limit of a function at a point,

identify points of continuity and discontinuity of a function,

obtain the derivative of a function at a point, find the tangent and normal to a given function at given points,

determine whether a function is decreasing or increasing,

solve some problems when it is required to minimize or maximize afunction, and

identify whether the derivative of a function can vanish once within aninterval.

1.2 REAL NUMBER SYSTEM

The set of all real numbers is denoted by R. The real number system is the foundation onwhich a large part of mathematics including calculus rests. You are familiar with theoperations of addition, subtraction, multiplication and division of real numbers and withinequalities. We shall recall some of their properties :

P1 - R is closed under addition.

P2 - Addition is associative and commutative.

P3 - R is closed under multiplication.

P4 - Multiplication is commutative and associative.

P5 - Multiplication is distributive over addition.P6 - For any two real numbers a and b, either a > b or a < b or a = b.

You are also familiar with the following subsets ofR.

(i) N, the set of natural numbers.

(ii) Z, the set of integers.

(iii) Q, the set of rational numbers.

Definition 1

If x is a real number, its absolute value, denoted by | x | is defined as

|x| =xifx0= xifx< 0

For example

| 5 | = 5

| 5 | = 5

The following theorems (without proof) gives some of the important propertiesof |x|.

Theorem 1

Ifxandyare real numbers, then(i) |x| = max {x,x}

(ii) |x| = | x|


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Differential Calculus(iii) |x|

2=x

2= | x|

2

(iv) |x+y| |x| + |y|

(v) |x y| | |x| |y| |

Definition 2

Let S be a non-empty subset of R. An element uR is said to be upper bound of S

if u x for all x S. If S has an upper bound, we say S is bounded above. Onsimilar lines we can define a lower bound for a non-empty set S to be a real

number v such that v x for all x S and we say that the set S is bounded below.

1.2.1 Intervals on the Real Line

Before we define an interval, let us see what is meant by a number line. The real numbersin setRcan be put into one-to-one correspondence with the points on a straight lineL. Inother words, we shall associate a unique point onLto each real number and vice-versa.

Consider a straight lineL(Figure 1.1(a)). Mark a point Oon it. The point Odivides thestraight line into two parts. We shall use the part of the left of Ofor representing negativereal numbers and the part of the right of Ofor representing positive real numbers. We

choose a pointAonLwhich is to the right of O. we shall represent the number 0 by Oand 1 byA. OAcan now serve as a unit. To each positive real number,xwe can associateexactly one point Plying to the right of OonL, so thatOP=xunits (=x). A negative real numberywill be represented by a point Qlying to theleft of Oon the straight lineLso that OQ=yunits (= y, sinceyis negative). We thusfind that to each real number we can associate a point on the line. Also, each point Sonthe line represents a unique real numberz, such thatz= OS. Further,zis positive if Sis tothe right of O, and is negative if Sis to the left of O.

This representation of real numbers by points on a straight line is often very useful.Because of this one-to-one correspondence between real numbers and the points of astraight line, we often call a real number a point ofR. Similarly,Lis called a number

line. Note that the absolute value or the modulus of any numberxis nothing but itsdistance from the point Oon the number line. In the same way,xydenotes the distancebetween the two numbersxandy(Figure 1.1(b)).

| x y |

y > x

| x y |

y < x

2 0 1 21

Q O A P Sx y

y x

(a) (b) Distance betweenxandyis |xy|

Figure 1.1

Now let us consider the set of real numbers which lie between two given real numbers aand b. Actually, there will be four different sets satisfying this loose condition. Theseare :

(i) (a, b) = {x| a


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this case we have drawn a hollow circle around aand bto indicate that they are notincluded in the graph. The set [a, b] contains both its end points and is called a closedinterval. In the representation of this closed interval, we have put thick black dots at aand bto indicate that they are included in the set.

The sets (a, b] and [a, b) are called half-open(or half-closed) intervals orsemi-open(or semi-closed) intervals, as they contain only one end point. This fact is

also indicated in their geometrical representation.Each of these intervals is bounded above by band bounded below by a.

Can we represent the set

I= {x: |x a| < }

on the number line? Yes, we can. We know thatx acan be thought of as the distancebetweenxand a. This meansIis set of all numbersx, whose distance from ais less than. Thus,Iis the open interval (a, a+ ). Similarly,

I1= {x: |x a| }

is the closed interval [a , a+ ]. Sometimes, we also come across sets like

I2= {x: | 0 < |x a| < }This means ifxI2, then the distance betweenxand ais less than , but is not zero. Wecan also say that the distance betweenxand ais less than, , butxa. Thus,

I2= (a , a+ ) {a} = (a , a) (a, a+ )

Apart from the four types of intervals listed above, there are a few more types. Theseare :

(a, ) = {x| a


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1.3 FUNCTIONS

Now let us move over to functions. Here we shall present some basic facts aboutfunctions which will help you refresh your knowledge. We shall look at variousexamples of functions and shall also define inverse functions. Let us start with thedefinition.

1.3.1 Definition and Examples

Definition 3

If X and Y are two non-empty sets, a function f from X to Y is a rule or a

correspondence which connects every member of X to a unique member

of Y. We write f : X Y (reads as f is a function from X to Y or f is a function

of X into Y). X is called thedomainand Y is called theco-domainof f. We shall

denote by f (x) that unique element of Y, which corresponds to x X.

The following examples will help you in understanding this definition better.

Example 1.1

Considerf:NRdefined byf(x) = x. Is f a function?

Solutionf is a function since the rulef(x) = xassociates a unique member (x) ofRtoevery memberxofN. The domain here isNand the co-domain isR.

Example 1.2

Considerf:NZ, defined by the rule2

)(x

xf = . Is f a function?

Solution

f is not a function fromNtoZas odd natural numbers like 1, 3, 5 . . . fromNcannot be associated with any member ofZ.

Example 1.3

Considerf:NN, under the rulef(x) = a prime factor ofx. Is f a function?

Solution

Here, since 6 = 2 3,f(6) has two values :f(6) = 2 andf(6) = 3. This rule doesnot associate a unique member in the co-domain with a member in the domain and,hence,f, as defined, is not a function ofNintoN.

Thus, you see, to describe a function completely we have to specify the following threethings :

(i) the domain,

(ii) the co-domain, and

(iii) the rule which assigns to each elementxin the domain, a single fullydetermined element in the co-domain.

Given a functionf:XY,f(x) Yis called the imageofxXunderfor thef-imageofx. The set off-images of all number ofX, i.e. {f(x) :xX} is called therangeoffand is denoted byf(X). It is easy to see thatf(X) Y.

Remark

(i) Throughout this unit we shall consider functions whose domain and co-domain are both subsets ofR. Such functions are often called real functionsor real value functions.

(ii) The variablexused in describing a function is often called a dummy

variable because it can be replaced by any other letter. Thus, for example,the rulef(x) = x,xNcan as well be written in the formf(t) = t, tN,or asf(u) = u, uN. The variablex(or tor u) is also called anindependent variableandf(x), which is dependent on this independentvariable, is called a dependent variable.


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Graph of a Function

A convenient and useful method for studying a function is to study it through itsgraph. To draw the graph of functionf:XY, we choose a system of co-ordinateaxes in the plane. For eachxX, the ordered pair (x,f(x)) determines a point inthe plane (Figure 1.2). The set of all the points obtained by considering all possiblevalues ofx, that is, (x,f(x)) :xX) is the graph off.

Figure 1.2

y

x

x))

(x)

x0

(x, f (

f

The role that the graph of a function plays in the study of the function will becomeclear as we proceed further. In the meantime let us consider some more examplesof functions and their graphs.

A Constant Function

A simple example of a function is a constant function. A constant functionsends all the elements of the domain to just one element of the co-domain.

For example, letf:RRbe defined byf(x) = 1.

Alternatively, we may write

f:x1 for allxRThe graph offis as shown in Figure 1.3.

y

1

0 x

f (x) = 1

Figure 1.3

It is the liney= 1.

In general, the graph of a constant functiony:xcis a straight line whichis parallel to thex-axis at a distance of cunits from it.

The Identity Function

Another simple but important example of a function is a function whichsends every element of the domain to itself.

LetXby any non-empty set, andfbe the function ofXdefined by settingf(x) =x, for allxX.

This function is known as the identity functiononXand is denoted byIX.


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Differential CalculusThe graph ofIR, the identity function ofR, is shown in Figure 1.4. It is the

liney=x.

y

x0

y = x

Figure 1.4

Absolute Value Function

Another interesting function is the absolute value function (or modulus

function) which can be defined by using the concept of the absolute value ofa real number as

0, a1.

This function is known as the general exponential function. A special caseof this function, where a= e, is often found useful. Figure 1.6 shows thegraph of the functionf:RRsuch thatf(x) = ex. This function is calledthe natural exponential function. Its range is the setR+of positive numbers.

y y = ex

1

2

3

(1, e)


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Figure 1.6

The Natural Logarithmic FunctionThis functionf:R+Ris defined on the setR+of all real numbers suchthatf(x) = ln (x). The range of this function isR. Its graph is shown inFigure 1.7.

x

(e, 1)

10

yy = ln (x)

2 3 4

Figure 1.7

The Greatest Integer Function

Take a real numberx. Either it is an integer, say n(so thatx= n), or it is notinteger. If it is not an integer, we can find an integer n, such thatn


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Differential CalculusConsider the function h:xx2, defined on the setR. Here h(2) = h(2) = 4; i.e.

2 and 2 are distinct members of the domainR, but their h-images are the same(Can you find some more members whose h-images are equal?). In general, thismay be expressed by saying : findx,ysuch thatxybut h(x) = h(y).

Now consider another function g:x2x+ 3. Here you will be able to see that ifx1andx2are two distinct real numbers, then g(x1) and g(x2) are also distinct.

For,)()(323222 21212121 xgxgxxxxxx ++ .

We have considered two functions here. While one of them, namely g, sendsdistinct members of the domain to distinct members of the co-domain, the other,namely h, does not always do so. We give a special name to functions like gabove.

Definition 4

A function f : X Y is said to be aone-one function(a 11 function or an

injective function) if images of distinct members of X are distinct members of Y.

Thus, the function gabove is one-one, whereas his not one-one.

Remark

The condition that the images of distinct members ofXare distinct members of Yin the above definition, can be replaced by either of the following equivalentconditions :

(i) For every pair of membersx,yofX,xyf(x) f(y).

(ii) For every pair of membersx,yofX,f(x) =f(y)x=y.

We have observed earlier that for a functionf:XY,f(X) Y.

This opens two possibilities :

(i) f(X) = Y, or

(ii) f(X) Y, that isf(X) is a proper subset of Y.

The function h:xx2, for allxRfalls in the second category. Since the squareof any real number is always non-negative, h(R) =R+(0) which is the set ofnon-negative real numbers. Thus h(R) R.

On the other hand, the function g:x2x+ 3 belongs to the first category. Given

anyyR(co-domain), if we take2

3

2

1

= yx , we find that g(x) =y. This

shows that every member of the co-domain is a g-image of some member of thedomain and, thus, is in the range g. From this, we get that g(R) =R. The following

definition characterises this property of the function.Definition 5

A function f : X Y is said to be anontofunction (or asurjectivefunction) if

every member of Y is the image of some members of X.

Thus his not an onto function, whereas gis an onto function. Functions which areboth one-one and onto (or bijective) are of special importance in mathematics. Letus see what makes them special.

Consider a functionf:XYwhich is both one-one and onto. Sincefis an ontofunction, eachyYis the image of somexX. Also, sincefis one-one,ycannot

be the image of two distinct members ofX. Thus, we find that to eachyYthere

corresponds a uniquexXsuch thatf(x) =y. Consequently,fsets up one-to-onecorrespondence between the members ofXand Y. It is the one-to-onecorrespondence between members ofXand Ywhich makes a one-one and ontofunction so special, as we shall soon see.


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Consider the functionf:NEdefined byf(x) = 2x, whereEis the set of evennatural numbers. We can see thatfis one-one as well as onto. In fact, to each

yEthere exists Ny

2such that y

yf =

2

. The correspondence

2

yy defines a function, say g, fromEtoNsuch that .

2)(

yyg =

The function g so defined is called as inverseoff. Since, to eachyEtherecorresponds a uniquexNsuch thatf(x) =yonly one such function gcan bedefined corresponding to a given functionf. For this reason, gis called the inverseoff.

As you will notice, the function gis also one-one and onto and therefore it willalso have an inverse. You must have already guessed that the inverse of gis thefunctionf.

From this discussion, we have the following :

Iffis one-one and onto function fromXto Y, then there exists a unique function

g: YXsuch that for eachyY, g(y) =xy=f(x). The function gso definedis called the inverse off. Further, if gis the inverse off, thenfis the inverse of g,and the two functionsfand gare said to be the inverse of each other. The inverseof a functionfis usually denoted byf 1.

To find the inverse of a given functionf, we proceed as follows :

Solve the equationf(x) =yforx. The resulting expression forx(in terms ofy)defines the inverse function.

Thus, if ,25

)(5

+=x

xf

we solve xyx for25

5

=+

This gives us .))2(5( 51

= yx

Hencef 1is the function defined by .))2(5()( 51

1 = yyf

1.3.3 Graphs of Inverse Functions

There is an interesting relation between the graphs of a pair of inverse functions becauseof which, if the graph of one of them is known, the graph of the other can be obtained

easily.Letf:XYbe a one-one and onto function, and let g: YXbe the inverse off. A

point (p, q) lies on the graph offq=f(p)p= g(q) (q,p) lies on the graph of g.Now, the point (p, q) and (q,p) are reflections of each other with respect to (w. r. t.) theliney=x. Therefore, we can say that the graphs offand gare reflections of each other w.r. t. the liney=x.

Therefore, it follows that if the graph of one of the functionsfand gis given, that of theother can be obtained by reflecting it w. r. t. the liney=x. As an illustration, the graphsof the functionsx3andy=x1/3are given in Figure 1.8.

x

y

0


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Figure 1.8

Do you agree that these two functions are inverses of each other? If the sheet of paper on

which the graph have been drawn is folded along the liney=x, the two graphs willexactly coincide.

If a given function is not one-one on its domain, we can choose a subset of the domain onwhich it is one-one and then define its inverse function. For example, consider thefunctionf:xsinx.

Since we know that sin (x+ 2) = sinx, obviously this function is not one-one

onR. But if we restrict it to the intervals

2,

2we find that it is one-one.

Thus, if

f(x) = sin (x), for all .2

,2

x

Then we can define

.sin,)(sin)( if11 xyyxxf ===

Similarly, we can define cos 1and tan 1functions as inverse of cosine and tangent

functions if we restrict the domains to [0, ] and

2,

2respectively.

SAQ 2

(a) Compare the graphs of lnxand exgiven in Figures 1.6 and 1.7 and verifythat they are inverses of each other.

(b) Which of the following functions are one-one?

(i) f:RRdefined byf(x) = |x|.

(ii) f:RRdefined byf(x) = 3x 1.

(iii) f:RRdefined byf(x) =x.

(iv) f:RRdefined byf(x) = 1.

(c) Which of the following functions are onto?

(i) f:RRdefined byf(x) = 3x+ 7.

(ii) f:R+Rdefined byf(x) = x .

(iii) f:RRdefined byf(x) =x2+ 1.

(iv) f:XRdefined byf(x) =x

1whereXstands for the set of non-zero

real numbers.

(d) Show that the functionf:XXsuch that 11

)( +

= xx

xf whereXis the set

of all real number except 1, is one-one and onto. Find its inverse.

(e) Give one example of each of the following :


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(i) a one-one function which is not onto?

(ii) onto function which is not one-one?

(iii) a function which is neither one-one nor onto?

1.3.4 New Functions from Given Functions

In this sub-section, we shall see how we can construct new functions from some givenfunctions. This can be done by operating upon the given functions in a variety of ways.We give a few such ways here.

1.3.5 Operations on Functions

Scalar Multiple of a Function

Consider the functionf:x3x2+ 1, for allxR. The functiong:x2 (3x2+ 1) for allxRis such that g(x) = 2f(x), for allxR. We saythat g= 2fand that gis a scalar multiple offby 2. In the above example, there isnothing special about the number 2. We could have taken any real number toconstruct a new function fromf. Also, there is nothing special about the particularfunction that we have considered. We could as well as have taken any otherfunction. This suggests the following definition : Letfbe a function with domain

Dand let kany real number. The scalar multiple offby kis a function with domainD. It is denoted by kfand is defined by setting (kf) (x) = kf(x).

Two special cases of the above definition are important.

(i) Given any functionf, if k= 0, the function kfturns out to be the zerofunction. That is, kf= 0.

(ii) If k= 1, the function kfis called the negative offand is denoted simplyby f.

Absolute Value Function (or Modulus Function) of a Given Function

Letfbe a function with domainD. The absolute value of functionfdenoted by |f|and read as modfis defined by setting (|f|) (x) = |f(x) | for allxD.

Since |f(x) | =f(x), iff(x) 0, thus |f| have the same graph for those values ofxfor whichf(x) 0.

Now let us consider those values ofxfor whichf(x) < 0. Here |f(x) | = f(x).Therefore, the graphs offand |f| are reflections of each other w. r. t. thex-axis forthose values ofxfor whichf(x) < 0.

As an example, consider the graph in Figure 1.9(a). The portion of the graph belowthex-axis, that is, the portion for whichf(x) < 0 has been shown as dotted.

To draw the graph of |f| we retain the undotted portion in Figure 1.9(a) as it is,and replace the dotted portion by its reflection w. r. t. thex-axis (Figure 1.9(b)).

y y


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x0 0 x

(a) (b)

Figure 1.9

Sum, Difference, Product and Quotient of Two Functions

If we are given two functions with a common domain, we can form several newfunctions by applying the four fundamental operations of addition, subtraction,multiplication and division on them.

(i) Define a function onDby setting (x) =f(x) + g(x). The functions iscalled the sum of the functionsfand g, and is denoted byf+ g. Thus,(f+ g) (x) =f(x) + g(x).

(ii) Define a function donDby setting d(x) =f(x) g(x). The function dis thefunction obtained by subtracting gfromf, and is denoted byf g. Thus, forallxD, (f g) (x) =f(x) g(x).

(iii) Define a functionponDby settingp(x) =f(x)g(x). The functionpiscalled the product of the functionsfand g, and is denoted byfg. Thus, for allxD, (fg) (x) =f(x) g(x).

(iv) Define a function qonDby setting)(

)()(

xg

xfxq = . The function q(x) exists

provided g(x) 0 for anyxD. The function qis called the quotient off

by gand is denoted by .g

fThus,

)(

)()(

xg

xfx

g

f= , (g(x) 0 for anyxD).

Remark

In case g(x) = 0 for somexD, we can consider the set, sayDof all those valuesofxfor which g(x) 0 and define D

g

fon by setting

)(

)()(

xg

xfx

g

f= for all

xD.

Example 1.4

Consider the functionsf:xx2and g:xx3. Find the functions which aredefined as

(i) f+ g,

(ii) f g,

(iii) fg,

(iv) (2f+ 3g), and

(v)g

f.

Solution

(i) (f+ g) (x) =x2+x3

(ii) (f g) (x) =x2x3

(iii) (fg) (x) =x5

(iv) (2f+ 3g) (x) = 2f(x) + 3g(x)= 2f(x) + 3g(x)

= 2x2+ 3x3.


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(v) Now, g(x) = 0x2= 0x= 0. Therefore, in order to define the function

g

f, we shall consider only non-zero values ofx. Ifx0,

xx

x

xg

xf 1

)(

)(3

2

== .

Therefore,g

fis the function ,

1:

xx

g

f wheneverx0.

1.3.6 Composition of Functions

We shall now describe a method of combining two functions which is somewhatdifferent from the ones studied so far. Uptil now, we have considered functions with thesame domain. We shall now consider a pair of functions such that the co-domain of oneis the domain of the other.

Letf:XYand g: YZbe two functions. We define a function h:XZby settingh(x) = g(f(x)).

To obtain h(x), we first take thef-imagef(x) of an elementxofX. Thusf(x) Y, whichis the domain of g. We then take the gimage off(x), that is, g(f(x)), which is an elementofZ. This scheme has been shown in Figure 1.10.

g[f(x)]

h

h

y

gf

f(x)

ZX

X

Figure 1.10The function h, defined above, is called the compositionoffand gand is written asgof. Note the order. We first find thef-image and then its g-image. Try to distinguishthe composite function goffrom the composite functionfogwhich will be defined onlywhenZis a subset ofX.

Example 1.5

For functionsf:xx2, xRand g:x8x+ 1, for allxR, obtain functionsgof and fog.

Solution

gof is a function fromRto itself, defined by

(gof ) (x) = g(f(x)) = g(x2) = 8x2+ 1 for allxR

fog is a function fromRto itself, defined by

(fog) (x) =f(g(x)) =f(8x+ 1) = (8x+ 1)2.

Thus gofandfogare both defined but are different from each other.

The concept of composite functions is used not only to combine functions, but also tolook upon a given function as made up of two simpler functions. For example, considerthe function

h:xsin (3x+ 7).

We can think of it as the composition (gof) of the functionsf:x3x+ 7, for allxRand g: usin u, for all uR.

Now let us try to find the composite functions fog and gof of the functions :


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f:x2x+ 3, for allxR, and g:x ,2

3

2

xfor allxR.

Note thatfand gare inverses of each other. Now

gof(x) = g(f(x)) = g(2x+ 3) = .2

3)32(

2

1xx =+

Similarly, fog (x)= f (g (x)) =f .323

22

23

2xxx =+

=

Thus, we see that gof(x) =xandfog(x) =xare the identity functions onR. What wehave observed here is true for any two functionsfand gwhich are inverses of each other.Thus, iff:XYand g: YXare inverses of each other, then gofandfogare identityfunctions. Since the domain of gofisXand that offogis Y, we can write this as :

gof=Ix,fog=Iy

This fact is often used to test whether two given functions are inverses of each other ornot.

1.3.7 Types of Functions

In this section, we shall talk about various types of functions, namely, even, odd,increasing and decreasing. In each case, we shall also try to explain the concept throughgraphs.

Even and Odd Functions

We shall first introduce two important classes of functions : even functions andodd functions.

Consider the functionfdefined onRby setting

f(x) =x2, for allxR.

You will notice that

f(x) = (x)2=x2=f(x), for allxR.

This is an example of an even function. Lets take a look at the graph(Figure 1.11) of this function. We find that the graph (a parabola) is symmetricalabout they-axis. If we fold the paper along they-axis, we shall see that the parts ofthe graph on both sides of they-axis completely coincide with each other. Suchfunctions are called even functions. Thus, a functionf, defined onRis evenif, foreachxR,f(x) =f(x).

y

x0

Figure 1.11

The graph of an even function is symmetric with respect to they-axis. We alsonote that if the graph of a function is symmetric with respect to they-axis, thefunction must be an even function. Thus, if we are required to draw the graph of aneven function, we can use this property to our advantage. We only need to drawthat part of the graph which lies to the right of they-axis and then just take its

reflection w. r. t. they-axis to obtain the part of the graph which lies on the left ofthey-axis.

Now let us consider the functionfdefined by settingf(x) =x3, for allxR. Weobserve thatf(x) = (x)3= x3= f(x), for allxR. If we consider another


16/44

20

function ggiven by g(x) = sinx, we shall be able to note again thatg(x) = sin (x) = sinx= g(x).

The functionsfand gabove are similar in one respect : the image of xis thenegative of the image ofx. Such functions are called oddfunctions. Thus, afunctionfdefined onRis said to be an oddfunction iff (x) = f(x), forallxR.

If (x,f(x)) is a point on the graph of an odd functionf, then (x, f(x)) is also apoint on it. This can be expressed by saying that the graph of odd function issymmetric with respect to the origin. In other words, if you turn the graph of anodd function through 180oabout the origin you will find that you get the originalgraph again. Conversely, if the graph of a function is symmetric with respect to theorigin, the function must be an odd function. The above facts are often usefulwhile handling odd functions.

While many of the functions that you will come across in this course will turn outto be either even or odd, there will be many more which will be neither even norodd. Consider, for example, the functionf:x(x+ 1)2.

Here f(x) = (x+ 1)2=x2 2x+ 1.

Is f(x) =f(x), for allxR?

The answer is no. Therefore,fis not an even function.

Further isf(x) = f(x), for allxR?

Again, the answer is no. Thereforefis not an odd function. The same conclusioncould have been drawn by considering the graph offwhich is given inFigure 1.12.

3

2

1

4

y

x13 2 2 310

Figure 1.12

You will observe that the graph is symmetric neither with respect to they-axis, nor with respect to the origin.

Now, there should be no difficulty in solving the exercise below.

SAQ 3

(a) Given below are two examples of even functions, alongwith their graphs.Try to convince yourself, by calculations as well as by looking at the graphs,that both the functions are, indeed, even functions.

(i) The absolute value function onR

f:x|x|

The graph offis shown in Figure 1.13.

x0

y


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21


Figure 1.13

(ii) The function gdefined on the set of non-zero real number by setting

.0,1

)(2

= xx

xg The graph of gis shown in Figure 1.14.

x

0

1

2

3

4

y

3

1 23 2 1

Figure 1.14

(b) We are giving two functions alongwith their graphs (Figures 1.15(a) and(b)). By calculations as well as by looking at the graphs, prove that each isan odd function.

(i) The identity function onR:

f:xx

(ii) The function gdefined on the set of non-zero real numbers by setting

.0,1

)( = xxxg

(a) (b)

Figure 1.15

(c) Which of the following functions are even, which are odd, and which areneither even nor odd?

(i) xx2+ 1, for allxR.

(ii) xx2 1, for allxR.

(iii) xcosx, for allxR.

(iv) x|x|, for allxR.

(v) .=

irrationalisif,1

rationalisif,0)(

x

xxf

x0

y

x2

2

4

4

2

4

2

4

1 3

113

3

1

3

0

y


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22

1.4 LIMITS

In Section 1.3, we introduced you to the concept of a function and discussed severaluseful types of functions. In this section, we consider the value to which

f(x) approaches asxgets closer and closer to some number a. The phrase in invertedcommas is to be understood intuitively and through practice. We do not give a formaldefinition here. We call such a value the limit off(x) and denote it by

Sometimes, this value may not exist, as you will see in an example later.

).(lim xfax

Example 1.6

Consider the functionf:RRgiven byf(x) =x+ 4. We want to find .)(lim2

xfx

Solution

Look at Tables 1.1 and 1.2. These give values off(x) asxgets closer and closer at2 through values less than 2 and through values greater than 2, respectively.

Table 1.1

x 1 1.5 1.9 1.99 1.999

f(x) 5 5.5 5.90 5.99 5.999

Table 1.2

x 3 2.5 2.1 2.01 2.001

f(x) 7 6.5 6.1 6.01 6.001

From the above tables, it is clear that asxapproaches 2,f(x) approaches 6. In fact,the nearerxis chosen to 2, the closerf(x) will be to 6. Thus, 6 is limit ofx+ 4 asxapproaches 2, that is, .6)4(lim

2=+

x

x

In the above example, the value of )4(lim2 + xx coincides with the valuex+ 4whenx= 2, that is, ).2()(lim

2fxf

x=

Numbersxnear 2 fall into two natural categories; those which are < 2, that is,those that lie to the left of 2, and those which are > 2, that is, those which lie to theright of 2.

We write

6)(lim2

=

xfx

to indicate that asxapproaches 2 from the left,f(x) approaches 6.We shall describe this limit as the left-hand limit off(x) asxapproaches(or tendsto) 2.


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23


6

6

6

Similarly,

)(lim2

=+

xfx

indicates that asxtends to 2 from the right,f(x) approaches 6.

We shall call this limit as the right-handed limit of f(x) asxapproaches2.

The left and right-hand limits are called one-sidedlimits.It is clear now that

)(lim2

=

xfx

if and only if, both

and)(lim2

=

xfx

6)(lim2

=+

xfx

.

In the above example, the value of )4(lim2+

x

xcoincides with the value of

x+ 4 whenx= 2, that is

).2()(lim2

fxfx

=

Likewise .1)13(lim 22

=+

xxx

as also .)13(lim1)13(lim 22

2

2+==+

+ xxxx

xx

Now consider another function given byRR }2{:f .2

4)(

2

=x

xxf This

function is not defined at the pointx= 2, since division by zero is undefined. Butf(x) is defined for values ofxwhich approach 2. So it makes sense to evaluate

.2

4lim

2

2

x

x

xAgain, we consider the following Tables 1.3 and 1.4 which give the

values off(x) asxapproaches 2 through values less than 2 and through valuesgreater than 2, respectively.

Table 1.3

x 1 1.5 1.9 1.99 1.999

f(x) 3 3.5 3.9 3.99 3.999

Table 1.4

x 3 2.5 2.1 2.01 2.001

f(x) 5 4.5 4.1 4.01 4.001

As you can see

and4)(lim2

=

xfx

.4)(lim2

=+

xfx

Hence, we shall say

.4)(lim2

=

xfx

Now we are in a position to define the limit of a function.Letfbe a function and let abe a real number. We do not require thatfbe defined at a,

but we do require thatfbe defined on a set of the form (ap, a) (a, a+p).


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24

(This guarantees that we can formf(x) for allxathat are sufficiently closeto a.)

Definition 6

f (x) is said to tend to the limit l as x approach a, written as

,)(lim lxfax

=

if, and only if,

,)(lim lxfax

=

and .)(lim lxfax

=+

There is another definition of the limit of a function, equivalent to the above definition.

Definition 7

Let f be a function defined on some set (a p, a)(a, a + p).

Then lxfax

=

)(lim

if, and only if, for each > 0 there exists > 0 such that | f (x) l | < when

0 < | x a | < . Note that we do not deny that f may be possibly defined at thepoint a. All we are saying is that the definition does not require it.

We illustrate the above definitions with the help of examples.

Example 1.7

Show that

.0;)(lim +=+

mnmanmxax

Solution

Let > 0. We seek a number > 0 such that, if 0 < |x a| < , then

| (mx

+n) (

ma+

n) | 0, there exists

> 0 such that if | x a | < , then | f (x) f (a)| < . Clearly, will dependboth on and a; if any of these are changed the same may not work.

Continuity of the function f (x)at an end point of an interval [a, b]of itsdomain is defined below :

(i) f (x)is continuous at the left end point a if = f (a).)(lim xfax +

(ii) f (x)is continuous at the right end point b if = f (b).)(lim xfbx

If a function is continuous at each point of an interval, it is continuous over

the interval.

Remark

If the interval is closed, the limit in the continuity test over the interval istwo-sided at an interior point and the appropriate one-sided at the end points.

Example 1.16

Prove thatf(x) = sinxis continuous atx= 0.Solution

(i) f(0) = sin 0 = 0

(ii) = = 0 and)(lim0

xfx

xx

sinlim0

(iii) =f(0) = 0)(lim0

xfx

Therefore,f(x) = sinxis continuous atx= 0. (In fact sinxand cosxarecontinuous at each realx.)

Example 1.17

Examine the continuity of the function .1

)(x

xf =

Solution


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30

The function is defined for all non-zero real value ofx. It is not defined atx= 0.

x

y

0

xy

1=

Figure 1.18

Also 0),(1

)(lim 000

if0

==

xxfx

xfxx

Therefore, the function is continuous for all realx0. It fails to be continuous atx= 0 (why)? (Figure 1.18). This function is continuous on any interval which doesnot includex= 0 as an element of it.

Theorem 4 : Properties of Continuous Functions

(a) If the functionsf(x) andg(x) are continuous atx=a, then

(i) f(x)

g(x),

(ii) f(x)g(x), and

(iii) 0,)(,)(

)(ag

xg

xf

are continuous atx=a.

(b)

Iff(x) is continuous atx=x0and ifg(y) is continuous aty=y0=g(x0),

then the composite functionF(x) =g(f(x)) is continuous atx=x0.

These results are in fact the immediate corollaries of the corresponding limit theoremsdiscussed in Section 1.4.

If a function is not continuous at a pointx0, it is said to be discontinuous atx0.

Example 1.18

Let )0(sin

)( = xx

xxf . Nowfis not defined atx= 0. If we definef(0) = 1

which is same as ,sinlim0 x

xx

thenfis continuous atx= 0.

x0

x

xy

sin=

y

Figure 1.19


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31

Differential CalculusExample 1.19

The functionfdefined by

>=

0for1

0for1)(

x

xxf

is discontinuous atx= 0.

Figure 1.20

x0

1

1y

Example 1.20Draw the graph of the function

=

0whenever

0whenever

xx

xxy

Solution

The function is continuous for


31/44

35


10=

=

h

hwhen his negative.

1)0()0(

lim0

=+

+ h

fhf

h

1)0()0(

lim0

=+

h

fhf

h

i.e.f (0) does not exist.

Remark

The above example will help you to conclude that a function can never havederivative at a point of discontinuity. In the light of this remark, look at thefollowing example.

Example 1.23

Examine the differentiability of the function


32/44

36

andh

fhf

h

)2()2(lim

0

+

hh

)23(1lim

0

=

00lim0 === h are not equal.

Therefore, the derivative does not exist atx= 2.

At all points (x0 andx2) the derivatives exist.

SAQ 6

(a) Show that the derivative of a constant function is zero.

(b) Show thaty= |x| is differentiable at all points except atx= 0.

(c) Find the derivative of the function

(i) y=x2

(ii) y= 3x2+ 5x 1.

1.6.2 Algebra of Derivatives

We now state some important rules regarding the derivatives of the sum, difference,

product and quotient of functions in terms of their derivatives.Theorem 6

Iff(x) andg(x) are differentiable functions at a pointx, then

(i)dx

xdg

dx

xdfxgxf

dx

d )()()]()([ =

(ii)dx

xdfxg

dx

xdgxfxgxf

dx

d )()(

)()())(.)(( +=

(iii)2)]([

)()(

)()(

)(

)(

xg

dx

xdgxf

dx

xdfxg

xg

xf

dx

d =

providedg(x) 0.

We shall give the proof of the result (ii) below. Other results we leave as an exercise foryou.

Proof

Let y=f(x) .g(x)

Thenh

xgxfhxghxf

dx

dy

h

)()()()(lim

0

++=

Now )()()()( xgxfhxghxf ++ )]()([)()]()([)( xfhxfxgxghxghxf ++++=


33/44

37


++

++=

h

xfhxfxg

h

xghxghxf

dx

dy

h

)]()([)(

)]()([)(lim

0

h

xghxghxf

dx

dy

hh

)()(lim)(lim

00

++=

h

xfhxfxg

h

)()(lim)(

0

++

.

That is,

,)()()()(dx

xdfxgdx

xdgxfdxdy +=

Use of the product rule (ii) and the result which states that derivative of a constantis zero gives

)()]([ xfdx

dcxcf

dx

d== ,

where cis a constant.

Example 1.24

Find the derivative ofy= 5x2sinx.

Solution

)](sin)(sin[5 2 2+= xdx

dxx

dx

dx

dx

dy

We derive )(sinxdx

das under :

h

xhxx

dx

d

h

sin)(sinlim)(sin

0

+=

+=

2

2sin2

coslim0

2

h

h

hxh

2

2sin

lim2

coslim0

20

2

h

h

hx

hh

+=

.

= cosx.1 = cosx

and xxdx

d

2)(2

=

Hence, ]sin2cos[5 2 xxxxdx

dy+= .

Example 1.25

Finddx

dy, if

2

cos

x

xy= .

Solution

By formula (iii)

4

22 )(cos)(cos

x

xdx

dxx

dx

dx

dx

dy

=


34/44

38

To obtain )(cosxdx

d, we have

h

xhxx

dx

d

h

cos)(coslim)(cos

0

+=

+=

2

2sinlim

2sinlim

02

02

h

h

hx

hh

2

2sin

lim2

sinlim0

20

2

h

h

hx

hh

+=

= sinx.1 = sinx

and xxdx

d

2)(2

=

Therefore,3

cos2sin

x

xxx

dx

dy +=

1.6.3 Derivative of a Composite Function (Chain Rule)

Theorem 7

Ify=f(u) is a differentiable function of uand u=g(x) is a differentiable

function ofx, theny=f(g(x)) as a function ofxis differentiable and

dx

du

du

dy

dx

dy.= .

Proof

Let u= g(x) be differentiable atx0andy=f(u) be differentiable at u0whereu0= g(x0).

Let h(x) =f(g(x)). The function g(x), being differentiable, is continuous atx0; soalso the functionf(u) at u0and, therefore, atx0.

u= g(x0+ x) g(x0)

where u0 as x0

and at u0

y=f(u0+ u) f(u0)=f(g(x0+ x)) f(g(x0))

= h(x0+ x) h(x0)

Assuming u0 (i.e. g(x) is not a constant function in neighbourhoodofx0), we write

x

u

u

y

x

y

=

.

In the limit, we get

x

u

u

y

x

y

xux

=

000limlimlim

ordx

du

du

dy

dx

dy=


35/44

39

Differential CalculusYou can also show that in the case of u= 0 the above formula holds true. In fact it

becomes an identity 0 = 0. Note that it cannot be divided by uin this case.

Example 1.26

Ify= sin (x2), finddx

dy.

Solution

Lety= sin (u) and u=x2. By the chain-rule

dx

duu

du

d

dx

dy.)(sin=

= (cos u) (2x)

= 2xcosx2.

1.6.4 Second Order Derivatives

You know that the derivative of a functiony=f (x), i.e.f (x) is also a function. Hence wecan calculate the derivative off (x) at a point of its domain, if it exists, by using the same

limiting procedure. That is, if h

xfhxf

h

)()(lim0

+ exists, it is called the second order

derivative off (x) with respect tox. We write it as )(or2

2

xfdx

yd . You can also calculate

higher order derivatives like 3rdorder, 4thorder etc., of a functionf(x).

Example 1.27

Find the (i) 2ndorder and (ii) 5thorder derivatives of .153 24 += xxxy

Solution

(i) 5643

+== xxydxdy

612)564( 232

2

=+=

= xxxdx

d

dx

dy

dx

d

dx

yd

(ii) 0,24,24 )5()4()3( === yyxy

1.6.5 Implicit Differentiation

If we are given an equation like , it is not easy to solve it foryin

terms ofx. However, it is often possible to calculate

0134 346 =++ yxyx

dx

dystraight from the equation by the

method of implicit differentiation. For example, if we differentiate both the sides of thegiven equation, we get

)0(0334446 2345dx

d

dx

dyy

dx

dyyxyx =+

++

0)916(46 2345 =++dx

dyyxyyx

Therefore,23

45

916

)46(

yxy

yx

dx

dy

+=

This holds for all points where .0916 23 yxy


36/44

40

SAQ 7

(a) Differentiate

(i)52

13

+

x

x

(ii) 83

13 + xx

(iii) (x2+x+ 6) sinx

(iv) (x+ 2) (x2+ 1) + cosx

(b) Find the derivatives of the functions

(i)5

322

+

x

x

(ii) tanx

(iii)x

x2cos1

)sin1(

+

(c) Find the 2ndorder derivatives of the functions

(i) sin2x

(ii) ax3+ bx2+ cx+ d

1.6.6 Derivatives of Some Elementary Functions

Power Functiony=x

( is real)

We have already proved that the power functiony=xn, where nis a positive

integer, has the derivative

1= nxndx

dy

We shall show now that the above rule also holds when qp

xy= (q,pare integerswith q> 0).

Let us assumep, q> 0.

Thenyq=xp. By the method of implicit differentiation, we get

11

=pq

xpdx

dy

yq


37/44

41


or nq

pnxx

q

p

xq

xp

yq

xp

dx

dy nqp

q

pp

p

q

p

=====

if,111

1

1

In fact the formula 1)( = xxdx

dholds true for any real . We shall accept

this result without proof.

Trigonometric Functions

We have already seen that

xxdx

dxx

dx

dsin)(cos,cos)(sin ==

By using the quotient rule you can show that

xxdx

dxx

dx

d 22 eccos)(cot,sec)(tan ==

xxxdx

d

xxxdx

d

coteccos)ec(cos,tansec)(sec ==

Logarithmic Function

We shall prove thatx

xdx

d 1)(log =

Let f(x) = logx

Then

+=+=+

x

h

hxhx

hh

xfhxf1log

1)](log)([log

1)()(

h

xhx

h

xfhxfxf

hh

log)(loglim

)()(lim)(

00

+=

+=

+=

x

h

hh1log

1lim

0

x

x

h

x

h

h

11log

lim0

.

+=

xxx

x

h

x

h

hh

111

1lim

1log

lim 00 ==

+

= .

=

+

1)1(log

lim0 y

y

y

Exponential Function (with respect to the basea)

Let y= axwith a> 0, but a1

Then logey=xlogea

By the method of implicit differentiation, we get

adx

dy

yelog

1=


38/44

42

Therefore, aaaydx

dye

xe loglog ==

In particular ify= ex, then xedx

dy= )1log( =ee

Inverse Function

Let y= sin 1x

Then siny=x

Differentiating both sides w. r. t.x, we have

1cos =dx

dyy

i.e.ydx

dy

cos

1=

=y2cos

1

y2sin1

1

=

21

1

x=

As we know that the range of

22

sin tois1 x , i.e.ylies between

22

to

,

so cosyis positive.

Similarly,2

1

1

1cos

xdx

dyxy

==

21

1

1tan

xdx

dyxy

+==

21

1

1cot

xdx

dyxy

+==

Example 1.28

If44

,)(tancos 1


39/44

43


)

=

=t

ty

tx,

)(

)(

Let us assume (t), (t) are differentiable andx= (t) has an inverse t= h(x). Then wecan consider the equation of the functiony=f(x) as a composite function.

(),( xhtty ==

Using the differentiation rule for a composite function (chain-rule), we get

dx

dh

dt

d

dx

dt

dt

dy

dx

dy == .

Since t= h(x) is inverse ofx= (t), we arrive at

dt

ddt

d

dx

dy

= .

Example 1.29

The parametric equation of a semi-circular arc of radius a with its centre at (0, 0)is

=

=t

tay

tax0,

sin

cos

Find4

at =t

dt

dy.

Solution

tta

ta

dt

dxdt

dy

dxdy cot

)sin(cos =

==

At 14

cot,4

=

==

dx

dyt

1.6.8 Logarithmic Differentiation

In some problems, it is easier to finddx

dyby first taking logarithmic and then

differentiating. Such process is called logarithmic differentiation. This is usually done in

two kinds of problems. First when the function is a product of many simpler functions. Inthis case logarithm converts the product into a sum and facilitates differentiation.Secondly, when the variablexoccurs in the exponent. In this case logarithm brings it to asimpler form.

Example 1.30

Differentiate sinxsin 2xsin 3x.

Solution

Lety= sinxsin 2xsin 3x

Then logy= log sinx+ log sin 2x+ log sin 3x.Differentiating both sides, we have


40/44

44

33cos3sin

122cos

2sin

1cos

sin

11..... x

xx

xx

xdx

dy

y++=

++=

x

x

x

x

x

xy

dx

dy

3sin

3cos3

2sin

2cos2

sin

cos

= sinxsin 2xsin 3x[cotx+ 2 cot 2x+ 3 cot 3x]Example 1.31

Differentiate (sinx)x

Solution

Let y= (sinx)x

Then logy=xlog sinx

Differentiating both the sides, we have

xx

xxdxdy

ycos

sin1sinlog1 ..+=

i.e. ]cotsin[log xxxydx

dy+=

= (sinx)x[log sinx+xcotx]

1.6.9 Differentiation by Substitution

Sometimes it is easier to differentiate by making substitution. Usually these examplesinvolve inverse trigonometric functions.

Example 1.32

Differentiate

+ xx21 1tan

Solution

Put x= tan

Then 1 +x2= sec2

=+ tansec1 2 xx

=

cossin

cos

1

=cos

sin1

2sin

2cos

2cos

2sin2

2sin

2cos

22

22

+

=

+

=

2sin

2cos

2sin

2cos

2sin2cos

2


41/44

45


=

+

=

+

=24

tan

2tan1

2tan1

2sin

2cos

2sin

2cos

=24

tantany

i.e.24

=y

)(tan2

1

2

1 1xdx

d

dx

d

dx

dy =

=

)1(2

12x+

=

SAQ 8

(a) A particle moves along a straight line. At any time tthe distance stravelledby the particle is given by the formula : s= 32t2+ 9. Find the velocity andacceleration of the particle at time, t= 2 units.

(b) Use the method of implicit differentiation to calculatedx

dy, if

8432 22 =+ yxyx

(c) Differentiate the following

(i) x2exsinx

(ii) (x+ 1)

2

(x+ 2)

3

(x+ 3)

4

(iii)2

1

1

2tan

x

x

(iv)2

21

1

1cos

x

x

+


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46


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ENGINEERING MATHEMATICS-I

This course on Engineering Mathematics-I (BME - 001) is in your hand. The wholecourse is comprises four blocks : Calculus, Vector Calculus, Matrices and Determinantsand Probability and Statistics.

In first block, you will be introduced to basic concepts of differential and integral

calculus. This includes limit, continuity, mean value theorem and application of calculusin engineering field. The matter of differential and integral calculus is presented in single

block because one has to continuously refer to the results of differential calculus whilestudying integral calculus.

Block 2 covers vector algebra, vector differential calculus, line, surface and volumeintegral. The usefulness of vector calculus in engineering mathematics results from thefact that many physical quantities for example, force, velocity, momentum etc. may

be represented by vectors.

Block 3 includes linear equations and Euclidean spaces, linear transformation and linearequations, matrices and determinants. Euclidean spaces are introduced as generalizationof concepts of two or three dimensional spaces. Linear transformation from oneEuclidean space to another Euclidean space are described in this block. Thecomputational aspects of linear transformation by using matrices and determinants have

been explained.

Probability and Statistics have relevance in our lives which have been introduced inBlock 4. But more than that, it is itself inherently interesting. Our need to cope withuncertainty leads us to study and use of probability theory. Under the topic of hypothesistesting, we are trying to determine when it is reasonable to conclude, from analysis of asample, that the entire population possesses a certain property, and when it is notreasonable to reach such a conclusion.

We believe that the course on engineering mathematics will provide insights which help

you in understanding other subjects in Diploma in Computer Integrated Manufacturing.

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CALCULUS : BASIC CONCEPTS

The knowledge and concepts of mathematics help in handling any engineering problem.In block of Mathematics, which is in your hands, you will be studying some concepts ofmathematics, which are a must for you. This block will be dealing with the concepts ofDifferentiation and Integration, and consists of four units.

Unit 1 is devoted to Differential Calculus. In fact, calculus was created to meet thepressing mathematical needs of 17th century for solving problems in Science andTechnology.

In Unit 1, the concept of function has been introduced and different types of functionsand inverse functions have been discussed. This unit also lays the foundations of calculusin terms of definitions and theorems centering around limit and continuity of functionsand algebra of derivatives. We will also be dealing with some of the applications of thederivatives to geometry and maxima and minima of functions. Some important theoremsnamely Rolles Theorem and Lagranges Mean-value Theorem have also been discussed.

Unit 2 deals in function of several variables. We have extended the notions of limit,continuity, partial derivatives, total differentiation, maxima and minima, Jacobians andthen applications for the functions of two or more variables in this unit.

Unit 3 deals with the concept of integrals and its applications. Indefinite integral of afunction as an antiderivative has been defined and the different techniques for findingintegrals of functions have been discussed. After introducing the concept of definiteintegral, some applications have been dealt such as finding the area of a curved surface.

The firm grounding in integrals provided by Unit 3 shall be helpful for easyunderstanding of Unit 4, dealing with ordinary differential equations.

For the want of clarity in concepts, number of solved examples has been introduced ineach unit. To help you check your understanding and to assess yourself, each unit

contains SAQs. The answers to these SAQs are given at the end of each unit. We suggestthat you look at them only after attempting the exercises.

At the end, we wish you all the best for your all educational endeavours.

unit 1_part 1.pdf engg math

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