mathematicsworksheet - fe success success (mechanical) math worksheet.pdfb m is the x axis intercept...

x axis

y axis

−3 −2 −1 0 1 2 3 4

−4

−3

−2

−1

1

2

3

4

y=x

A(4,-3)

M(a,a)

B(p,q)

x

y

odd functionf(-x) = - f(x)

-aa

Mathematics Worksheet

UNSOLVED Mathematics problems (with answers)for practice for FE-Mechanical exam

Prof. Prashant Morewww.FEsuccess.com

Written, typeset and published by

Prof. Prashant More

www.FEsuccess.com

[email protected]

First Edition November 2019.

ISBN No.

All rights reserved. No part of this book can be reproduced in paper or

electronic format without prior written permission of the publisher.

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I took the Fundamentals of Engineering (FE) Exam in April 2001. I could

manage to pass it in the first attempt mostly because I used to work as a

tutor during my graduate studies at the Student Development Center of

University of Toledo, Ohio. I assisted and tutored almost every subject of

Mechanical Engineering. In October 2003, I helped a Civil Engineering

graduate pass this exam in his first attempt. During last 15 years of my

teaching career I have taught Pneumatics and Hydraulics, Manufacturing

Operations Management, Entrepreneurship and Management, Engineer-

ing Mechanics, Statics and Strength of Materials, Finite Element Analysis,

Mechanical and Electrical Equipment for building, Engineering Graphics

with AutoCAD, Solidworks, Autodesk Inventor, CAD/CAM, Developmen-

tal as well as Applied Technical Math. I had my own tutoring center for

Engineering Mathematics for Mumbai University students. At high school

level, I have taught Computer Math, Geometry and Algebra at Front Royal,

Virginia.

So basically, I had a large collection of hand-written notes and near-

infinite numbers of question papers created which I felt will be helpful if I

compile them in a nice ready-to-publish format.

This book is a humble attempt to encompass the topics required for study

for the FE Exam. This book is primarily written for test takers who have

been out of academic environment for a long time.

I would like to offer my sincere thanks to my previous supervisor, Dr. Alex

Echeverria (PE). Working under his guidance made me realize the true

spirit of teaching. His guidance at early stage of my career and the con-

fidence he had in me made me select teaching as my lifelong passion.

Having the material ready and compiling it to be published is altogether

a different story. I had to learn Latex typesetting language for the same.

For this purpose, I would like to thank Mr. Durga Prasad Rao who was

my colleague at University of Mumbai and Mr. Aneez Kundukulathil, who

worked with me at Muscat, Oman. Mr. Sampat Dhanawade maintained

the necessary documentation updated.

It took me almost eight years to prepare this book as it was done mostly in

my spare time while working full time as an Assistant Professor. However,

I assure that this will not prevent me from listening your suggestions and

incorporating fresh ideas into the content and presentation of the book.

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Contents

1 Analytic Geometry 6

1.1 Straight Lines: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Conics: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.3 Circle: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1.4 Ellipse: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1.5 Parabola: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

1.6 Hyperbola: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

1.7 Distance Formula: . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

2 Calculus 78

2.1 Limits: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

2.2 Derivatives: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

2.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

2.4 Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

2.5 Integration by Partial Fractions: . . . . . . . . . . . . . . . . . . . 112

2.6 Integration Techniques for Trigonometric Functions: . . . . . . . 114

2.7 Definite Integrals: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

2.8 Area Bounded by a Curve: . . . . . . . . . . . . . . . . . . . . . . . 121

2.9 Volume of Revolution of Solids: . . . . . . . . . . . . . . . . . . . . 127

3 Linear Algebra 134

3.1 Types of Matrices: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

3.2 Determinant of a Matrix: . . . . . . . . . . . . . . . . . . . . . . . . 138

3.3 Matrix Operations: . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

4 Vector Analysis 156

4.1 Unit Vector: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

4.2 Direction Ratios and Direction Cosines: . . . . . . . . . . . . . . . 160

4.3 Addition/Subtraction of Two Vectors: . . . . . . . . . . . . . . . . 162

4.4 Dot Product: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

4.5 Cross Product: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

4.6 The Del Operator: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

4.7 Gradient: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

4.8 Divergence: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

4.9 Curl: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

5 Differential Equations 196

5.1 General Terminology: . . . . . . . . . . . . . . . . . . . . . . . . . 197

5.2 First Order Differential Equations: . . . . . . . . . . . . . . . . . . 201

5.3 Higher Order Differential Equations: . . . . . . . . . . . . . . . . 213

5.4 Finding Differential Equation from Solution: . . . . . . . . . . . . 220

5.5 Applying Boundary Conditions: . . . . . . . . . . . . . . . . . . . 228

6 Numerical Methods 241

6.1 Newton’s Method of Root Extraction: . . . . . . . . . . . . . . . . 242

6.2 Newton’s Method of Minimization: . . . . . . . . . . . . . . . . . . 247

6.3 Forward Rectangular Rule: . . . . . . . . . . . . . . . . . . . . . . 250

6.4 Trapezoidal Rule: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252

6.5 Simpson’s Rule: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

7 Answer Keys- Mathematics 258

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1 Analytic Geometry

Introduction

Analytic Geometry studies the representation of shapes by algebraic equa-

tions. These equations indicate related key points and distance between

them. In this module, geometrical shapes obtained by section of a cone

are studied in details.

x axis

y axis

0 1 2 3 40

1

2

3

4

run

rise

Figure 1.1: Slope of a line as a ratio of Riseand Run

1.1 Straight Lines:

L E A R N I N G O B J E C T I V E S

After successful completion of this module, you will be able

to:

• Find and interpret equation of a straight line in various

forms.

• Perform slope calculations including parallel and

perpendicular lines.

• Find angle between two coplanar, non-parallel lines.

Slope of a straight line gives you an idea about its inclination with ref-

erence to x-axis. Slope is also referred as gradient.

Slope of a Straight Line:

Slope of a straight line = Rise

Run

= ∆y

∆x

m = y2 − y1

x2 −x1(1.1)

Equation of a Straight Line:

Slope Intercept Format:

y = mx +b (1.2)

where m = slope and b = y-intercept

For the above line, y-intercept = 1, and

slope =Ri se

Run=

2

4=

1

2

So the equation will be,

y = 0.5x +1

2y = x +2

x −2y +2 = 0

Slope Point Format:

y − y1 = m(x −x1) (1.3)

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where x1 and y1 are the coordinates of the point through which the line

passes.

Double Intercept Format:

x

a+ y

b= 1 (1.4)

where a = x intercept and b = y intercept

For parallel lines slopes are equal i.e. m1 = m2

For perpendicular lines

m1 ∗m2 = -1

or,

m2 = −1

m1(1.5)

Angle between Two Straight Lines:

Angle between two straight lines is given by:

α= tan−1(

m2 −m1

1+m1m2

)(1.6)

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Solved Examples:

S O LV E D E X A M P L E - A A A - 0 1

A N G L E B E T W E E N T W O S T R A I G H T L I N E S

1. The angle between lines 2x- 9y + 16=0 and x + 4y + 5 = 0

is given by:

(A) tan−1(

2

9

)

(B) tan−1(−1

4

)

(C) tan−1(−1

2

)(D) The lines are parallel to each other.

Solution:

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G R A P H O F A B S O L U T E VA L U E F U N C T I O N

2. A graph of the equation, y =| x−3 | would consist of which

one of the following?

(A) One straight line.

(B) Two parallel straight lines.

(C) One curved line.

(D) Two straight lines.

Solution:

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X A N D Y- I N T E R C E P T S O F A S T R A I G H T L I N E

3. Given the slope-intercept form of a line as y = mx + b,

which one of the following is true?

(A)m

bis the y axis intercept

(B) b is the x axis intercept

(C) y - b is the slope of the line

(D) -b

mis the x axis intercept

Solution:

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R E FL E C T I O N O F A P O I N T

4. A ray of light coming from the point (1, 2) is reflected at

a point A on the x-axis and then passes through the point

(5, 3). The coordinates of the point A are:

(A) (13

5,0)

(B) (5

13,0)

(C) (- 7, 0)

(D) None of these

Solution:

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M I D P O I N T O F I N T E R C E P T B E T W E E N C O O R D I N AT E

A X E S

5. If the co-ordinates of the middle point of the portion of

a line intercepted between coordinate axes (3,2), then the

equation of the line will be:

(A) 2x+3y=12

(B) 3x+2y=12

(C) 4x-3y=6

(D) 5x-2y=10

Solution:

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I M A G E O F A P O I N T W I T H R E S P E C T T O A L I N E

6. The image of the point (4, -3) with respect to the line y =

x is:

(A) (-4, -3)

(B) (3, 4)

(C) (-4, 3)

(D) (-3, 4)

Solution:

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A R E A O F T R I A N G L E F O R M E D B Y L I N E S

7. A line L passes through the points (1, 1) and (2, 0) and

another line L passes through (1

2,0) and perpendicular to

L. Then the area of the triangle formed by the lines L, L

and y- axis, is:

(A)15

8

(B)25

4

(C)25

8

(D)25

16

Solution:

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D O U B L E - I N T E R C E P T E Q U AT I O N

8. The area of triangle formed by the lines

x = 0,

y = 0 andx

a+ y

b= 1,

is:

(A) ab

(B)ab

2

(C) 2ab

(D)ab

3

Solution:

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E Q U AT I O N O F D I A G O N A L S O F A PA R A L L E L O G R A M

9. The diagonals of a parallelogram PQRS are along the lines

x+3y=4 and 6x-2y=7. Then PQRS must be a:

(A) Rectangle

(B) Square

(C) Cyclic quadrilateral

(D) Rhombus

Solution:

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A B S O L U T E VA L U E F U N C T I O N A C R O S S QU A D R A N T S

10. The area enclosed within the curve |x|+|y|=1 is:

(A)p

2

(B) 1

(C)p

3

(D) 2

Solution:

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P E R P E N D I C U L A R B I S E C T O R O F A L I N E

11. The line 3x +2y =24 meets y -axis at A and x-axis at B. The

perpendicular bisector of AB meets the line through (0,-1)

parallel to x-axis at C. The area of the triangle ABC is:

(A) 182 sq.units

(B) 91 sq.units

(C) 48 sq.units

(D) None of these

Solution:

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D I S TA N C E O F A P O I N T F R O M A L I N E

12. The equation of the line joining the point (3, 5) to the

point of intersection of the lines 4x +y -1 =0 and 7x -3y

-35 =0 is equidistant from the points (0, 0) and (8, 34).

(A) True

(B) False

(C) Nothing can be said

(D) None of these

Solution:

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A N G L E B E T W E E N D I A G O N A L S O F A QU A D R I L AT E R A L

13. The sides AB, BC, CD and DA of a quadrilateral are

x +2y = 3,

x = 1,

x −3y = 4,

5x + y +12 = 0

respectively. The angle between diagonals AC and BD is:

(A) 45◦

(B) 60◦

(C) 90◦

(D) 30◦

Solution:

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P E R P E N D I C U L A R T O I N T E R I O R B I S E C T O R O F A N

A N G L E

14. Given vertices

A(1,1),

B(4,−2) and

C (5,5)

of a triangle, then the equation of the perpendicular

dropped from C to the interior bisector of the angle A is:

(A) y - 5 =0

(B) x - 5 =0

(C) y + 5 =0

(D) x + 5 =0

Solution:

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15. A line 4x+y=1 passes through the point A(2,-7) meets the

line BC whose equation is 3x-4y+1=0 at the point B. The

equation to the line AC so that AB = AC, is:

(A) 52x + 89y + 519 =0

(B) 52x + 89y - 519 =0

(C) 89x + 52y + 519 =0

(D) 89x + 52y - 519 =0

Solution:

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16. If the equation of base of an equilateral triangle is 2x-y=1

and the vertex is (-1, 2), then the length of the side of the

triangle is:

(A)

√20

3

(B)2p15

(C)

√8

15

(D)

√15

2

Solution:

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17. Let PS be the median of the triangle with vertices:

P (2,2),

Q (6,1) and

R (7,3)

The equation of the line passing through (1,-1) and paral-

lel to PS is:

(A) 2x-9y-7=0

(B) 2x-9y-11=0

(C) 2x+9y-11=0

(D) 2x+9y+7=0

Solution:

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18. The equation of perpendicular bisectors of the sides AB

and AC of a triangle ABC are x - y+5=0 and x+2y=0 respec-

tively. If the point A is (1,-2) , then the equation of line

BC is:

(A) 23x + 14y - 40=0

(B) 14x - 23y + 40=0

(C) tan−1(2)

(D) 14x + 23y - 40=0

Solution:

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Figure 1.2: Various conic sections, Source:Stackexchange, Latex code by Marmot

1.2 Conics:



to:

• Understand origin of a conic section as a section of cone

by a cutting plane at various angles.

• Recognize the conic sections from their functions in

standard from and from their graphs.

• Convert a function of a conic section to standard form to

determine whether it yields a circle, a parabola, an ellipse,

or a hyperbola.

• Graph each of the conic sections from its function in

standard form.

• Create the function of a conic section, given information

on its important points and distances.

When a cone is cut by a plane, the resulting section of cone has differ-

ent possibilities:

Eccentricity:

• For Ellipse e < 1

• For parabola e = 1

• For hyperbola e > 1

Conic Section Equation:

The general form of the conic section equation is:

Ax2 +B x y +C y2 +Dx +E y +F = 0 (1.7)

where both A and C are non-zero.

• If B2 - 4AC <0, an ellipse is defined.

• If B2 - 4AC >0, a hyperbola is defined.

• If B2 - 4AC = 0, the conic is a parabola.

• If A = C and B = 0, a circle is defined.

• If A = B = C = 0, a straight line is defined.

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x2 + y2 +2ax +2by + c = 0 (1.8)

is the normal form of the conic section equation, if that conic section has

a principal axis parallel to a coordinate axis.

h = a;k = b

r =√

a2 +b2 − c

• If a2 + b2 - c is positive, a circle, center (-a, -b).

• If a2 + b2 - c equals zero, a point at (-a, -b).

• If a2 + b2 - c is negative, locus is imaginary

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Solved Examples:

S O LV E D E X A M P L E - A A B - 0 2

I D E N T I F Y I N G C O N I C S E C T I O N - I

19. 6x2 + 12x + 6y2 - 8y =100 is an example of a:

(A) circle

(B) parabola

(C) ellipse

(D) hyperbola

Solution:

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I D E N T I F Y I N G C O N I C S E C T I O N - I I

20. 4x2 + 12x + y2 - 8y = 64 is an example of a:

(A) circle

(B) parabola

(C) ellipse

(D) hyperbola

Solution:

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I D E N T I F Y I N G C O N I C S E C T I O N - I I I

21. 6y = 3x2 - 15 is an example of a:

(A) circle

(B) parabola

(C) ellipse

(D) hyperbola

Solution:

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x

y

−2 −1 0 1 2

−2

−1

0

1

2

Figure 1.3: Circle with center at Origin

x

y

−2 −1 0 1 2 3

−2

−1

0

1

2

3

Figure 1.4: Circle with center at a pointother than Origin

1.3 Circle:



to:

• Find equation of circle in standard and non-standard

form.

• Given the equation of a circle, find its center and radius

or diameter.

Equation of circle with center at Origin:

Equation of a circle with center at (0,0) and radius = r

x2 + y2 = r 2

Equation of circle with center at point other that Origin:

Equation of a circle with center at (h,k) and radius = r

(x −h)2 + (y −k)2 = r 2 (1.9)

or in another form:

x2 + y2 +2g x +2 f y + c = 0

which has center at (-g,-f) and

radius =√

g 2 + f 2 − c

A given equation in x and y can represent a circle, if it follows following

three conditions:

1. The highest power of x is 2. Also, the highest power of y term is 2.

2. Coefficient of x2 is same as coefficient of y2.

3. There is NO product term xy.

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Solved Examples:

S O LV E D E X A M P L E - A AC - 0 1

22. Consider an ant crawling along the curve

(x −2)2 + y2 = 4

where x and y are in meters. The ant starts at the point

(4,0) and moves counter-clockwise with a speed of 1.57

meters per second. The time taken by the ant to reach

the point (2,2) is (in seconds):

(A) 1

(B) 2

(C) 4

(D) π/2

Solution:

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23. What is the radius of the circle

x2 + y2 −6y = 0?

(A) 2

(B) 3

(C) 4

(D) 5

Solution:

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24. What are the coordinates of the center of the curve

x2 + y2 −2x −4y −31 = 0?

(A) (-1, -1)

(B) (-2, -2)

(C) (1, 2)

(D) (2, 1)

Solution:

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25. A circle whose equation is:

x2 + y2 +4x +6y −23 = 0

has its center at:

(A) (2, 3)

(B) (3, 2)

(C) (-3, 2)

(D) (-2, -3)

Solution:

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26. What is the radius of a circle with the equation:

x2 −6x + y2 −4y −12 = 0

(A) 3.46

(B) 7

(C) 5

(D) 6

Solution:

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D I A M E T E R O F A C I R C L E

27. The diameter of a circle described by

9x2 +9y2 = 16

(A)4

3

(B)16

9

(C)8

3

(D) 4

Solution:

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D I S TA N C E B E T W E E N C E N T E R O F C I R C L E A N D Y- A X I S

28. How far from the y-axis is the center of the curve

2x2 +2y2 +10x −6y −55 = 0

(A) 2.5

(B) 3.0

(C) 2.75

(D) 3.25

Solution:

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D I S TA N C E B E T W E E N C E N T E R O F C I R C L E S

29. What is the distance between the centers of the circles

x2 + y2 +2x +4y −3 = 0 and

x2 + y2 −8x −6y +7 = 0?

(A) 7.07

(B) 7.77

(C) 8.07

(D) 7.87

Solution:

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S H O R T E S T D I S TA N C E F R O M A P O I N T T O A C I R C L E

center

30. The shortest distance from A (3, 8) to the circle

x2 + y2 +4x −6y = 12

is equal to?

(A) 2.1

(B) 2.3

(C) 2.5

(D) 2.7

Solution:

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P R O P E R T I E S F R O M E Q U AT I O N O F A C I R C L E

31. The equation circle

x2 + y2 −4x +2y −20 = 0

describes:

(A) A circle of radius 5 centered at the origin.

(B) An eclipse centered at (2, -1).

(C) A sphere centered at the origin.

(D) A circle of radius 5 centered at (2, -1).

Solution:

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A R E A O F C I R C L E F R O M I T S E Q U AT I O N

32. Find the area (in square units) of the circle whose equa-

tion is

x2 + y2 = 6x −8y

(A) 20 π

(B) 22 π

(C) 25 π

(D) 27 π

Solution:

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E Q U AT I O N O F C I R C L E F R O M G I V E N C O N D I T I O N S

33. Determine the equation of the circle whose radius is 5,

center on the line x = 2 and tangent to the line 3x - 4y +

11 = 0.

(A) (x −2)2 +(y −2)2 = 5

(B) (x −2)2 + (y +2)2 = 25

(C) (x −2)2 + (y +2)2= 5

(D) (x −2)2 +(y −2)2 = 25

Solution:

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E Q U AT I O N O F A C I R C L E G I V E N I T S C E N T E R A N D

TA N G E N T S

34. Find the equation of the circle with the center at (-4, -5)

and tangent to the line 2x + 7y - 10 = 0.

(A) x2 + y2 + 8x - 10y - 12 = 0

(B) x2 + y2 + 8x - 10y + 12 = 0

(C) x2 + y2 + 8x + 10y - 12 = 0

(D) x2 + y2 8x + 10y + 12 = 0

Solution:

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C O N D I T I O N F O R P O I N T C I R C L E

35. Find the value of k for which the equation

x2 + y2 +4x −2y −k = 0

represents a point circle.

(A) 5

(B) 6

(C) -6

(D) -5

Solution:

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E Q U AT I O N O F S P H E R E F R O M I T S C E N T E R A N D

R A D I U S

36. The equation of a sphere with center at (-3, 2, 4) and of

radius 6 units is?

(A) x2 + y2 +z2+6x - 4y - 8z = 36

(B) x2 +y2+ z2 +6x - 4y - 8z = 7

(C) x2 +y2 + z2 +6x - 4y + 8z = 6

(D) x2 + y2 + z2 +6x - 4y + 8z = 36

Solution:

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x

y

−3 −2 −1 0 1 2 3

−2

−1

0

1

2

Figure 1.5: Ellipse with center at a pointother that Origin

x

y

−3 −2 −1 0 1 2 3

−2

−1

0

1

2

Figure 1.6: Ellipse with center at Origin

1.4 Ellipse:



to:

• Find equation of ellipse types, regular and rectangular,

eccentricity, directrix, focus, latus rectum, relation

between a and b.

Equation of an ellipse when center is at (h,k) is:

(x −h)2

a2 + (y −k)2

b2 = 1 (1.10)

Equation of an ellipse when center is at (0,0) is:

x2

a2 + y2

b2 = 1 (1.11)

Foci: (±ae,0) = (± c, 0) where e is the eccentricity of the ellipse.

c2 = a2 −b2 (1.12)

Here, a is called the semi major axis, and b is called semi minor axis.

(Assuming a > b) For the ellipse shown in the figure, the semi major axis=

3 and semi minor axis = 1.

Eccentricity:

e =√

1− b2

a2 = c

a(1.13)

Also, for an ellipse,

e < 1 (1.14)

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A A D - 0 1

G E N E R A L E Q U AT I O N O F A C O N I C S E C T I O N

37. The general equation of a conic section is given by the

following equation:

Ax2 +B x y +C y2 +Dx +E y +F = 0

A curve maybe identified as an ellipse by which of the fol-

lowing conditions?

(A) B 2- 4AC <0

(B) B 2 - 4AC = 0

(C) B 2 - 4AC >0

(D) B 2 - 4AC = 1

Solution:

FEsuccess.com


E Q U AT I O N O F L O C U S G I V E N A C O N D I T I O N

38. Point P (x, y) moves with a distance from point (0, 1) one-

half of its distance from line y = 4. The equation of its lo-

cus is?

(A) 2x2 - 4y2 = 5

(B) 4x2 + 3y2 = 12

(C) 2x2 + 5y3 = 3

(D) x2 + 2y2 = 4

Solution:

FEsuccess.com


L E N G T H O F L AT U S R E C T U M O F A N E L L I P S E

39. What is the length of the latus rectum of

4x2 +9y2 +8x −32 = 0?

(A) 2.5

(B) 2.7

(C) 2.3

(D) 2.9

Solution:

FEsuccess.com


D I S TA N C E B E T W E E N F O C I O F A N E L L I P S E

40. The lengths of the major and minor axes of an ellipse are

10 m and 8 m, respectively. Find the distance between

the foci.

(A) 3

(B) 4

(C) 5

(D) 6

Solution:

FEsuccess.com


C E N T E R O F A N E L L I P S E

41. The equation

25x2 +16y2 −150x +128y +81 = 0

has its center at?

(A) (3, -4)

(B) (3, 4)

(C) (4, -3)

(D) (3, 5)

Solution:

FEsuccess.com


M A J O R A X I S O F A N E L L I P S E

42. Find the major axis of the ellipse

x2 +4y2 −2x −8y +1 = 0

(A) 2

(B) 10

(C) 4

(D) 6

Solution:

FEsuccess.com


L E N G T H O F L AT U S R E C T U M O F A N E L L I P S E

43. An ellipse with an eccentricity of 0.65 and has one of its

foci 2 units from the center. The length of the latus rec-

tum is nearest to?

(A) 3.5 units

(B) 3.8 units

(C) 4.2 units

(D) 3.2 units

Solution:

FEsuccess.com


E C C E N T R I C I T Y O F A N E L L I P S E

44. The earth’s orbit is an ellipse with the sun at one of the

foci. If the farthest distance of the sun from the earth is

105.5 million km and the nearest distance of the sun from

the earth is 78.25 million km, find the eccentricity of the

ellipse.

(A) 0.15

(B) 0.25

(C) 0.35

(D) 0.45

Solution:

FEsuccess.com


E Q U AT I O N O F D I R E C T R I X O F A N E L L I P S E

45. An ellipse with center at the origin has a length of major

axis 20 units. If the distance from center of ellipse to its

focus is 5, what is the equation of its directrix?

(A) x = 18

(B) x = 20

(C) x = 15

(D) x = 16

Solution:

FEsuccess.com

−3 −2 −1 1 2 3

2

4

6

8

10

x

f (x) = x2

Figure 1.7: Parabola: Plot of y = x2

−3 −2 −1 1 2 3

−10

−8

−6

−4

−2

x

f (x) =−x2

Figure 1.8: Parabola: Plot of y = −x2

1.5 Parabola:



to:

• Find equation of parabola- Types of parabolas,

eccentricity directrix, focus, latus rectum, asymptote.

The standard form of a parabola’s equation is generally expressed:

y2 = 4ax

The role of ’a’

If a >0, the parabola opens towards right

if a <0, it opens towards left.

x2 = 4by

The role of ’b’

If b >0, the parabola opens upwards

if b <0, it opens downwards.

Key Properties of a parabola:

• Vertex is the minimum / maximum point of the parabola.

• Focus is a point on the axis of symmetry of the parabola that is a set

distance from the vertex of the parabola.

• Directrix is a line (not a point) that is perpendicular to the axis of sym-

metry of the parabola and does not intersect with the parabola.

• Eccentricity of a parabola is always equal to 1 because any point on

parabola is equidistant from focus and directrix.

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Solved Examples:

S O LV E D E X A M P L E - A A E - 0 1

F I N D I N G C U R V E F R O M A N E Q U AT I O N

46. 3x2 + 2x - 5y + 7 = 0. Determine the curve.

(A) Parabola

(B) Ellipse

(C) Circle

(D) Hyperbola

Solution:

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F O C U S O F A PA R A B O L A

47. The focus of the parabola y2 = 16x is at

(A) (4, 0)

(B) (0, 4)

(C) (3, 0)

(D) (0, 3)

Solution:

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E Q U AT I O N O F V E R T E X O F A PA R A B O L A

48. Where is the vertex of the parabola x2 = 4(y - 2)?

(A) (2, 0)

(B) (0, 2)

(C) (3, 0)

(D) (0, 3)

Solution:

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E Q U AT I O N O F D I R E C T R I X O F A PA R A B O L A

49. Find the equation of the directrix of the parabola y2 =

16x.

(A) x = 2

(B) x = -2

(C) x = 4

(D) x = -4

Solution:

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V E R T E X O F A PA R A B O L A

50. Given the equation of a parabola

3x +2y2 −4y +7 = 0

Locate its vertex.

(A) (5/3, 1)

(B) (5/3, -1)

(C) (-5/3, -1)

(D) (-5/3, 1)

Solution:

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F I N D I N G FA C I N G D I R E C T I O N O F A C U R V E

51. In the equation

y =−x2 +x +1

where is the curve facing?

(A) Upward

(B) Facing left

(C) Facing right

(D) Downward

Solution:

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F O C U S O F PA R A B O L A

52. Find the location of the focus of the parabola

y2 +4x −4y −8 = 0

(A) (2.5, -2)

(B) (3, 1)

(C) (2, 2)

(D) (-2.5, -2)

Solution:

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E Q U AT I O N O F A X I S O F S Y M M E T RY

53. Find the equation of the axis of symmetry of the function

y = 2x2 −7x +5

(A) 7x + 4 = 0

(B) 4x + 7 = 0

(C) 4x - 7 = 0

(D) x - 2 = 0

Solution:

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E Q U AT I O N O F PA R A B O L A F R O M D I R E C T R I X A N D

F O C U S

54. A parabola has its focus at (7, -4) and directrix y = 2. Find

its equation.

(A) x2 + 12y - 14x + 61 = 0

(B) x2 - 14y + 12x + 61 = 0

(C) x2 - 12x + 14y + 61 = 0

(D) none of the above

Solution:

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E Q U AT I O N O F D I R E C T R I X O F A PA R A B O L A

55. Given a parabola

(y −2)2 = 8(x −1)

What is the equation of its directrix?

(A) x = -3

(B) x = -1

(C) y = -3

(D) y = 3

Solution:

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−4 −2 2 4

−4

−2

2

4

x

y

Figure 1.9: Hyperbola. The dotted lines areasymptotes.

`= b2

a

x2

a2− y2

b2= 1

a

b c

cF2 F1V

Center

Figure 1.10: Hyperbola. The foci pointsare shown with their relationship with theasymptotes. F1 and F2 are the Foci points,` is the Semi-Latus Rectum, V is the Vertex

1.6 Hyperbola:



to:

• Find equation of hyperbola- regular and rectangular,

eccentricity, directix, focus, latus rectum, asymptote,

relation between a and b.

Equation of a hyperbola when center is at (h,k)

(x −h)2

a2 − (y −k)2

b2 = 1 (1.15)

Equation of a hyperbola when center is at (0,0)

x2

a2 − y2

b2 = 1 (1.16)

• Focus of hyperbola: The two points on the transverse axis. These points

are what controls the entire shape of the hyperbola since the hyper-

bola’s graph is made up of all points, P, such that the distance between

P and the two foci are equal. To determine the foci you can use the for-

mula: a2 +b2 = c2

• Transverse axis: This is the axis on which the two foci are.

• Asymptotes: The two lines that the hyperbolas come closer and closer

to touching. The equation of the asymptotes is always:

y =±a

bx

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Solved Examples:

S O LV E D E X A M P L E - A A F - 0 1

E Q U AT I O N O F H Y P E R B O L A F R O M V E R T I C E S

56. Find the equation of the hyperbola having vertices:

(±5,0), foci: (±8,0)

(A)x2

39− y2

25= 1

(B)x2

14− y2

25= 1

(C)x2

25+ y2

39= 1

(D)x2

25− y2

39= 1

Solution:

FEsuccess.com


E Q U AT I O N O F T H E A S Y M P T O T E O F H Y P E R B O L A

57. What is the equation of the asymptote of the hyperbolax2

9− y2

4= 1?

(A) 2x - 3y = 0

(B) 3x - 2y = 0

(C) 2x - y = 0

(D) 2x + y = 0

Solution:

FEsuccess.com


E Q U AT I O N O F H Y P E R B O L A G I V E N A S Y M P T O T E S

58. Find the equation of the hyperbola whose asymptotes are

y = ± 2x and which passes through (5/2, 3).

(A) 4x2 + y2 + 16 = 0

(B) 4x2 + y2 - 16 = 0

(C) x2 - 4y2 - 16 = 0

(D) 4x2 - y2 = 16

Solution:

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I D E N T I F Y I N G C U R V E R E P R E S E N T E D B Y A N E Q U AT I O N

59. 4x2 - y2 = 16 is the equation of a/an?

(A) parabola

(B) hyperbola

(C) circle

(D) ellipse

Solution:

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E C C E N T R I C I T Y O F C U R V E

60. Find the eccentricity of the curve:

9x2 −4y2 −36x +8y = 4

(A) 1.80

(B) 1.92

(C) 1.86

(D) 1.76

Solution:

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D I S TA N C E F R O M F O C U S O F A H Y P E R B O L A

61. How far from the x-axis is the focus F of the hyperbola

x2 −2y2 +4x +4y +4 = 0?

(A) 4.5

(B) 3.4

(C) 2.7

(D) 2.1

Solution:

FEsuccess.com

1.7 Distance Formula:


• Calculate distance between two points in space.

Distance between two points in the space P1 (x1, y1, z1) and P2 (x2, y2, z2)

is given by:

√(x2 −x1)2 + (y2 − y1)2 + (z2 − z1)2 (1.17)

This can be proved by repeated application of the Pythagorean Theo-

rem.

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Solved Examples:

S O LV E D E X A M P L E - A AG - 0 1

D I S TA N C E B E T W E E N T W O P O I N T S I N 3 - D

62. The distance between the points (-5,-5,7) and (3,0,5) is:

(A) 4.531

(B) 5.657

(C) 9.643

(D) 93

Solution:

FEsuccess.com

2 Calculus

Introduction

Calculus is the study of how things change. It provides a framework for

modeling systems in which there is change, and a way to deduce the pre-

dictions of such models. It is the branch of mathematics that deals with

the finding and properties of derivatives and integrals of functions. The

two main types are differential calculus and integral calculus.

2.1 Limits:



to:

• Understand concepts of limits- Algebraic, trigonometric,

involving infinity, using L-Hospitals Rule.

A limit is the value that a function or sequence "approaches" as the

input or index approaches some value.

Sometimes, a function is undefined if it takes some exact value of ’x’. In

that case, we can find out where the function was approaching. This can

be done by removing the factors causing undefined nature or sometimes

by using specific mathematical identities.

L Hopital’s rule:

If a limit is in indeterminate form (such as0

0or

∞∞ ) then,

limx→c

f (x)

g (x)= limx→c f ′(x)

limx→c g ′(x)= limx→c f ′′(x)

limx→c g ′′(x)(2.1)

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A B A - 0 1

1. If a function is continuous at a point, then:

(A) The limit of function may not exist at that point

(B) The function must be derivative at that point.

(C) Limit of the function at that point tends to infinity

(D) The limit must exist at that point and the value of this

limit should be same as value of the function at that

point.

Solution:

FEsuccess.com


2. Let x denote a real number. Find out the INCORRECT

statement.

(A) S = x :x > 3 represents the set of all real numbers

greater than 3

(B) S = x : x2 <0 represents the empty set.

(C) S = x :x ∈ A and x ∈ B represents the union of set A and

set B.

(D) S = x :a < x < b represents the set of all real numbers

between a and b, where a and b are real numbers.

Solution:

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3. Consider the function f(x) = | x | in the interval -1 ≤ x ≤ 1.

At the point x = 0, f (x) is:

(A) continuous and differentiable

(B) non-continuous and differentiable

(C) continuous and non-differentiable

(D) neither continuous nor differentiable

Solution:

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4.

limx→0

(1−cos x

x2

)is:

(A)1

4

(B)1

2

(C) 1

(D) 2

Solution:

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5. What is

limθ→0

sinθ

θ

equal to?

(A) θ

(B) sinθ

(C) 0

(D) 1

Solution:

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6.

limx→0

sin2 x

x

is equal to:

(A) 0

(B) 3

(C) 1

(D) -1

Solution:

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7. Which of the following functions is not differentiable in

the domain [-1,1]?

(A) f (x) = x2

(B) f (x) = x - 1

(C) f (x) = 2

(D) f (x) = maximum (x, - x )

Solution:

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8. At x = 0, the function f (x) = x3+ 1 has

(A) a maximum value

(B) a minimum value

(C) a singularity

(D) a point of inflection

Solution:

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9. The minimum value of function y = x2 in the interval [1,

5] is:

(A) 0

(B) 1

(C) 25

(D) undefined

Solution:

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10. The value of:

limx→8

x13 −2

x −8

(A)1

16

(B)1

12

(C)1

8

(D)1

4

Solution:

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11. If

f (x) = 2x2 −7x +3

5x2 −12x −9

then l i mx→3 f (x) will be:

(A)−1

3

(B)5

18

(C) 0

(D)2

5

Solution:

FEsuccess.com


12.

limx→1

(1−x) tan(πx

2

)=?

(A)π

2

(B) π

(C)2

π

(D) 0

Solution:

FEsuccess.com


13. Let f: R → R be defined by:

(3x2 +4)cos x

Then

limh→0

f (h)+ f (−h)−8

h2

is equal to :

(A) 0

(B) 2

(C)π

2

(D) π

Solution:

FEsuccess.com

x

y

x x + h

B

Af(x)

f(x + h)

Figure 2.1: First Principle of Derivative

2.2 Derivatives:



to:

• Calculate derivative using basic formulae- algebraic,

trigonometric, inverse trigonometric, logarithmic and

exponential.

• Calculate derivative using rules of derivatives such as -

Addition/subtraction, product, quotient and chain rule.

• Apply differentiation to calculate tangent and normal, rate

of change, maxima/minima and errors/approximation.

First Principle of Derivative

f ′(x) = lim∆x→0

f (x +∆x)− f (x)

∆x(2.2)

Some Basic Formulae:

d

d xc = 0 (2.3)

d

d x(u ± v ±w ± ....) = du

d x± d v

d x± d w

d x± ..... (2.4)

d

d x(cu) = c

du

d x(2.5)

d

d x(uv) = u

d v

d x+ v

du

d x(2.6)

d

d x(

u

v) =

v dud x −u

d v

d xv2 (2.7)

Chain Rule :

d y

d x= d y

du

du

d x(2.8)

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List of Derivative Formulae:

Algebraic Group

d

d xxn = nxn−1 (2.9)

Trigonometric Group

d

d xsin x = cos x (2.10)

d

d xcos x =−sin x (2.11)

d

d xtan x = sec2 x (2.12)

d

d xcot x =−csc2 x (2.13)

d

d xsec x = sec x tan x (2.14)

d

d xcsc x =−csc x cot x (2.15)

Logarithmic and Exponential Group

d

d xex = ex (2.16)

d

d xax = ax log a (2.17)

d

d xlog x = 1

x(2.18)

Inverse Trigonometric Group

d

d xsin−1 x = 1p

1−x2(2.19)

d

d xcos−1 x = −1p

1−x2(2.20)

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d

d xtan−1 x = 1

1+x2 (2.21)

d

d xcot−1 x = −1

1+x2 (2.22)

d

d xsec−1 x = 1

xp

x2 −1(2.23)

d

d xcsc−1 x = −1

xp

x2 −1(2.24)

FEsuccess.com

Solved Problems

S O LV E D E X A M P L E - A B B - 0 1

14. Equation of the line normal to function f (x) = (x −8)23 +1

at P(0, 5) is:

(A) y = 3x - 5

(B) y = 3x + 5

(C) 3y = x + 15

(D) 3y = x - 15

Solution:

FEsuccess.com


15. If

x = a(θ+ sinθ)

and

y = a(1−cosθ)

thend y

d xwill be equal to:

(A) sin

(θ

2

)

(B) cos

(θ

2

)

(C) tan

(θ

2

)

(D) cot

(θ

2

)

Solution:

FEsuccess.com


16. For the following parametric function, what is thed 2 y

d t 2 at

t= 0?

x = 1− t 2

y = t − t 3

(A) ∞

(B) 0

(C) -1

(D)2

3

Solution:

FEsuccess.com


17. The minimum point of the function

f (x) =(

x3

3

)−x

is at:

(A) x = 1

(B) x = - 1

(C) x = 0

(D) x =1p3

Solution:

FEsuccess.com


18. The right circular cone of largest volume that can be en-

closed by a sphere of 1m radius has a height of:

(A)1

3m

(B)2

3m

(C)2p

2

3m

(D)4

3m

Solution:

FEsuccess.com


19. If

In = d n

d xn (xn log x),

then:

In −nIn−1 =?

(A) n

(B) n -1

(C) n!

(D) (n -1)!

Solution:

FEsuccess.com

2.3 Partial Derivatives



to:

• Define a partial derivative.

• Compute partial derivatives by viewing certain variables

as constant.

• Understand notation and computation for higher-order

partial derivatives.

When working with functions of more than one variable, the question

in calculus becomes: how can we evaluate the rate of change? The answer

is called a partial derivative. Given a function f(x, y) or f(x, y, z), the partial

derivative of f with respect to x, ∂ f = fx , is found by treating all variables

other than x as constants.

Technically, were finding

limh→0

f (x +h, y)− f (x, y)

h

The partial derivatives ∂ f = fy and ∂ f = fz have analogous definitions.

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A B C - 0 1

20. Let f = y x . What is:∂2 f

∂x∂y

at x = 2, y = 1?

(A) 0

(B) ln2

(C) 1

(D)1

ln2

Solution:

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21. The function f (x, y) = 2x2 +2x y − y3 has:

(A) only one stationary point at (0, 0)

(B) two stationary points at (0, 0) and

(1

6,−1

3

)(C) two stationary points at (0, 0) and (1, - 1)

(D) no stationary point

Solution:

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22. If u =x + y

x − y, then

∂u

∂x+ ∂u

∂y= ??

(A)1

x − y

(B)2

x − y

(C)1

x − y2

(D)2

x − y2

Solution:

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23. If u = log(x2 + y2), then∂2u

∂x2 + ∂2u

∂y2 =??

(A)1

x2 + y2

(B) 0

(C)x2 − y2

(x2 + y2)2

(D)y2 −x2

(x2 + y2)2

Solution:

FEsuccess.com

2.4 Indefinite Integrals



to:

• Calculate integration of a function using basic formulae -

algebraic, trigonometric, inverse trigonometric,

logarithmic and exponential.

• Calculate integration by substitution, by partial fractions,

by trigonometric substitution, by parts

Integration Formulae:

Algebraic Group

∫xnd x = xn+1

n +1+C (2.25)

Trigonometric Group

∫cos x d x = sin x +C (2.26)

∫sin x d x =−cos x +C (2.27)

∫sec2 x d x = tan x +C (2.28)

∫csc2 x d x =−cot x +C (2.29)

∫sec x tan x d x = sec x +C (2.30)

∫csc x cot x d x =−csc x +C (2.31)

Logarithmic and Exponential Group

∫ex d x = ex +C (2.32)

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∫ax d x = ax

log a+C (2.33)

∫1

xd x = log x +C (2.34)

Inverse Trigonometric Group

∫1p

1−x2d x = sin−1 x +C (2.35)

∫ −1p1−x2

d x = cos−1 x +C (2.36)

∫1

1+x2 d x = tan−1 x +C or −cot−1 x +C (2.37)

∫1

|x|p

x2 −1d x = sec−1 x +C or −csc−1 x +C (2.38)

Integration by Substitution method

There are occasions when it is possible to perform an apparently difficult

piece of integration by first making a substitution.

I =∫

x sin(x2 +1)d x

Substitute

(x2 +1) = u

Then

2xd x = du

or

xd x = 1

2du

Then the integral becomes

I = 1

2

∫sinudu

=−1

2cosu

=−1

2cos(x2 +1)+C

FEsuccess.com

Integration by Parts:

I =∫

x cos x d x

=∫

u d v

= uv −∫

v du

= x sin x −∫

sin x d x

= x sin x +cos x +C

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A B D - 0 1

24. ∫sec x

sec x + tan xd x

(A) tan x + sec x +C

(B) sec x − tan x +C

(C) tan x − sec x +C

(D) tan x −cot x +C

Solution:

FEsuccess.com

S O LV E D E X A M P L E - A B D - 0 2

25. Evaluate ∫ p1+ sin5x d x

(A)2

5

[sin

5x

2−cos

5x

2

]+C

(B)2

5

[sin

5x

2+cos

5x

2

]+C

(C)2

3

[sin

3x

2−cos

3x

2

]+C

(D)2

3

[sin

3x

2+cos

3x

2

]+C

Solution:

FEsuccess.com

2.5 Integration by Partial Fractions:



to:

• Distinguish between proper and improper fractions.

• Express an algebraic fraction as the sum of its partial

fractions.

For partial fractions, the given fraction must be proper, i.e the highest

degree of denominator polynomial must be higher than the highest degree

of numerator polynomial. if not, first division must be carried out to make

it a proper polynomial.

Rules on how to set Partial Fractions (with examples):

1.3

(x +2)(x +3)= A

(x +2)+ B

(x +3)

2.3

(x +2)2 = A

(x +2)+ B

(x +2)2

3.3

(x2 +2)(x +3)= Ax +B

(x2 +2)+ C

(x +3)

4.3

(x2 +2)(x +3)2 = Ax +B

(x2 +2)+ C

(x +3)+ D

(x +3)2

FEsuccess.com

Solved Examples

S O LV E D E X A M P L E - A B E - 0 1

26. Evaluate: ∫1

x2 −4d x

(A)1

4ln |x −2|/|x +2|+C

(B)1

2ln |x −2|/|x +2|+C

(C)1

4ln |x +2|/|x −2|+C

(D)1

2ln |x +2|/|x −2|+C

Solution:

FEsuccess.com

2.6 Integration Techniques for Trigonometric Functions:



to:

• Use trigonometric identities to simplify and solve

integration problems.

1. ∫cos2 xd x =

∫1+cos2x

2d x

=∫

1

2d x + 1

2

∫cos2xd x

= x

2+ sin2x

2×2+C

2. ∫sin2 xd x =

∫1−cos2x

2d x

=∫

1

2d x − 1

2

∫cos2xd x

= x

2− sin2x

2×2+C

3. ∫x sin xd x = x(−cos x)− (1)(−sin x)+C

=−x cos x + sin x +C

4. ∫x cos xd x = x(sin x)− (1)(−cos x)+C

= x sin x +cos x +C

5. ∫sin x cos xd x =

∫2sin x cos x

2d x

=∫

sin2x

2d x

= 1

2

(−cos2x

2

)+C

= −cos2x

4+C

Alternatively, the same integration can be calculated by the method of

substitution. ∫sin x cos xd x

Put

sin x = t

cos xd x = d t

FEsuccess.com

∫td t = t 2

2+C

= sin2 x

2+C

FEsuccess.com

x

y

y = f(x)

a b

f(a)

f(b)

x

y

even functionf(-x) = f(x)

-a a

x

y

odd functionf(-x) = - f(x)

-aa

2.7 Definite Integrals:



to:

• Find the definite integral of algebraic,

trigonometric/inverse trigonometric, exponential and

logarithmic functions.

• Use properties of definite integrals.

∫ b

af (x)d x = [F (x)]b

a = F (b)−F (a)

In definite integrals, there is no constant of integration, C.

Properties of Definite Integrals:

1. Change of variable has no effect on definite integration.∫ b

af (x)d x =

∫ b

af (y)d y

2. If limits of definite integration are reversed, then the answer changes

sign. (because now the area under the curve given by f(x) is calculated

backwards.) ∫ b

af (x)d x =−

∫ a

bf (x)d x

3. Definite integration can be split into two sub-definite integration as

long as the point of division is in the interval of original limits of def-

inite integration.∫ b

af (x)d x =

∫ c

af (x)d x +

∫ b

cf (x)d x.....a < c < b

4. special cases for odd and even functions.

For even function, f(-x) = f(x) i.e the graph is symmetrical about y-axis.

In such case, ∫ a

−af (x)d x = 2

∫ a

0f (x)d x

For odd functions, f(-x) = -f(x) and graph of f(x) is mirrored about x-axis

AND y-axis. In such case, ∫ a

−af (x)d x = 0

This is because the ’negative’ area balances the ’positive’ area after the

origin.

5. ∫ 2a

0f (x)d x =

∫ a

0f (x)d x +

∫ a

0f (2a −x)d x

FEsuccess.com

Solved Examples

S O LV E D E X A M P L E - A B G - 0 1

27. ∫ a

−a(sin6 x + sin7 x)d x

is equal to:

(A) 2∫ a

0 (sin6 x)d x

(B) 2∫ a

0 (sin7 x)d x

(C) 2∫ a

0 (sin6 x + sin7 x)d x

(D) zero

Solution:

FEsuccess.com


28. The value of the integral:∫ ∞

−∞d x

1+x2

is:

(A) −π

(B)−π2

(C)π

2

(D) π

Solution:

FEsuccess.com


29. If f (x) is an even function and a is a positive real number,

then∫ a−a f (x)d x equals:

(A) 0

(B) a

(C) 2a

(D) 2∫ a

0 f (x)d x

Solution:

FEsuccess.com


30. Let f be a continuous and positive real valued function on

[0, 1]. Then ∫ π

0f (sin x)cos x d x

is equal to:

(A) 0

(B) 1

(C) -1

(D) ∞

Solution:

FEsuccess.com

2.8 Area Bounded by a Curve:



to:

• Apply integration to find area under a curve.

Area bounded by a curve is given by:

A =∫ b

af (x)d x

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A B H - 0 1

31. The area bounded by two curves y = x2 and y2 = x be-

tween the lines x = 0.5 and x = 0.7 is:

(A) 0.345

(B) 0.399

(C) 0.921

(D) 0.082

Solution:

FEsuccess.com


32. What is the area (in square units) bounded by the curve

y2 = x and the line x - 4 = 0?

(A)30

3

(B)31

3

(C)32

3

(D)29

3

Solution:

FEsuccess.com


33. Find the area bounded by the parabolas y = 6x -x2and y

=x2 - 2x. Note: The parabolas intersect at points (0,0) and

(4,8).

(A) 44/3 square units

(B) 64/3 square units

(C) 74/3 square units

(D) 54/2 square units

Solution:

FEsuccess.com


34. Find the area bounded by the parabola x2= 4y and y = 4.

(A) 21.33

(B) 33.21

(C) 31.32

(D) 13.23

Solution:

FEsuccess.com


35. The area enclosed between the straight line y = x and the

parabola y = x2 in the x -y plane is:

(A)1

6

(B)1

4

(C)1

3

(D)1

2

Solution:

FEsuccess.com

2.9 Volume of Revolution of Solids:



to:

• Apply integration to find volume of a solid.

Volume of solid of revolution about x axis is given by,

V =π∫ b

a

[f (x)

]2 d x

Or, if the function is given in terms of x = f(y), then Volume of solid of rev-

olution is given by,

V =π∫ d

c

[f (y)

]2 d y

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A B I - 0 1

36. Determine the volume of the solid obtained by rotating

the region bounded by 3p

x, and the x-axis about the x-

axis.

(A)128π

5

(B)144π

5

(C)96π

5

(D)72π

5

Solution:

FEsuccess.com


37. The volume of an object expressed in spherical co-

ordinates is given by:∫ 2π

0

∫ π3

0

∫ 1

0r 2 sinφdr dφdθ

The value of the integral is:

(A)π

3

(B)π

6

(C)π

32

(D)π

4

Solution:

FEsuccess.com


38. The area enclosed by the ellipse (x2)/9 + (y2)/4 = 1 is re-

volved about the line x = 3. What is the volume gener-

ated?

(A) 355.3

(B) 360.1

(C) 370.3

(D) 365.1

Solution:

FEsuccess.com


39. The parabolic arc y =p

x, 1 ≤ x ≤ 2 is revolved around the

x-axis. The volume of the solid of revolution is:

(A)π

4

(B)π

2

(C)3π

4

(D)3π

2

Solution:

FEsuccess.com


40. Find the volume (in cubic units) generated by rotating a

circle x2 + y2 + 6x + 4y + 12 = 0 about the y-axis.

(A) 39.48

(B) 47.23

(C) 59.22

(D) 62.11

Solution:

FEsuccess.com

References:

1. Paul’s Online Math Notes:

http://tutorial.math.lamar.edu

2. http://www.math.umd.edu/~tjp/

131 09.2 lecture notes.pdf

FEsuccess.com

3 Linear Algebra

3.1 Types of Matrices: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

3.2 Determinant of a Matrix: . . . . . . . . . . . . . . . . . . . . . . . . 138

3.3 Matrix Operations: . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

Introduction

A matrix is a n-dimensional array of numbers or expressions arranged in

a set of rows and columns. A basic understanding of elementary matrix

algebra is essential for the analysis of engineering systems.

Matrices can be used to solve simultaneous equations.

3.1 Types of Matrices:



to:

• Identify types of matrices such as Square, Diagonal,

Scalar, Row, Column, Identity.

Row Matrix

A matrix which has only one row.

[2 −6 4

]

Column Matrix:

A matrix which has only one column.

3

4

−1

Square Matrix:

A matrix having same no. of rows and columns.3 6 1

4 5 2

2 0 −2

Diagonal Matrix:

A square matrix where only diagonal elements are present, non-diagonal

elements are zero. 3 0 0

0 5 0

0 0 −2

Scalar Matrix:

Diagonal matrix where all diagonal elements are identical.3 0 0

0 3 0

0 0 3

FEsuccess.com

Identity Matrix:

Scalar Matrix where the diagonal elements are one.1 0 0

0 1 0

0 0 1

Singular Matrix:

A Matrix whose determinant is equal to zero.[2 3

−6 −9

]

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - AC A - 0 1

1. For which value of x will the matrix given below become

singular?

8 x 0

4 0 2

12 6 0

(A) 4

(B) 6

(C) 8

(D) 12

Solution:

FEsuccess.com

3.2 Determinant of a Matrix:



to:

• Find determinant of a matrix.

• Understand the conditions for zero determinant.

Only square matrices have determinant. (i.e. have the same number of

rows as columns).

For a 2 X 2 matrix, the determinant can be calculated as:

∣∣∣∣∣1 2

3 4

∣∣∣∣∣= 1×4−2×3 =−2

For a 3 X 3 matrix, the determinant can be calculated as:

∣∣∣∣∣∣∣a b c

d e f

g h i

∣∣∣∣∣∣∣• Multiply a by the determinant of the 2 × 2 matrix that is not in a’s row

or column.

• Likewise for b, and for c

• Add them up, making sure that b has a negative sign.

∣∣∣A∣∣∣= a

∣∣∣∣∣e f

h i

∣∣∣∣∣−b

∣∣∣∣∣d f

g i

∣∣∣∣∣+ c

∣∣∣∣∣d e

g h

∣∣∣∣∣Conditions for Zero Determinant:

Sometimes, just by inspection, you may be able to predict that the deter-

minant of a matrix is zero. e.g.

1. If two rows are identical

2. If two columns are identical

3. if two rows (or two columns) are multiples of corresponding elements.

4. if all the elements in any row (or any column) are zero.

In the above example, -6 and -9 are (-3) times their counterparts from the

first row.

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - AC B - 0 1

2. If A and B are square matrices of size n x n, then which of

the following statement is not true?

(A) det. (AB) = det (A) det (B)

(B) det (kA) = kn det (A)

(C) det (A + B) = det (A) + det (B)

(D) det (AT ) =1/det (A−1)

Solution:

FEsuccess.com

3.3 Matrix Operations:



to:

• Add, subtract and multiply two matrices.

• Find inverse of a matrix

• Find rank of a matrix and understand its importance.

Scalar multiplication of a Matrix:

If a matrix is multiplied by scalar, then ALL the elements get multiplied by

that scalar. e.g.

3×

3 6 1

4 5 2

2 0 −2

=

9 18 3

12 15 6

6 0 −6

Transpose of a Matrix:

If rows of a matrix are interchanged as column of the matrix, then the re-

sulting matrix is the transpose of the original one. e.g. the transpose of

: 3 6 1

4 5 2

2 0 −2

will be :

3 4 2

6 5 0

1 2 −2

Multiplication of Two Matrices:

A is m X n matrix(means having m rows and n columns), B is n X s matrix

(means having n rows and s columns), then they can be multiplied and the

result matrix C will be a m X s matrix.

In other words, columns of first matrix should match row of second ma-

trix. Also, the common dimension gets eliminated and the product matrix

does not have common dimension, neither as row nor as column. e.g. A

3 X 4 matrix ( 3 rows and 4 columns) can be multiplied with a 4X6 matrix,

because the ’inner’ dimensions (4 and 4) are matching, which gets ’elimi-

nated’ and the answer will be 3 X 6 matrix.

C = ci j =(

n∑l=1

ai l∗bl j

)FEsuccess.com

e.g., 3 4 2

6 5 0

1 2 −2

×

1 0 2

2 −2 0

3 2 −1

=

3+8+6 0−8+4 6+0−2

6+10+0 0−10+0 12+0+0

1+4−6 0−4−4 2+0+2

=

17 −4 4

16 −10 12

−1 −8 4

In matrix multiplication, the order in which the matrices are multiplied

is important. (In other words, matrix multiplication is NOT commutative.)

A×B 6= B × A

You may take the same matrices mentioned above and verify that the an-

swer is NOT the same.

Inverse of a Matrix :

For a square matrix A, the inverse is written A−1. When A is multiplied by

A−1 the result is the identity matrix I. Non-square matrices do not have

inverses.

Note: Not all square matrices have inverses. A square matrix which has

an inverse is called invertible or non singular, and a square matrix with-

out an inverse is called non-invertible or singular. A square matrix whose

determinant is zero, does not have inverse.

A A−1 = A−1 A = I

(AB)−1 = B−1 A−1

For a 2 X 2 matrix, the inverse can be easily found out by the following

formula:

if A = [a b

c d

]

Then A−1 = 1

det A

[d −b

−c a

]A−1 = 1

ad −bc

[d −b

−c a

]

These matrix operations (transpose, determinant, multiplication and

inverse) can be performed using a scientific calculator, so you must be

comfortable with the MATRIX or MAT mode of your calculator.

FEsuccess.com

Solved Examples

S O LV E D E X A M P L E - AC C - 0 1

3. Multiplication of matrices E and F is G. matrices E and G

are:

E =

cosθ −sinθ 0

sinθ cosθ 0

0 0 1

G =

1 0 0

0 1 0

0 0 1

What is the matrix F ?

(A)

cosθ −sinθ 0

sinθ cosθ 0

0 0 1

(B)

cosθ cosθ 0

−cosθ sinθ 0

0 0 1

(C)

cosθ sinθ 0

−sinθ cosθ 0

0 0 1

(D)

sinθ −cosθ 0

cosθ sinθ 0

0 0 1

Solution:

FEsuccess.com

FEsuccess.com

Rank of a Matrix

The rank of a matrix is the maximum number of independent rows (or, the

maximum number of independent columns). A square matrix A n × n is

non-singular only if its rank is equal to n.


4. Consider the system of simultaneous equations:

x +2y + z = 6

2x + y +2z = 6

x + y + z = 5

This system has:

(A) unique solution

(B) infinite number of solutions

(C) no solution

(D) exactly two solutions

Solution:

FEsuccess.com

FEsuccess.com


5. Which of the following is the inverse of the matrix?[−5 7

3 −4

]

(A)

[−4 −7

−3 −5

]

(B)

[4 7

3 5

]

(C)

[5 −7

−3 4

]

(D) The inverse does not exist.

Solution:

FEsuccess.com


6. Compute the product AB of the two matrices below.3 1

2 4

6 5

1 2

∗[−3 1

4 2

]

(A) −12 1

10 10

2 16

5 5

(B)

−5 5

10 10

2 16

5 5

(C)

−5 5

10 10

2 16

−5 5

(D)

−1 12

10 10

2 16

5 5

Solution:

FEsuccess.com

FEsuccess.com


7. If A =

[4 −1

5 −2

]and B =

[1 2

−2 1

]

(A)

(AB)T =[

6 7

9 8

]

(B)

(AB)T =[

6 9

7 8

]

(C)

(AB)T =[

7 6

8 9

]

(D)

(AB)T =[

9 8

6 7

]

Solution:

FEsuccess.com


8. If A =

[4 −1

5 −2

]and (AB)T =

[6 9

7 8

]then B = ??

(A) [1 2

−2 1

]

(B) [−1 2

−2 1

]

(C) [−1 2

−2 1

]

(D) [1 2

2 1

]

Solution:

FEsuccess.com


9. What is the value of matrix element c(2,3), given C = A +

B + A, where:

A =

0 2 −1

1 3 −1

2 2 1

B =

2 −1 −1

1 0 5

−2 1 1

(A) 9

(B) 6

(C) 3

(D) 15

Solution:

FEsuccess.com


10. The solution to the system of equations[2 5

−4 3

][x

y

][2

−30

]is:

(A) 6,2

(B) -6,2

(C) -6,-2

(D) 6,-2

Solution:

FEsuccess.com


11. If A and B be real symmetric matrices of size n x n, then:

(A) AAT = 1

(B) A = A−1

(C) AB = BA

(D) (AB)T = BA

Solution:

FEsuccess.com


12.

x +2y + z = 4

2x + y +2z = 5

x − y + z = 1

The system of algebraic equations given above has:

(A) a unique solution of x = 1, y = 1 and z = 1.

(B) only the two solutions of (x = 1, y = 1, z = 1) and (x = 2,

y = 1, z = 0)

(C) infinite number of solutions

(D) no feasible solution

Solution:

FEsuccess.com

References:

1. http://home.scarlet.be/math/matr.htm

2. http://linear.ups.edu/

version3/pdf/fcla-draft-solutions.pdf

FEsuccess.com

4 Vector Analysis

4.1 Unit Vector: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

4.2 Direction Ratios and Direction Cosines: . . . . . . . . . . . . . . . 160

4.3 Addition/Subtraction of Two Vectors: . . . . . . . . . . . . . . . . 162

4.4 Dot Product: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

4.5 Cross Product: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

4.6 The Del Operator: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

4.7 Gradient: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

4.8 Divergence: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

4.9 Curl: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

Introduction

A vector has both magnitude as well as direction in space. Force, velocity,

acceleration are some of the examples of vector quantities. Vector opera-

tions are used to find moments, to define electromagnetic forces, and to

study rotational effects on a body.

Figure 4.1: Length of a Vector

4.1 Unit Vector:



to:

• To be able to find length of a vector as well as unit

normal vector.

Length (Magnitude) of a Vector:

Imagine a box with length, width and height equal to the i,j,k components

of a vector. Then the length of the vector will be diagonal of this box, and

it can be calculated using Pythagorean theorem.

If ~a = a1 i +a2 j +a3k then length of the vector =

|~a| =√

a21 +a2

2 +a23

e.g. if a = 2i + 3j + 4k then,

|a| =√

22 +32 +42 =p29

Calculating Unit Vector:

Unit vector has a magnitude (length) = 1 and is given by:

a =± a

|a|e.g. if a = 2i + 3j + 4k then,

|a| =√

22 +32 +42 =p29

a =±2i +3 j +4kp29

The ± sign is because there can be two unit vectors opposite to each other.

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A D A - 0 1

1. Determine the magnitude of the force vector F = 20i +

60 j - 90k (N).

(A) 130 N

(B) 120 N

(C) 100 N

(D) 110 N

Solution:

FEsuccess.com

S O LV E D E X A M P L E - A D A - 0 2

2. For the spherical surface x2 + y2 + z2 = 1, the unit outward

normal vector at the point (1p2

,1p2

,0) is given by:

(A)1p2

i + 1p2

j

(B)1p2

i − 1p2

j

(C) k

(D)1p2

i + 1p2

j + 1p2

k

Solution:

FEsuccess.com

4.2 Direction Ratios and Direction Cosines:



to:

• Understand what is meant by the terms direction ratios

and direction cosines of a vector.

• Calculate these quantities given a vector in cartesian

form.

For any vector r = ai + b j + ck its direction ratios are a : b : c. Its direc-

tion cosines are

l = apa2 +b2 + c2

m = bpa2 +b2 + c2

n = cpa2 +b2 + c2

where,

l 2 +m2 +n2 = 1

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A D B - 0 1

3. If the direction ratios of a line are 1, 1, 2, find the direc-

tion cosines of the line.

(A)1

6,

1

6,

2

6

(B)2

6,

2

6,

1

6

(C)2p6

,2p6

,1p6

(D)1p6

,1p6

,2p6

Solution:

FEsuccess.com

4.3 Addition/Subtraction of Two Vectors:



to:

• Add and subtract vectors analytically as well as

graphically.

Add/subtract corresponding i, j and k components with similar com-

ponents of another vector.

e.g. if a = 2i + 3j + 4k and

b = 3i - 8j + k then,

a + b = (2+3)i + (3+(-8))j + (4+1)k = 5i -5j +5k

Similarly,

a - b = (2-3)i + (3-(-8))j + (4-1)k = -i +11j +3k

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A D C - 0 1

4. Given the vectors A = i - 2j + 4k and B = 3i + j - 2k, find R

= A + B .

(A) 4i - j + 2k

(B) 4i + j + 2k

(C) 4i + j - 2k

(D) -4i - j + 2k

Solution:

FEsuccess.com

4.4 Dot Product:



to:

• Evaluate dot product.

• Understand physical interpretation of dot product and its

applications.

• Understand properties and identities related to dot

product.

Calculation of Dot Product:

a.b = a1b1 +a2b2 +a3b3

Let us take the same example vectors again.

e.g. if a = 2i + 3j + 4k and

if b = 3i - 8j + k then,

a.b = 2×3+3× (−8)+4×1 =−14

Dot product can be positive or negative but since it is a scalar it will not

have i,j and k components.

Application of Dot Product:

Dot product is used to calculate scalar results such as:

• Work done by a force.

• Projection of a vector on another vector.

• Angle between two vectors.

Properties of Dot Product:

a.b = |a| ∣∣b∣∣cosθ

where |a| and∣∣b∣∣ represent the magnitudes of individual vectors and θ is

the angle between the two vectors.

The order of vectors while taking dot product is not important.

a.b = b.a

Now, consider the dot product of same two unit vectors,

i .i = j . j = k.k = 1

This is because the angle between two same unit vectors is zero, hence the

cosθ becomes one and also the magnitudes of unit vectors is one.

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Now, consider the dot product of dissimilar unit vectors.

i . j = j .k = k.i = 0

Again, here the magnitudes of unit vectors are one, but since the dissimi-

lar unit vectors i,j and k are always perpendicular to each other, the cosθ

becomes zero and hence, the dot products will become zero.

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Solved Examples:

S O LV E D E X A M P L E - A D D - 0 1

5. The angle between two unit-magnitude coplanar vectors

P(0.866, 0.500, 0) and Q(0.259, 0.966, 0) will be:

(A) 0◦

(B) 30◦

(C) 45◦

(D) 60◦

Solution:

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6. Given the 3-dimensional vectors: A = i(xy) + j(2yz) +

k(3zx) and B = i(yz) + j(2zx) + k(3xy). Determine the scalar

product at the point (1,2,3).

(A) 144

(B) 138

(C) 132

(D) 126

Solution:

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7. What is the angle between two vectors A and B if A = 4i -

12 j + 6k and B = 24i - 8 j + 6k?

(A) 168.45◦

(B) 51.22◦

(C) 86.32◦

(D) -84.64◦

Solution:

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8. Determine the dot product of the two vectors U = 8i - 6 j

+ 4k and V = 3i + 7 j + 9k.

(A) 18

(B) 16

(C) 14

(D) 12

Solution:

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9. Two perpendicular vectors are given in terms of their

components by U = Ux i - 4 j + 6k and V = 3i + 2 j - 3k.

Determine the component Ux .

(A) 5.67

(B) 6.67

(C) 7.67

(D) 8.67

Solution:

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10. A force−→F = −K (yi + x j ) (where K is a positive constant)

acts on a particle moving in the x-y plane. Starting from

the origin, the particle is taken along the positive x- axis

to the point (a, 0) and then parallel to the y-axis to the

point (a, a). The total work done by the forces−→F on the

particle is:

(A) −2 K a2

(B) 2 K a2

(C) − K a2

(D) K a2

Solution:

FEsuccess.com

4.5 Cross Product:



to:

• Evaluate cross product.

• Understand physical interpretation of cross product and

its applications.

• Understand properties and identities related to cross

product.

When two vectors are multiplied in such a way that their multiplication

is also a vector, then it is referred as cross product.

Cross products are mostly useful when we are studying phenomenons in-

volving rotational effects, such as moment of a force.

In order to evaluate cross product, create a 3X3 determinant with first row

as i,j,k and then write down i,j,k component values of individual vectors.

~a ×~b =

∣∣∣∣∣∣∣i j k

a1 a2 a3

b1 b2 b3

∣∣∣∣∣∣∣Let us take the same example vectors again,

e.g. if a = 2i + 3 j + 4k and

b = 3i - 8 j + k then,

~a ×~b =

∣∣∣∣∣∣∣i j k

2 3 4

3 −8 1

∣∣∣∣∣∣∣Then, ~a ×~b = 3-(-32)i - (2-12) j + (-16-9)k = 35i +10 j -25k

Properties of Cross Product:∣∣∣~a ×~b∣∣∣= |a| ∣∣b∣∣sinθ

The order of vectors while taking cross product is important.

a×b =−(b×a)

Now consider the cross product of two similar unit vectors,

FEsuccess.com

i× i = 0

j× j = 0

k×k = 0

This is because the angle between two similar unit vectors is zero, hence

sinθ becomes zero.

Alternatively, there is another way to look at this property. For cross

product of two same vectors, the two rows of the determinant (which was

mentioned above) will become zero. And from the properties of determi-

nants, if two rows (or even two columns) are identical, then the value of

that determinant becomes zero.

i× j = k

j×k = i

k× i = j

The results are negative if the order of vectors is reversed.

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Solved Examples:

S O LV E D E X A M P L E - A D E - 0 1

11. Cross product of two unit vectors has a magnitude = 1.

The angle between the vectors will be:

(A) 60◦

(B) 120◦

(C) 45◦

(D) 90◦

Solution:

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12. What is the cross product A x B of the vectors, A = i + 4 j

+ 6k and B = 2i + 3 j + 5k ?

(A) i - j -k

(B) i + j +k

(C) 2i + 7 j - 5k

(D) 2i + 7 j + 5k

Solution:

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13. Find the magnitude of the following vector:

A× B

where

A = (-2, -5, 2)

B = (-5, -2, -3)

(A) 32.53

(B) 21.56

(C) 26.55

(D) 24.03

Solution:

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14. The area of a triangle formed by the tips of vectors a, b

and c is:

(A) 12 (a − b).(a − c)

(B) 12 | (a − b)× (a − c) |

(C) 12 | (a × b × c) |

(D) 12 (a × b).c

Solution:

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Figure 4.2: The Del Operator

4.6 The Del Operator:



to:

• Define Del operator in terms of partial derivatives.

The Del (∇) operator is defined as:

∇=(∂

∂xi+ ∂

∂yj+ ∂

∂zk)

The Del operator itself is a vector. Depending upon whether it is operated

on scalar or a vector, the following three possibilities will arise:

1. Operated on a scalar - Gradient

2. Operated on a vector with a dot product - Divergence

3. Operated on a vector with a cross product - Curl

We will now see each of these cases individually in next sections.

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Solved Examples:

S O LV E D E X A M P L E - A D F - 0 1

15. The Del operator is called as:

(A) Gradient

(B) Curl

(C) Divergence

(D) Vector differential operator

Solution:

FEsuccess.com

4.7 Gradient:



to:

• Calculate gradient.

• Understand physical interpretation and applications of

gradient.

Gradient is a vector with the magnitude and direction of the maximum

change of the function in space. Gradient is calculated for scalars and the

result of calculation (the gradient itself) is a vector.

∇φ=(∂

∂xi+ ∂

∂yj+ ∂

∂zk)φ

Let us evaluate gradient of a scalar function φ described below.

φ= x3z

Then,∂φ

∂x= 3x2z

∂φ

∂y= 0

∂φ

∂z= x3

∇φ= (3x2zi+x3k

)If you want to calculate ∇φ at certain point, let’s say at (1,2,-1) then sub-

stitute these coordinates instead of x, y and z.

∇φ= (3(1)2(−1)i+ (1)3k

)=−3i+k

Physical Interpretation of Gradient:

The gradient represents the direction of greatest change. The gradient

points to the maximum of the function; follow the gradient, and you will

reach the local maximum. It is a vector, so it points towards the direction

of greatest change.

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Solved Examples:

S O LV E D E X A M P L E - A D G - 0 1

16. The directional derivative of the scalar function

f (x, y , z) = x2 +2y2 + z

at the point P = (1, 1, 2) in the direction of the vector a =

3i - 4j is:

(A) -4

(B) -2

(C) -1

(D) 1

Solution:

FEsuccess.com

S O LV E D E X A M P L E - A D G - 0 2

17. The vector which is normal to the surface

2xz2 −3x y −4x = 7

at point (1,-1,2) is:

(A) 2i - 3 j + 8k

(B) 2i + 3 j + 4k

(C) 7i - 3 j + 8k

(D) 7i - 5 j + 8k

Solution:

FEsuccess.com

4.8 Divergence:



to:

• Calculate divergence.


divergence.

Divergence is calculated for a vector, but since it is a dot product be-

tween the del (∇) operator (which is a vector) and another vector, the final

result is a scalar.

∇.V =(∂

∂xi+ ∂

∂yj+ ∂

∂zk)

.(v1i + v2 j + v3k)

Let us take an example of a variable vector which keeps on changing de-

pending upon the location (coordinates).

e.g. if

a = x3zi +x y2 j + y zk

∇.V =(∂

∂xi+ ∂

∂yj+ ∂

∂zk)

.(x3zi +x y2 j + y zk)

∇.V = (3x2z +2x y + z)

If you want to calculate at some location, let’s say, at (1,2-1) then simply

substitute the coordinates,

∇.V at (1,2−1) = (3(1)2(−1)+2(1)(2)+ (−1)) = 0

Physical Interpretation of the Divergence:

• Consider a vector field F that represents a fluid velocity: The divergence

of F at a point in a fluid is a measure of the rate at which the fluid is

flowing away from or towards that point.

• A positive divergence is indicating a flow away from the point.

• Physically divergence means that either the fluid is expanding or that

fluid is being supplied by a source external to the field.

• The lines of flow diverge from a source and converge to a sink.

• If there is no gain or loss of fluid anywhere then div F = 0. Such a vector

field is said to be solenoidal.

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Solved Example:

S O LV E D E X A M P L E - A D H - 0 1

18. The divergence of vector i = xi + yj + zk is:

(A) i + j + k

(B) 3

(C) 0

(D) 1

Solution:

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19. The divergence of the vector field

x2 yi+x yj+ z2k

at P (1, 1, 1) is:

(A) 5

(B) 1

(C) 4

(D) 2

Solution:

FEsuccess.com


20. Determine the divergence of the vector: V = (x2)i +(−x y) j + (x y z)k at the point (3,2,1).

(A) 9

(B) 11

(C) 13

(D) 7

Solution:

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21. The divergence of the vector field (x−y)i+(y−x)j+(x+y+z)k is:

(A) 0

(B) 1

(C) 2

(D) 3

Solution:

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22. The divergence of the vector field,

3xzi +2x y j − y z2k

at a point (1,1,1) is equal to:

(A) 7

(B) 4

(C) 3

(D) 0

Solution:

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23. Let f be a scalar field, and let F be a vector field. Which

of the following expressions is meaningful:

(A) ∇ f +∇. f

(B) ∇F +∇. f

(C) ∇ f +∇.F

(D) ∇ f +∇× F

Solution:

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24. The divergence of vector i = xi + yj + zk is:

(A) i + j + k

(B) 3

(C) 0

(D) 1

Solution:

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25. A vector is said to be solenoidal when its:

(A) Divergence is zero

(B) Divergence is unity

(C) Curl is zero

(D) Curl is unity

Solution:

FEsuccess.com

4.9 Curl:



to:

• Calculate curl.


curl.

∇×V =(∂

∂xi+ ∂

∂yj+ ∂

∂zk)× (v1i+ v2j+ v3k)

Let us take same example of a variable vector.

e.g. if

a = x3zi +x y2 j + y zk

∇×V =

∣∣∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂zx3z x y2 y z

∣∣∣∣∣∣∣∣∣∇×V = ((z −0)i − (0−x3) j + (y2 −0)k)

= (zi +x3 j + y2k)

If you want to calculate at some location, let’s say, at (1,2-1) then simply

substitute the coordinates,

∇×V at (1,2−1) = (zi +x3 j + y2k)

= (−1)(i + (1)3 j + (2)2k)

=−i + j +4k

Physical Interpretation of the Curl:

• Consider a vector field F that represents a fluid velocity, then the curl of

F at a point in a fluid is a measure of the rotation of the fluid.

• If there is no rotation of fluid anywhere then

∇×F = 0

Such a vector field is said to be irrotational or conservative.

• For a 2D flow with F represents the fluid velocity, ∇×F is perpendicular

to the motion and represents the direction of axis of rotation.

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Solved Examples

S O LV E D E X A M P L E - A D I - 0 1

26. The vector field is F = xi - yj (where i and j are unit vec-

tor) is:

(A) divergence free, but not irrotational

(B) irrotational, but not divergence free

(C) divergence free and irrotational

(D) neither divergence free nor irrational

Solution:

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S O LV E D E X A M P L E - A D I - 0 2

27. A vector is said to be irrotational when its:

(A) Divergence is zero

(B) Divergence is unity

(C) Curl is zero

(D) Curl is unity

Solution:

FEsuccess.com

References:

1. Prof. Rama Bhat’s Notes:

http://users.encs.concordia.ca/~rbhat/ENGR233/

2. http://betterexplained.com/articles/vector-calculus-understanding-the-

gradient/

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5 Differential Equations

5.1 General Terminology: . . . . . . . . . . . . . . . . . . . . . . . . . 197

5.2 First Order Differential Equations: . . . . . . . . . . . . . . . . . . 201

5.3 Higher Order Differential Equations: . . . . . . . . . . . . . . . . 213

5.4 Finding Differential Equation from Solution: . . . . . . . . . . . . 220

5.5 Applying Boundary Conditions: . . . . . . . . . . . . . . . . . . . 228

Introduction

A differential equation is a mathematical equation that relates some func-

tion of one or more variables with its derivatives. Many laws governing

natural phenomena are relations (equations) involving rates at which things

happen (derivatives). Equations containing derivatives are differential equa-

tions. Newton’s 2nd law can be expressed as a second order differential

equation. Diffusion and wave equations are also differential equations.

5.1 General Terminology:



to:

• Identify degree and order of a differential equation.

Differential Equations are equations that involve dependent variables

and their derivatives with respect to the independent variables.

Ordinary differential equations involve only one independent variable.,

whereas partial differential equations involve two or more independent

variables.

The order of a differential equation is the order of the highest order deriva-

tive present in the equation.

The degree of a differential equation is the power of the highest order

derivative in the equation.

e.g. (d 2 y

d x2

)3

+ d y

d x= si nx

has order = 2 and degree = 3

bnd n y(x)

d xn + ....+b1d y(x)

d x+b0 y(x) = f (x)

where bn , ....b1,b0 are constants are referred as ordinary linear differen-

tial equation.

It’s characteristics polynomial is:

bnr n +bn−1r n−1 + ...+b1r +b0 = 0

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Solved Examples:

S O LV E D E X A M P L E - A E A - 0 1

1. Determine the order and degree of the differential equa-

tion

2xd 4 y

d x4 +5x2(

d y

d x

)3

−x y = 0

(A) Fourth order, first degree

(B) Third order, first degree

(C) First order, fourth degree

(D) First order, third degree

Solution:

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2. The partial differential equation:

∂2φ

∂x2 + ∂2φ

∂y2 = ∂2ψ

∂x2 + ∂2ψ

∂y2 = 1

has:

(A) degree 1 order 2

(B) degree 1 order 1

(C) degree 2 order 1

(D) degree 2 order 2

Solution:

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3. The Blasius equation,

d 3 f

dη3 + f

2

d 2 f

dη2 = 0

is a

(A) second order nonlinear ordinary differential equation

(B) third order nonlinear ordinary differential equation

(C) third order linear ordinary differential equation

(D) mixed order nonlinear ordinary differential equation

Solution:

FEsuccess.com

5.2 First Order Differential Equations:



to:

• Use analytical methods to solve first order differential

equation by direct integration and separation of variables.

Variable Separable Differential Equations:

Bring all y terms on one sides with dy and keep all x terms with dx on the

other side.

Integrate to get the final solution.

f1(x)g1(y)d x + f2(x)g2(y)d y = 0

f1(x)

f2(x)d x + g2(y)

g1(y)d y = 0

∫f1(x)

f2(x)d x +

∫g2(y)

g1(y)d y = c

Let’s see an example of this:

Solve the following first order differential equation using variable sepa-

rable method:d y

d x= 3x2 y

First, bring all similar terms on one side.

d y

y= 3x2d x

Now integrate both sides with respect their respective variables.∫d y

y=

∫3x2d x +C

where C is the constant of integration.

ln y = x3 +C

or in another format,

y = e(x3+C )

Now the exponential can be split as multiplication of two exponents. How-

ever, eC will also be a constant, let’s say c.

y = ex3.eC = ex3

.c

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y = c.ex3

If boundary conditions are given, those can be substituted to eliminate c.

Let’s say if it is given then y(0) = -1, which means at x = 0, y = -1, then

substituting,

(−1) = ce03

which means, c = -1, So the final solution will be,

y =−ex3

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Solved Examples:

S O LV E D E X A M P L E - A E B - 0 1

4. 9 grams of bacteria grows at the rate of1

5gram per day

per gram. At this rate how long will it take to reach 17

gram?

(A) 40 days

(B) 4.44 days

(C) 3.83 days

(D) 3.18 days

Solution:

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5. The population of a country doubles in 50 years. How

many years will it be five times as much? Assume that the

rate of increase is proportional to the number of inhabi-

tants.

(A) 100 years

(B) 116 years

(C) 120 years

(D) 98 years

Solution:

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6. What is the solution of the first order differential equation

y(k+1) = y(k) + 5.

(A) y(k) = 4 - 5/k

(B) y(k) = 20 + 5k

(C) y(k) = C - k, where C is constant

(D) The solution is non-existent for real values of y

Solution:

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7. Solve xy’ (2y - 1) = y (1 - x)

(A) ln (xy) = 2 (x - y) + C

(B) ln (xy) = x - 2y + C

(C) ln (xy) = 2y - x + C

(D) ln (xy) = (x + 2y) + C

Solution:

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8. Find the general solution of y’ = y sec x.

(A) y = C (sec x + tan x)

(B) y = C (sec x - tan x)

(C) y = C sec x tan x

(D) y = C (sec2x tan x)

Solution:

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9. Which of the following equations is a variable separable

DE?

(A) (x + x2 y) dy = (2x + x y2) dx

(B) (x + y) dx - 2y dy = 0

(C) 2y dx = (x2 + 1) dy

(D) y2 dx + (2x - 3y) dy = 0

Solution:

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10. Radium decomposes at a rate proportional to the amount

present. If half of the original amount disappears after

1000 years, what is the percentage lost in 100 years?

(A) 6.70%

(B) 4.50%

(C) 5.36%

(D) 4.30%

Solution:

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11. According to Newton’s law of cooling, the rate at which a

substance cools in air is directly proportional to the dif-

ference between the temperature of the substance and

that of air. If the temperature of the air is 30◦ and the

substance cools from 100◦ to 70◦ in 15 minutes, how long

will it take to cool 100◦ to 50◦?

(A) 33.85 min.

(B) 43.50 min.

(C) 35.39 min.

(D) 45.30 min.

Solution:

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12. The solution of the differential equation

d y

d x+ y2 = 0

is:

(A) y =1

x + c

(B) y =−x3

3x + c

(C) cex

(D) unsolvable as equation is nonlinear

Solution:

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13. Consider the differential equationd y

d x= (

1+ y2)

x The

general solution with constant c is:

(A) y = tanx2

2+ tanc

(B) y = tan2( x

2+ c

)(C) y = tan2

( x

2

)+ c

(D) y = tan

(x2

2

)+ c

Solution:

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5.3 Higher Order Differential Equations:



to:

• Find the complete solution of a nonhomogeneous

differential equation as a linear combination of the

complementary function and a particular solution.

a0d 2 y

d x2 +a1d y

d x+a2 y = f (x)

where a0, a1 and a2 are constants.

The general solution of such a differential equation contains two parts:

the Complementary Function and the Particular Integral.

Complementary Function (CF):

Neglect the right hand side function f(x) and write the auxiliary equation

in the form:

a0D2 +a1D +a2 = 0

where D represents derivative operatord

d x.

Solving the auxiliary equation, a quadratic, will yield two roots, say m1

and m2.

The Complementary Function (CF) is then written based on the roots.

• If m1 and m2 are real and different, then the CF is

y = Aem1x +Bem2x

• If m1 and m2 are real and equal, ie m1 = m2 = m, then the CF is

y = (A+B x)emx

• If m1 and m2 are complex roots of the form α+ iβ and α− iβ, then the

CF is

y = eαx (A cosβx +B sinβx)

where A and B are arbitrary constants.

Let us take one example to illustrate complementary function.

Find the complementary function for the following second order differen-

tial equation.d 2 y

d x2 − d y

d x−2y = e5x

Here, to find the complementary function, you have to neglect the RHS

after equal to sign.d 2 y

d x2 − d y

d x−2y = 0

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Or using the simplified notation, D =d

d x

D2 −D −2 = 0

This is a quadratic equation with roots, D1 = 2 and D2 = -1

The roots are NOT repeated, so this is a simple case. the CF will be

y =C1e2x +C2e−x

Particular Integral (PI):

Consider RHS now, which is f(x).

PI = 1

Auxiliary equationf (x)

= 1

f (D)f (x)

for emx , following situations will arise

• If m 6= 0, then1

f (D)emx = emx

f (m)

• If m = 0 and f’(m) 6= 0 then1

f (D)emx = xemx

f ′(m)

• If m = 0 and f’(m) = 0 and f”(m) 6= 0 then1

f (D)emx = x2emx

f ′′(m)

Now, consider the same example,

Find the Particular Intregal, for the above example.

d 2 y

d x2 − d y

d x−2y = e5x

PI = 1

f (D)e5x

= 1

D2 −D −2e5x

= 1

(D −2)(D +1)e5x

Now, we have to substitute the coefficient of power of e in place of D.

PI = 1

(5−2)(5+1)e5x

= 1

18e5x

Total Solution

The total solution is = CF + PI So for the above example, total solution is

y =C1e2x +C2e−x + 1

18e5x

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Solved Examples:

S O LV E D E X A M P L E - A E C - 0 1

14. If roots of the auxiliary equation are1

2± i

p3

2what is the

solution of the differential equation?

(A) y = e

p3

2x(

A cos1

2x +B sin

1

2x

)

(B) y = e

1

2x(

A cos

p3

2x +B sin

p3

2x

)

(C) y = e

p3

2x(

A cos1

2x +B sin

1

2x

)

(D) y = e

p3

2x(

A cos1

2x +B sin

1

2x

)

Solution:

FEsuccess.com


15. The Complementary Function (C.F.) for the differential

equation:

(D4 +4)y = 0

where, D =d

d xi s :

(A) e−x (C1 cos x +C2 sin x)

(B) ex (C1 cos x +C2 sin x)

(C) e−x (C1 cos x +C2 sin x)+ex (C3 cos x +C4 sin x)

(D) e−x2(C1 cos x +C2 sin x)

Solution:

FEsuccess.com


16. The particular solution for the differential equation:

d 2 y

d x2 +3d y

d x+2y = 5cos x

(A)1

2cos x + 3

2sin x

(B)3

2cos x + 1

2sin x

(C)3

2sin x

(D)1

2cos x

Solution:

FEsuccess.com


17. Which of the following is a solution of the differential

equation:d 2 y

d x2 +pd y

d x+ (q +1)y = 0

(A) e−3x

(B) xe−x

(C) xe−2x

(D) x2e−2x

Solution:

FEsuccess.com


18. Ford 2 y

d x2 +4d y

d x+3y = 3e2x

the particular integral is:

(A)1

15e2x

(B)1

5e2x

(C) 3e2x

(D) C1e−x +C2e−3x

Solution:

FEsuccess.com

5.4 Finding Differential Equation from Solution:



to:

• Finding the differential equation, given a solution by

multiple differentiation and eliminating constants.

An expression with n arbitrary constants will yield a differential equa-

tion of order n. So to get the nth order derivative you’ll have to differentiate

the expression n times, and in that process obtain n more relations to have

a total of n+1 relations from which you can eliminate the n arbitrary con-

stants to obtain the differential equation.

The constants will be eliminated by successive differentiation. For exam-

ple, consider

y = ax2 +bx + c y

There are 3 arbitrary constants a, b and c so just differentiate 3 times to

obtain the DE y”’=0

Now consider

y2 = 4ax

Since there is only one constant a, differentiate once to get 2yy’=4a. Now

eliminate 4a to obtain the DE 2xy’=y

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A E D - 0 1

19. Find the differential equation whose general solution is y

= C1x + C2ex .

(A) (x - 1) y" - xy’ + y = 0

(B) (x + 1) y" - xy’ + y = 0

(C) (x - 1) y" + xy’ + y = 0

(D) (x + 1) y" + xy’ + y = 0

Solution:

FEsuccess.com


20. Find the equation of the curve at every point of which the

tangent line has a slope of 2x.

(A) x = -y2 + c

(B) y = -x2 + c

(C) y = x2 + c

(D) x =y2 + c

Solution:

FEsuccess.com


21. Find the differential equation of family of straight lines

with slope and y-intercept equal.

(A) xydy = x3/4

(B) ydx=(x+1)dy

(C) x2y = x(x+1)dx

(D) y = x3/4

Solution:

FEsuccess.com


22. Find the differential equations of the family of lines pass-

ing through the origin.

(A) ydy - xdx = 0

(B) xdy - ydx = 0

(C) xdx + ydy = 0

(D) ydx + xdy = 0

Solution:

FEsuccess.com


23. What is the differential equation of the family of parabo-

las having their vertices at the origin and their foci on the

x-axis.

(A) 2xdy - ydx = 0

(B) xdy + ydx = 0

(C) 2ydx - xdy = 0

(D) dy/dx - x = 0

Solution:

FEsuccess.com


24. Determine the differential equation of the family of lines

passing through (h, k).

(A) (y - k)dx - (x - h)dy = 0

(B) (y - h) + (y - k) = dy/dx

(C) (x - h)dx - (y - k)dy = 0

(D) (x + h)dx - (y - k)dy = 0

Solution:

FEsuccess.com


25. Determine the differential equation of the family of circles

with center on the origin.

(A) (y ′′)3 - xy + y = 0

(B) y” - xyy = 0

(C) x + yy’= 0

(D) (y ′)3 + (y ′′)2 + xy = 0

Solution:

FEsuccess.com

5.5 Applying Boundary Conditions:



to:

• Solve application problems requiring the use of

higher-order differential equations with boundary

conditions.

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A E E - 0 1

26. Solve the differential equation: x (y - 1) dx + (x + 1) dy =

0. If y = 2 when x = 1, determine y when x = 2.

(A) 1.80

(B) 1.48

(C) 1.55

(D) 1.63

Solution:

FEsuccess.com


27. If dy = x2 dx; what is the equation of y in terms of x if the

curve passes through (1,1)?

(A) x2 - 3y + 3 = 0

(B) x3 - 3y + 2 = 0

(C) x3 + 3y2 + 2 = 0

(D) 2y + x3 + 2 =0

Solution:

FEsuccess.com


28. Find the solution ofd 2 y

d x2 = y which passes through the

origin and the point (ln 2,3

4)

(A)

y = 1

2ex −e−x

(B)

y = 1

2(ex +e−x )

(C)

y = 1

2

(ex −e−x)

(D)

y = 1

2ex +e−x

Solution:

FEsuccess.com


29. The equation y2 = cx is the general solution of:

(A) y = 2y/x

(B) y = 2x/y

(C) y = y/(2x)

(D) y = x/(2y)

Solution:

FEsuccess.com


30. Solve the differential equation dy - xdx = 0, if the curve

passes through (1,0)?

(A) 3x2 + 2y - 3 = 0

(B) 2y + x2- 1 = 0

(C) x2 - 2y - 1 = 0

(D) 2x2 + 2y - 2 = 0

Solution:

FEsuccess.com


31. y = f(x) is the solution ofd 2 y

d x2 = 0. Boundary conditions

are y =5,d y

d x= 2 at x = 10, then the value of f(15) is:

(A) 10

(B) 12

(C) 15

(D) 18

Solution:

FEsuccess.com


32. It is given that

y ′′+2y ′+ y = 0,

y(0) = 0,

y(1) = 0

What is y(0.5)?

(A) 0

(B) 0.37

(C) 0.62

(D) 1.13

Solution:

FEsuccess.com


33. Ford 2 y

d x2 +4d y

d x+3y = 3e2x

the particular integral is:

(A)1

15e2x

(B)1

5e2x

(C) 3e2x

(D) C1e−x +C2e−3x

Solution:

FEsuccess.com


34. The complete solution of the ordinary differential equa-

tion:d 2 y

d x2 +pd y

d x+q y = 0

is y = c1e−x + c2e−3x Then p and q are:

(A) p = 3, q = 3

(B) p = 3, q = 4

(C) p = 4, q = 3

(D) p = 4, q = 4

Solution:

FEsuccess.com


35. Which of the following is a solution of the differential

equationd 2 y

d x2 +pd y

d x+ (q +1)y = 0

if, p= 4, q= 3?

(A) e−3x

(B) xe−x

(C) xe−2x

(D) x2e−2x

Solution:

FEsuccess.com


36. Given that x+ 3x = 0, and x(0) = 1, x(0) = 0, what is x(1) ?

(A) -0.99

(B) -0.16

(C) 0.16

(D) 0.99

Solution:

FEsuccess.com

References:

1. Paul’s Online Math Notes:

http://tutorial.math.lamar.edu

FEsuccess.com

6 Numerical Methods

6.1 Newton’s Method of Root Extraction: . . . . . . . . . . . . . . . . 242

6.2 Newton’s Method of Minimization: . . . . . . . . . . . . . . . . . . 247

6.3 Forward Rectangular Rule: . . . . . . . . . . . . . . . . . . . . . . 250

6.4 Trapezoidal Rule: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252

6.5 Simpson’s Rule: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

Introduction

Numerical methods are iterative techniques used to find out approximate

solution of differential equations, roots of an equation or value of an inte-

gral.

These are repetitive processes to find approximate solutions for problems

which are difficult to solve using traditional methods.

Even though numerical methods are sometimes time consuming, they are

more efficient to code, and solutions can be obtained with the help of a

computer.

Numerical methods can give reasonably accurate answers by same method

for a wide range of problems that are otherwise, by analytical methods, too

difficult to generalize.

Answers given by numerical methods depend upon initial guess value.

Accuracy of a numerical method can be improved by reducing the interval

spacing (thereby increasing the no. of iterations) and considering more

number of significant digits.

6.1 Newton’s Method of Root Extraction:



to:

• Understand how the Newton-Raphson method works,

• Be able to apply the Newton-Raphson method to certain

problems.

a j+1 = a j − f (x)f ′(x)

∣∣∣x=a j

(6.1)

Here the steps are as follows:

1. Begin with some trial value close to the actual answer.

2. Substitute this value in the above formula to get updated value of the

iteration.

3. Repeat this till the value of x stabilizes.

4. In rare circumstances, the value of x will not stabilize. It may diverge

away from the actual answer or simply keep on toggling between cur-

rent value and previous value. In that case, select different trial value or

totally reject this method and try solving problem with other method.

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A FA - 0 1

1. Let f(x) be an equation such that f(a) >0, f(b) >0 for two

real numbers a and b. Then:

(A) at least one root of f(x) = 0 lies in (a, b)

(B) no root lies in (a, b)

(C) either no root or an even number of roots lie in (a, b)

(D) None of these

Solution:

FEsuccess.com


2. The real root of the equation 5x - 2cos x - 1 = 0 (upto two

decimal accuracy) is:

(A) 0.51

(B) 0.53

(C) 0.54

(D) 0.55

Solution:

FEsuccess.com


3. Find the root of the equation f(x) = 10 cos(x) using the

Newton-Raphson method. The initial guess is x =π

4

(A) 1.51

(B) 1.53

(C) 1.57

(D) 1.59

Solution:

FEsuccess.com


4. Starting from x0 = 1, one step of Newton-Raphson

method in solving the equation x3 + 3x − 7 = 0 gives the

next value (x1) as:

(A) x1 = 0.5

(B) x1 = 1.406

(C) x1 = 1.5

(D) x1 = 2

Solution:

FEsuccess.com

6.2 Newton’s Method of Minimization:



to:

• Use Newton’s method of minimization to minimize a

twice differentiable function f.

In one dimension:

xn+1 = xn − f ′(xn)

f ′′(xn)

In multi-dimension:

xk+1 = xk −[∂2h

∂x2

∣∣∣x=xk

]−1∂h

∂x

∣∣∣x=xk

(6.2)

where,

∂h

∂x=

∂h

∂x1∂h

∂x2...

...∂h

∂xn

and

∂2h

∂x21

∂2h

∂x1∂x2... ...

∂2h

∂x1∂xn

∂2h

∂x1∂x2

∂2h

∂x22

... ...∂2h

∂x1∂xn

... ... ... ... ...

... ... ... ... ...∂2h

∂x1∂xn

∂2h

∂x2∂xn

∂2h

∂x2n

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A F B - 0 1

5. According to the Newton’s method of minimization, at

what value of x, 0.14 < x < 0.15 the function f (x) =7x − ln(x) has a minimum value?

(A) 0.142857

(B) 0.145728

(C) 0.146673

(D) 0.149157

Solution:

FEsuccess.com

S O LV E D E X A M P L E - A F B - 0 2

6. Find the minimum value of the function

f (x, y) = x2 + y2 +2x +4

using Newton’s method of minimization. Use (2,1) as an

initial guess.

(A) 0

(B) -27

(C) -19

(D) 3

Solution:

FEsuccess.com

Figure 6.1: Forward Rectangular Rule

6.3 Forward Rectangular Rule:



to:

• Evaluate definite integral using Forward rectangular rule.

In forward rectangular rule, the area under integration is divided into

rectangles, where each rectangle has width = ∆x and height = value of f(x)

at that point. Area of each rectangle = length × width. We have to sum all

such sub-areas (n-1) times to get the following formula.∫ b

af (x)d x ≈∆x

n−1∑k=0

f (a +k∆x)

The error in rectangular method is more as compared to other methods,

as we ’conveniently’ ignoring some area in each calculation.

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A F C - 0 1

7. Using forward rectangular rule, calculate the value of

ln(1.2) using,

ln(1.2) =∫ 1.2

1

1

xd x

Take step value ∆ x = 0.1

(A) 0.20202

(B) 0.190909

(C) 0.212121

(D) 0.20505

Solution:

FEsuccess.com

a = x0 x1

y = f (x)

x

y

Figure 6.2: Area under a curve is dividedinto several trapezoids

6.4 Trapezoidal Rule:



to:

• Use the multiple-segment trapezoidal rule of integration

to solve problems.

∫ b

af (x)d x ≈ 4x

2

(y0 +2y1 +2y2 +2y3 + .....+2yn−1 + yn

)(6.3)

for n= 1 the above equation modifies to,

∫ b

af (x)d x ≈∆x

∣∣∣∣ f (a)+ f (b)

2

∣∣∣∣ (6.4)

In trapezoidal rule, the steps are as follows:

1. the area under the curve is divided into n intervals.

2. The width of each interval is ∆x = b −a

nwhere a and b are limits of

integration and n is no. of intervals.

3. The area of each trapezoid is =1

2×∆ x × sum of parallel sides =

1

2×∆ x

× (f (xi−1)+ f (xi )

)4. If you add all such trapezoids, all vertical sides, except first and last one

are added twice. The first and last vertical lines are added only once.

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A F D - 0 1

8. In order to estimate∫ 6

22t d t using the Trapezoidal rule

with six subintervals. Then the width of each subinterval

is:

(A)2

3

(B) 1

(C) -3

4

(D)1

6

Solution:

FEsuccess.com


9. Match the CORRECT pairs.

Integration Scheme Order of Polynomial

P. Simpson’s 38 Rule 1. First

Q. Trapezoidal Rule 2. Second

R. Simpson’s 13 Rule 3. Third

(A) P-3, Q-1, R-2

(B) P-2, Q-1, R-3

(C) P-1, Q-2, R-3

(D) P-3, Q-2, R-1

Solution:

FEsuccess.com


10. Evaluate∫ 6

0

1

1+xd x using trapezoidal method, using 6

intervals.

(A) 1.9661

(B) 1.9587

(C) 2.0214

(D) 1.9459

Solution:

FEsuccess.com

a a+b2

b

y = f (x)

y = p2(x)

x

y

Figure 6.3: Simpson’s Rule,Source: tex.stackexchange.com,answered by user percusse

6.5 Simpson’s Rule:



to:

• Given an even integer n, compute an approximate value

of an integral using Simpson’s rule.

Simpson’s rule is a numerical method that approximates the value of a

definite integral by using quadratic polynomials.∫ b

af (x)d x

≈ 4x

3

(y0 +4y1 +2y2 +4y3 +2y4 + .....+4yn−1 + yn

)

FEsuccess.com

Solved Examples:

S O LV E D E X A M P L E - A F E - 0 1

11. Evaluate∫ 6

0

1

1+xd x using Simpson’s

1

3

r d

rule, using 6 in-

tervals.

(A) 1.9661

(B) 1.9587

(C) 2.0214

(D) 1.9459

Solution:

FEsuccess.com

7 Answer Keys- Mathematics

Analytic Geometry

Straight

Lines:

AAA-01: D

AAA-02: C

AAA-03: D

AAA-04: A

AAA-05: A

AAA-06: D

AAA-07: D

AAA-08: B

AAA-09: B

AAA-10: D

AAA-11: B

AAA-12: A

AAA-13: C

AAA-14: B

AAA-15: A

AAA-16: A

AAA-17: D

AAA-18: D

Conics:

AAB-01: A

AAB-02: A

AAB-03: C

AAB-04: B

Circle:

AAC-01: B

AAC-02: B

AAC-03: C

AAC-04: D

AAC-05: C

AAC-06: C

AAC-07: A

AAC-08: A

AAC-09: A

AAC-10: D

AAC-11: C

AAC-12: B

AAC-13: C

AAC-14: D

AAC-15: B

Ellipse:

AAD-01: A

AAD-02: B

AAD-03: B

AAD-04: D

AAD-05: A

AAD-06: C

AAD-07: A

AAD-08: A

AAD-09: B

Parabola:

AAE-01: A

AAE-02: A

AAE-03: B

AAE-04: D

AAE-05: D

AAE-06: D

AAE-07: C

AAE-08: C

AAE-09: A

AAE-10: B

Hyperbola:

AAH-01: A

AAH-02: A

AAH-03: D

AAH-04: B

AAH-05: A

AAH-06: C

Distance

Formula:

AAG-01: C

Calculus

Limits:

ABA-01: D

ABA-02: C

ABA-03: C

ABA-04: B

ABA-05: D

ABA-06: A

ABA-07: D

ABA-08: D

ABA-09: B

ABA-10: B

ABA-11: B

ABA-12: C

ABA-13: B

Derivatives:

ABB-01: B

ABB-02: C

ABB-03: A

ABB-04: A

ABB-05: D

ABB-06: D

Partial

Derivatives:

ABC-01: C

ABC-02: B

ABC-03: B

ABC-04: B

Indefinite

Integrals

ABD-01: C

ABD-02: A

Integration

by Partial Fractions:

ABE-01: A

Definite

Integrals:

ABG-01: A

ABG-02: D

ABG-03: D

ABG-04: A

Area

Bounded by

a Curve:

ABH-01: D

ABH-02: C

ABH-03: B

ABH-04: A

ABH-05: A

Volume

of Revolution

of Solids:

ABI-01: C

ABI-02: A

ABI-03: A

ABI-04: D

ABI-05: C

FEsuccess.com

Linear Algebra

Types of Matrices:

ACA-01: A

Determinant of a Matrix:

ACB-01: C

Matrix Operations:

ACC-01: C

ACC-02: C

ACC-03: B

ACC-04: B

ACC-05: B

ACC-06: A

ACC-07: C

ACC-08: D

ACC-09: D

ACC-10: C

FEsuccess.com

Vector Analysis

Unit Vector:

ADA-01: D

ADA-02: A

Direction Ratios

and

Direction Cosines:

ADB-01: D

Addition

Subtraction

of Vectors:

ADC-01: A

Dot

Product

ADD-01: C

ADD-02: B

ADD-03: B

ADD-04: A

ADD-05: D

ADD-06: C

Cross

Product:

ADE-01: D

ADE-02: C

ADE-03: A

ADE-04: B

The Del

Operator:

ADF-01: D

Gradient:

ADG-01: B

ADG-02: C

Divergence:

ADH-01: B

ADH-02: A

ADH-03: A

ADH-04: D

ADH-05: C

ADH-06: D

ADH-07: B

ADH-08: A

Curl:

ADI-01: C

ADI-02: C

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Differential Equations

General

Terminology

AEA-01: A

AEA-02: A

AEA-03: B

First Order

Differential

Equations

AEB-01: D

AEB-02: B

AEB-03: B

AEB-04: D

AEB-05: A

AEB-06: C

AEB-07: A

AEB-08: A

AEB-09: A

AEB-10: D

Higher Order

Differential

Equations

AEC-01: B

AEC-02: C

AEC-03: A

AEC-04: C

AEC-05: B

Finding

Differential Equation

from Solution

AED-01: A

AED-02: C

AED-03: B

AED-04: B

AED-05: A

AED-06: A

AED-07: C

Applying

Boundary

Conditions

AEE-01: C

AEE-02: B

AEE-03: C

AEE-04: C

AEE-05: C

AEE-06: C

AEE-07: A

AEE-08: B

AEE-09: C

AEE-10: C

AEE-11: D

FEsuccess.com

Numerical Methods

Newton′s Method

of Root Extraction

AFA-01: C

AFA-02: C

AFA-03: C

AFA-04: C

Newton′s Method

of Minimization

AFB-01: A

Forward

Rectangular

Rule

AFC-01: B

Trapezoidal

Rule

AFD-01: A

AFD-02: A

AFD-03: C

Simpson′s Rule

AFE-01: B

FEsuccess.com

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