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Std. XII Commerce

Mathematics and Statistics – I

Written according to the New Text book (2013-2014) published by the Maharashtra State

Board of Secondary and Higher Secondary Education, Pune.

Third Edition: April 2016

Salient Features :

• Precise Theory for every Topic.

• Exhaustive coverage of entire syllabus.

• Topic-wise distribution of all textual questions and practice problems at thebeginning of every chapter.

• Relevant and important formulae wherever required.

• Covers answers to all Textual Questions.

• Practice problems based on Textual Exercises and Board Questions(March 08 March 16) included for better preparation and self evaluation.

• Multiple Choice Questions at the end of every chapter.

• Two Model Question papers based on the latest paper pattern.

• Includes Board Question Papers of March and October 2014, 2015 andMarch 2016.

No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

Printed at: Repro India Ltd., Mumbai

P.O. No. 16590

10124_10570_JUP

Preface Mathematics is not just a subject that is restricted to the four walls of a classroom. Its philosophy and applications are to be looked for in the daily course of our life. The knowledge of mathematics is essential for us, to explore and practice in a variety of fields like business administration, banking, stock exchange and in science and engineering. With the same thought in mind, we present to you "Std. XII Commerce: Mathematics and Statistics-I" a complete and thorough book with a revolutionary fresh approach towards content and thus laying a platform for an in depth understanding of the subject. This book has been written according to the revised syllabus and includes two model question papers based on the latest paper pattern. At the beginning of every chapter, topic-wise distribution of all textual questions including practice problems have been provided for simpler understanding of various types of questions. Every topic included in the book is divided into sub-topics, each of which are precisely explained with the associated theories. We have provided answer keys for all the textual questions and miscellaneous exercises. In addition to this, we have included practice problems based upon solved exercises which not only aid students in self evaluation but also provide them with plenty of practice. We've also ensured that each chapter ends with a set of Multiple Choice Questions so as to prepare students for competitive examinations. We are sure this study material will turn out to be a powerful resource for students and facilitate them in understanding the concepts of Mathematics in the most simple way. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you. Please write to us on: [email protected]

Best of luck to all the aspirants! Yours faithfully

Publisher

BOARD PAPER PATTERN Time: 3 Hours Total Marks: 80 1. One theory question paper of 80 marks and duration for this paper will be 3 hours.

2. For Mathematics and Statistics, (Commerce) there will be only one question paper and two answer papers. Question paper will contain two sections viz. Section I and Section II. Students should solve each section on separate answer books.

Section – I Q.1. This Question will have 8 sub-questions, each carring two marks. [12 Marks] Students will have to attempt any 6 out of the given 8 sub-questions. Q.2. This Question carries 14 marks and consists of two sub parts (A) and (B) as follows: [14 Marks] (A) It contains 3 sub-questions of 3 marks each. Students will have to attempt any 2 out of the given 3 sub-questions. (B) It contains 3 sub-questions of 4 marks each. Students will have to attempt any 2 out of the given 3 sub-questions. Q.3. This Question carries 14 marks and consists of two sub parts (A) and (B) as follows: [14 Marks] (A) It contains 3 sub-questions of 3 marks each. Students will have to attempt any 2 out of the given 3 sub-questions (B) It contains 3 sub-questions of 4 marks each. Students will have to attempt any 2 out of the given 3 sub-questions.

Section – II Q.4. This Question will have 8 sub-questions, each carring two marks. [12 Marks] Students will have to attempt any 6 out of the given 8 sub-questions. Q.5. This Question carries 14 marks and consists of two sub parts (A) and (B) as follows: [14 Marks] (A) It contains 3 sub-questions of 3 marks each. Students will have to attempt any 2 out of the given 3 sub-questions. (B) It contains 3 sub-questions of 4 marks each. Students will have to attempt any 2 out of the given 3 sub-questions. Q.6. This Question carries 14 marks and consists of two sub parts (A) and (B) as follows: [14 Marks] (A) It contains 3 sub-questions of 3 marks each. Students will have to attempt any 2 out of the given 3 sub-questions (B) It contains 3 sub-questions of 4 marks each. Students will have to attempt any 2 out of the given 3 sub-questions.

Evaluation Scheme for Practical i. Duration for practical examination for each batch will be one hour. ii. Total marks : 20

MARKWISE DISTRIBUTION

Unitwise Distribution of Marks Section - I

Sr.No. Units Marks with Option 1 2 3 4 5 6 7

Mathematical Logic Matrices Continuity Differentiation Application of Derivative Integration Definite Integrals

08 08 08 08 10 08 08

Total 58

Unitwise Distribution of Marks

Section - II Sr. No. Units Marks with Option

1.

Commercial Arithmetic: Ratio, Proportion, Partnership Commission, Brokerage, Discount Insurance, Annuity

13

2. Demography 08 3. Bivariate Data Correlation 08 4. Regression Analysis 07 5. Random Variable and Probability Distribution 08

6. Management Mathematics 14

Total 58

Weightage of Objectives Sr. No. Objectives Marks Marks with Option Percentage

1 2 3 4

Knowledge Understanding Application Skill

08 22 32 18

13 32 45 26

10.00 27.50 40.00 22.50

Total 80 116 100.00

Weightage of Types of Questions Sr. No. Types of Questions Marks Marks with Option Percentage

1 2 3

Objective Type Short Answer Long Answer

24 24 32

32 36 48

30 30 40

Total 80 116 100.00

No. Topic Name Page No.

1 Mathematical Logic 1 2 Matrices 41 3 Continuity 121 4 Differentiation 150 5 Applications of Derivative 188 6 Integration 218 7 Definite Integrals 281 Model Question Paper - I 323 Model Question Paper - II 325 Board Questions Paper – March 2014 327 Board Questions Paper – October 2014 329 Board Questions Paper – March 2015 331 Board Questions Paper – October 2015 333 Board Questions Paper – March 2016 335

121

Chapter 03: Continuity

Type of Problems Exercise Q. Nos.

Continuity of Standard Function

3.1 Q.1

Practice Problems

(Based on Exercise 3.1) Q.1

Examine the Continuity of a Function at a given point

3.1 Q.2, 3, 10

Practice Problems

(Based on Exercise 3.1) Q.2, 3, 9, 12, 13, 14, 15, 16, 17

Miscellaneous Q.1

Practice Problems

(Based on Miscellaneous) Q.1, 2, 8

Types of Discontinuity (Removable Discontinuity/ Irremovable Discontinuity)

3.1 Q.4

Practice Problems

(Based on Exercise 3.1) Q.4

Miscellaneous Q.2, 10

Practice Problems

(Based on Miscellaneous) Q.3

Find the value of Function if it is Continuous at given point

3.1 Q.5, 7

Practice Problems

(Based on Exercise 3.1) Q.5, 7, 10, 24

Miscellaneous Q.3, 4

Practice Problems

(Based on Miscellaneous) Q.4

Find the value of k/a/b if the Function is Continuous at a given point/points.

3.1 Q.6, 8, 9

Practice Problems

(Based on Exercise 3.1) Q.6, 8, 11, 18, 19, 20, 21, 22, 23, 25, 26

Miscellaneous Q.5, 6, 7, 8

Practice Problems

(Based on Miscellaneous) Q.5, 6, 7

Find the points of Discontinuity for the given Functions

Miscellaneous Q.9

Continuity03

122

Std. XII : Commerce (Maths ‐ I)

Syllabus: 3.1 Continuity of a function at a point 3.2 Algebra of continuous functions 3.3 Types of discontinuity 3.4 Continuity of some standard functions Introduction Continuity is ‘the state of being continuous’ and continuous means ‘without any interruption or disturbance’. For example, the price of a commodity and its demand are inversely proportional. The graph of demand curve of a commodity is a continuous curve without any breaks or gaps. Note: A graph consisting of jumps is not a graph of continuous function. 3.1 Continuity of a function at a point Definition: A function f is said to be continuous at a point x = a in the domain of f, if

alimx

f(x) exists and a

limx

f(x) = f (a).

i.e. if a

limx

f(x) = a

limx

f(x) = f (a)

If any of the above conditions is not satisfied by the function, then it is discontinuous at that point. The point is known as point of discontinuity. eg., Consider the function, f(x) = 2x + 7, x 4 = 5x 5, x 4 Since, f(x) has different expressions for the value of x left hand and right hand limits have to be found

out.

4lim

x

f(x) = 4

limx

(5x 5) = 5 4 5 = 15

Also, f (4) = 5 (4) 5 = 15 and

4lim

x

f(x) = 4

limx

(2x + 7) = 2 4 + 7 = 15

4

limx

f(x) = 4

limx

f(x) = f (4)

f(x) is continuous at x = 4. Continuity of a function on its domain Definition: A real valued function f : D R where D R is said to be a continuous function on D, if it is continuous at every point in the domain D. eg., Consider the functions, i. f(x) = 3x4 + x2 + 3x ii. f(x) = sin x These two functions are continuous on every domain D, where D R. 3.2 Algebra of continuous functions If f and g are two real valued functions defined on the same domain, which are continuous at x = a, then 1. the function kf is continuous at x = a, for any

constant k R. 2. the function f g is continuous at x = a 3. the function f . g is continuous at x = a

4. the function f

g is continuous at x = a, when

g (a) 0 5. composite functions, f[g(x)] and g[f(x)], if

well defined are continuous functions at x = a. 3.3 Types of discontinuity 1. Removable discontinuity: A real valued

function f is said to have removable discontinuity at x = a in its domain, if

alimx

f(x) exists but a

limx

f(x) f (a)

i.e. a

limx

f(x) =a

limx

f(x) f(a)

This type of discontinuity can be removed by redefining the function f at x = a as

f (a) = a

limx

f(x).

eg., Consider the function,

f(x) = 5 6

2

x x

x , x 2

= 2 , x = 2

Here, 2

limx

f(x) = 2

limx

2 5 6

2

x x

x

= 2

limx

3 2

2

x x

x

= 2

limx

x 3

.... [ x 2, x 2, x 2 0] = 2 3 = 1

2limx

f(x) exists

Y

XO

X

YDemand

Pri

ce

123


Also, f (2) = 2 …. (given)

2limx

f(x) f (2)

function f is discontinuous at x = 2, This discontinuity can be removed by

redefining f as follows:

f(x) = 2 5 6

2

x x

x , x 2

= 1 , x = 2 x = 2 is a point of removable discontinuity. 2. Irremovable discontinuity: A real valued

function f is said to have irremovable discontinuity at x = a, if

alimx

f(x) does not

exist i.e. a

limx

f(x) a

limx

f(x) or one of the

limits does not exist. Such a function can not be redefined as

continuous function. eg., Consider the function, f(x) = x2 + 2x + 3 , x 3 = x2 1 , x 3 Here,

3lim

x

f(x) = 3

limx

x2 + 2x + 3

= (3)2 + 2(3) + 3 = 18 and

3lim

x

f(x) = 3

limx

x2 1= (3)2 1 = 8

3

limx

f(x) 3

limx

f(x)

limit of the function does not exist. f has irremovable discontinuity at x = 3 3.4 Continuity of some standard functions 1. Constant function: The constant function

f(x) = k (where k R is a constant). The function is continuous for all x belonging to its domain.

eg., f(x) = 10, f(x) = log10 100 , f(x) = e7 2. Polynomial function: The function

f(x) = a0 + a1x + a2x2 + …. + anx

n, where n N, a0, a1 …. an R is continuous for all x belonging to domain of x.

eg., f(x) = x2 + 5x + 9, f(x) = x3 5x + 9,

f(x) = x4 16, x R. 3. Rational function: If f and g are two

polynomial functions having same domain

then the rational function f

g is continuous in its

domain at points where g(x) 0.

eg.,

Consider the function, 2

2

5 6

9

x x

x

Here, f(x) = x2 + 5x + 6 and g(x) = x2 9 Given function is continuous on its domain, where x2 9 0 i.e., (x + 3) (x 3) 0 i.e., x + 3 0, x 3 0 i.e., x 3, x 3 The function is continuous on its domain

except at x = 3, 3. 4. Trigonometric function: sin (ax + b) and

cos (ax + b), where a, b R are continuous functions for all x R.

eg., sin (5x + 2), cos (7x 11) x R. Note: Tangent, cotangent, secant and cosecant

functions are continuous on their respective domains.

5. Exponential function: f(x) = ax , a > 0, a 1, x R is an exponential function, which is continuous for all x R.

eg.,

f(x) = 3x , f(x) = 1

2

x

, f(x) = ex x R,

where a > 0, a 1. 6. Logarithmic function: f(x) = loga x where

a > 0, a 1 is a logarithmic function which is continuous for every positive real number i.e. for all x R+

eg., f(x) = loga 7x , f(x) = loga 9x2 x R, where

a > 0, a 1. Some Important Formulae Algebra of limits: If f(x) and g(x) are any two functions, 1.

alimx

[f(x) + g(x)] = a

limx

f(x) + a

limx

g(x)

2. a

limx

[f(x) g(x)] = a

limx

f(x) a

limx

g(x)

3. a

limx

[f(x)g(x)] = a

limx

f(x)a

limx

g(x)

4. a

f ( )lim

g( )

x

x

x = a

a

lim f ( )

lim g( )

x

x

x

x, where

alimx

g(x) 0

5. a

limx

[k.f(x)] = ka

limx

f(x), where k is a constant.

124


Limits of Algebraic functions: 1.

alimx

x = a

2. a

limx

k = k, where k is a constant.

3. a

limx

n na

a

x

x= nan 1

Limits of Trigonometric functions:

1. 0

limx

sin x

x= 1

2. 0

limx

tan x

x = 1

3. 0

limx

cos x = 1

Limits of Exponential functions:

1. 0

limx

a 1x

x= log a, where (a > 0, a 1)

2. 0

limx

(1 + x)1

x = e

Limits of Logarithmic functions:

1. 0

limx

log 1 x

x= 1

Exercise 3.1 1. Are the following functions continuous on

the set of real numbers? Justify your answers. i. f(x) = 7 Solution: Given, f(x) = 7 It is a constant function. f(x) is continuous on the set of real

numbers i.e., x R ii. f(x) = e Solution: Given, f (x) = e

It is a constant function …. [ e = 2.71828]

f(x) is continuous on the set of real numbers i.e., x R.

iii. f (x) = log 19 Solution: Given, f(x) = log 19 Here, log 19 is a constant f(x) is a constant function f(x) is continuous on the set of real

numbers i.e., x R.

iv. f(x) = 7x4 5x3 3x + 1 Solution: Given, f(x) = 7x4 5x3 3x + 1 It is a polynomial function f(x) is continuous on the set of real

numbers i.e., x R v. g(x) = sin (4x 3) Solution: Given, g(x) = sin (4x 3) It is a sine function f(x) is continuous on the set of real

numbers i.e., x R

vi. h(x) =2

3 2

5 +7 +2

+ + +3

x x

x x x

Solution:

Given, h(x) = 2

3 2

5 7 2

3

x + x

x x x+

It is a rational function and is discontinuous if x3 + x2 + x + 3 = 0 But, x R, x3 + x2 + x + 3 0 h(x) is continuous on the set of real

numbers, except when x3 + x2 + x + 3 = 0

vii. g(x) = 2

2

13 16 19

2 1

x x

x

Solution:

Given, g(x) = 2

2

13 16 19

2 1

x x

x

It is a rational function and is discontinuous, if 2x2 + 1 = 0 But x R, 2x2 + 1 0 g (x) is continuous on the set of real

numbers i.e., x R viii. f(x) = 5x Solution: Given, f(x) = 5x It is an exponential function It is continuous on the set of real

numbers i.e., x R ix. f(x) = 32x 15x Solution: Given, f(x) = 32x 15x It is the difference of two exponential functions It is continuous on the set of real

numbers i.e., x R

125


x. f(x) = e(5x + 7) Solution: Given, f(x) = e(5x + 7) It is an exponential function It is continuous on the set of real

numbers i.e., x R 2. Examine the continuity of the following

functions at the given point. (All functions are defined on R R) i. f(x) = x2 – x + 9, for x 3 = 4x + 3, for x > 3; at x = 3.

[Mar 15] Solution:

3lim

x

f(x) = 3

limx

(x2 x + 9)

= (3)2 3 + 9 = 15 …. (i) and

3lim

x

f(x) = 3

limx

(4x + 3)

= 4(3) + 3 = 15 …. (ii) Also, f (3) = (3)2 3 + 9 = 15 …. (iii)

3lim

x

f(x) = 3

limx

f(x) = f(3)

…. [From (i), (ii) and (iii)] f is continuous at x = 3.

ii. f(x) =2 16

4

x

x

, for x 4

= 8, for x = 4; at x = 4. [Oct 15] Solution:

4

limx

f(x) = 4

limx

2 16

4

x

x

= 4

limx

2 24

4

x

x

= 4

limx

( 4) ( 4)

( 4)

x x

x

=

4limx

(x + 4)

….[ x 4, x 4, x 4 0]

4

limx

f(x) = 4 + 4 = 8 …. (i)

Also, f (4) = 8 …. (ii)(given)

4limx

f(x) = f (4) …. [From (i) and (ii)]

f is continuous at x = 4.

iii. f(x) =x x

x x

2

3

3 2 4

7 9 2

, for x 2

=1

13, for x = 2 ; at x = 2

Solution: Consider, x3 + 7x 9 2 By synthetic division, we get

2 1 0 7 9 2

2 2 9 2

1 2 9 0

x3 + 7x 9 2 = (x 2 ) (x2+ 2 x+ 9)

2

limx

f(x) =2

limx

2

2

2 2 2 4

2 2 9

x x x

x x x

=2

limx

2

2 2 2 2

2 2 9

x x x

x x x

=2

limx

2

2 2 2

2 2 9

x x

x x x

=2

limx 2

2 2

2 9

x

x x

[x 2 , x 2 , x 2 0]

=2

2 2 2

( 2) 2 2 9

= 2

2 2 9

2

2lim f

13

xx ….(i)

Also, f( 2 ) = 1

13 ….(ii)(given)

2

limx

f(x) f ( 2 )

….[From (i) and (ii)] f is discontinuous at x = 2 .

iv. f(x) =sin 5 x

x, for x 0

= 1, for x = 0; at x = 0. Solution:

0

limx

f(x) = 0

limx

sin5x

x=

0limx

sin5

5

x

x 5

= 5 0

limx

sin5

5

x

x= 5 (1)

.…[ x0, 5x0, 0

limx

sin x

x = 1]

0

limx

f(x) = 5

Also, f(0) = 1 …. (given)

0limx

f(x) f (0)

f is discontinuous at x = 0.

126


v. f(x) =

3 2 7

1

x

x, for x 1

= –1

3, for x = 1; at x = 1. [Oct 14]

Solution:

1

limx

f(x) = 1

limx

3 2 7

1

x

x

= 1

limx

3 2 7

1

x

x

3 2 7

3 2 7

x

x

= 1

limx

22(3) 2 7

1 3 2 7

x

x x

= 1

limx

9 2 7

1 3 2 7

x

x x

= 1

limx

2 2

1 3 2 7

x

x x

= 1

limx

2 1

1 3 2 7

x

x x

= 1

limx

2

3 2 7

x

…. [x 1, x 1, x 1 0]

= 2

3 2 1 7

= 2

6

=

1

3

1

limx

f(x) = 1

3

…. (i)

Also, f(1) = 1

3

…. (ii)(given)

1

limx

f(x) = f(1) ….[From (i) and (ii)]


vi. f(x) =3 5

1 5

x

x

, for x 4

=1

8, for x = 4; at x = 4.

Solution:

4

limx

f(x) = 4

limx

3 5

1 5

x

x

= 4

limx

3 5

1 5

x

x

1 5

1 5

x

x

3 5

3 5

x

x

= 4

limx

22

22

3 5 1 5

1 5 3 5

x x

x x

= 4

limx

9 5 1 5

1 5 3 5

x x

x x

= 4

limx

4 1 5

4 3 5

x x

x x

= 4

limx

4 1 5

4 3 5

x x

x x

= 4

limx

1 5

3 5

x

x

....[x 4, x 4, x 4 0]

= 1 5 4

3 5 4

= 1 1

3 3

= 2

6

4

limx

f(x) = 1

3

…. (i)

Also, f(4) = 1

8 …. (ii)(given)

4

limx

f(x) f(4) …. [From (i) and (ii)]


vii. f(x) = x2 cos

1

x, for x 0

= 0 , for x = 0; at x = 0

[Oct 15]

Solution:

0

limx

f(x) = 0

limx

x2cos 1 x

cos x [ 1, 1] for all x R,

also, when x 0, x 0 cos 1

x

exists.

Let 1

cos x

= finite number = k (say)

0

limx

x2 1

cos x

= 0

limx

x2k

where k [1, 1]

0

limx

f(x) = 0 …. (i)

127


Also, f(0) = 0 …. (ii)(given)

0

limx

f(x) = f(0) …. [From (i) and (ii)]


viii. f(x) =

tan sin

sin 3 3sin

x x

x x, for x < 0

= 2 2

2

3 sin 2 sin

3

x x

x,

for x 0; at x = 0

Solution:

Consider,

0

limx

f(x) = 0

limx

tan sin

sin 3 3 sin

x x

x x

= 0

limx

3

tan sin

3sin 4sin 3sin

x x

x x x

…[sin3 = 3sin – 4sin3]

= 0

limx

3

tan sin

4sin

x x

x

= 0

limx

3

sinsin

cos4sin

xx

xx

= 0

limx

3

sin sin cos

4sin

x x x

x

= 0

limx

3

sin 1 cos

4 sin

x x

x

= 0

limx

2

3

sin 2sin2

4sin

xx

x

…[1 – cos x = 2sin2

2

x]

= 1

2

0

limx

2

3

3

3

sin .sin2

sin

xx

xx

x

=

2

20

3

30

sin1 sin 2lim2

44

sinlim

x

x

xx

xx

x

x

=

2

0 0

3

0

sin1 sin 1 2lim lim2 4

2

sinlim

x x

x

xx

xx

xx

=

2

3

1 11 1

2 41

= 1

8

…[x0,2

x 0,0

limx

sin x

x = 1]

0

limx

f(x) = 1

8

…. (i)

Also, 0

limx

f(x) = 0

limx

2 2

2

3sin 2sin

3

x x

x

= 0

limx

22

2 2

2sin3sin

3 3

xx

x x

= 0

limx

22

2 20

sinsin 2lim

3 x

xx

x x

=0

limx

22

20

sinsin 2lim

3 x

xx

x x

= (1)2 –

2

3 (1)

.…[0

limx

sin1

x

x]

= 1 – 2

3

0

limx

f(x) = 1

3 …. (ii)

0

limx

f(x) 0

limx

f(x) ….[From (i) and (ii)]

f(x) does not exist

f is discontinuous at x = 0

128


ix. f(x) =

x x x

x x

3 2

3

9 2

6

, for x < 2

=

3 2

1 4

2 2x x x, for x > 2

= 4 for x = 2; at x = 2 Solution: Consider,

2

limx

f(x) =2

limx

3 2

3

9 2

6

x x x

x x

=

3 2

3

2 + 2 9 2 2

2 2 6

=8

0

2

limx

f(x) does not exist.

But f(2) = 4

2lim

xf(x) f(2)


x. f(x) = 6 3 2 1x x x

x, for x < 0

= 2

4 4 2x x

x, for x > 0

= 1, for x = 0, at x = 0 Solution:

0

limx

f(x) = 0

limx

6 3 2 1 x x x

x

= 0

limx

6 3 2 1 1 1 x x x

x

= 0

limx

6 1 3 1 2 1 x x x

x

= 0

limx

6 1 3 1 2 1 x x x

x

=0

limx

6 1 3 1 2 1

x x x

x x x

= 0 0 0

6 1 3 1 2 1lim lim lim

x x x

x x xx x x

= log 6 + log 3 – log 2

…[0

limx

a 1log a

x

x]

= log 6 3

2

0

limx

f(x) = log 9 …. (i)

0

limx

f(x) =0

limx 2

4 4 2 x x

x

=0

limx 2

14 2

4 x

x

x

=0

limx

2

2

4 2 4 1

4

x x

xx

=0

limx

2

2

4 1

4

x

xx

=0

limx

2

0

4 1 1lim

4

x

xxx

= (log 4)2 . 0

1

4

…[0

limx

a 1log a

x

x]

0

limx

f(x) = (log 4)2 .... (ii)

0

limx

f(x) 0

limx

f(x) .... [From (i) and (ii)]

0

limx

f(x) does not exist

f is discontinuous at x = 0 3. Discuss the continuity of the following

functions.

i. f(x) =2a 1x

x, x 0, a 0 & a 1

= 2 log a, x = 0; at x = 0. Solution:

0

limx

f(x) =0

limx

2a 1x

x

=0

limx

2a 12

2

x

x=

02lim

x

2a 1

2

x

x

0

limx

f(x) = 2 log a …. (i)

….[x0,2x0, 0

limx

a 1x

x = log a]

Also, f(0) = 2 log a …. (ii)(given)

0

limx

f(x) = f(0) ….[From (i) and (ii)]

f is continuous at x = 0

ii. f(x) =

5 3

4 3

x x

x x, x 0

= log5

4, x = 0; at x = 0.

Solution:

0limx

f(x) =0

limx

5 3

4 3

x x

x x

129


=0

limx

5 3

4 3

x x

x xx

x

=0

0

5 3lim

4 3lim

x x

x

x x

x

x

x

= 0

0

5 1 3 1lim

4 1 3 1lim

x x

x

x x

x

x

x

=0

0

5 1 3 1lim

4 1 3 1lim

x x

x

x x

x

x x

x x

=0 0

0 0

5 1 3 1lim lim

4 1 3 1lim lim

x x

x x

x x

x x

x x

x x

= log 5 log 3

log 4 log 3

....[0

limx

a 1x

x= log a]

0

limx

f(x) =

5log

34

log3

…. (i)

Also, f (0) = log5

4 …. (ii) (given)

0

limx

f(x) f(0)

f is discontinuous at x = 0

iii. g(x) =

2

51

2

xx , x 0

= e5/2, x = 0; at x = 0. Solution:

0

limx

g(x) =0

limx

2

51

2

xx

=0

limx

52

551

2

xx

0

limx

g(x) = e5 …. (i)

....[ x 0, 5

2

x 0 0

limx

1

1 xx = e]

Also g(0) = 52e …. (ii)(given)

0

limx

g(x) g(0) ….[From (i) and (ii)]

g is discontinuous at x = 0

iv. h(x) = log 1 + 2 x

x, x 0

= 2, x = 0; at x = 0. Solution:

0

limx

h(x) =0

limx

log 1 2 x

x

=0

limx

log 1 22

2

x

x

=0

2 limx

log 1 2

2

x

x

= 2(1)

....[x0, 2x0, 0

limx

log 11

x

x]

0

limx

h(x) = 2 …. (i)

Also, h(0) = 2 …. (ii)(given)

0

limx

h(x) = h(0) ….[From (i) and (ii)]

h is continuous at x = 0

v. f(x) =32 1

tan

x

x, x < 0

=

2

e e 2x x

x, x > 0

= 2, x = 0 at x = 0. Solution:

0

limx

f(x) =0

limx

32 1

tan

x

x=

0limx

32 13

3tan

x

xx

x

=

3

0

0

2 13lim

3tan

lim

x

x

x

xx

x

= 3 log 2

1

....[x0, 3x0,

0limx

a 1x

x= log a,

0

tanlimx

x

x = 1]

0

limx

f(x) = 3 log 2 …. (i)

0

limx

f(x) =0

limx

2

e e 2 x x

x

=0

limx

2

1e 2

e x

x

x

130


=0

limx

2

2e 2e 1

e

x x

x

x

=0

limx 2

2

1

e 1 1

e

x

xx

=2

0 0

1

e 1 1lim lim

e

x

xx xx

= 2

0

11

logee

....[0

limx

e 1x

x= log e]

0

limx

f(x) = 1 .... (ii)

0

limx

f(x) 0

limx


f(0) does not exist f is discontinuous at x = 0 4. Discuss the continuity of the following

functions at the points given against them. If the function is discontinuous, determine whether the discontinuity is removable. In that case, redefine the function, so that it becomes continuous.

i. f(x) = x2 – 2x – 1, for x < 2 = 3x – 1, for x ≥ 2; at x = 2. Solution:

2

limx

f(x) =2

limx

(x2 – 2x – 1)

= (2)2 – 2(2) – 1 = 4 – 4 – 1 = 1

2lim

xf(x) = 1 …. (i)

and2

limx

f(x) =2

limx

(3x – 1)

= 3(2) – 1 = 6 1

2lim

xf(x) = 5 …. (ii)

2

limx

f(x) 2

limx


limit of the function does not exist. f has irremovable discontinuity at x = 2

ii. f(x) =

3 27

3

x

x, for x < 3

= 8x, for x 3; at x = 3. Solution:

3

limx

f(x) =3

limx

3 27

3

x

x

=3

limx

3 33

3

x

x

=3

limx

23 3 9

3

x x x

x

....[a3 – b3 = (a – b)(a2 + ab + b2)]

=3

limx

(x2 + 3x + 9)

[ x 3, x 3 x – 3 0]

= (3)2 + 3(3) + 9

3lim

xf(x) = 27 ….(i)

and3

limx

f(x) =3

limx

8x = 8(3)

3

limx

f(x) = 24 ….(ii)

3

limx

f(x) 3

limx


limit of the function does not exist f has irremovable discontinuity at x = 3

iii. f(x) =sin9

2

x

x, for x 0

=1

2, for x = 0; at x = 0.

Solution:

0

limx

f(x) =0

limx

sin9

2

x

x

=0

1 sin 9lim 9

2 9

x

x

x

=0

9 sin 9lim

2 9x

x

x

= 91

2

[x0, 9x0, 0

limx

sin x

x= 1]

0

limx

f(x) = 9

2 …. (i)

Also, f(0) = 1

2 …. (ii) (given)

0

limx

f(x) f(0) ….[From (i) and (ii)]

131


f has removable discontinuity at x = 0 This discontinuity can be removed by

redefining the function as:

f(x) = sin9

2

x

x, for x 0

=9

2, for x = 0 ; at x = 0

iv. f(x) =2e 1

5

x

x, for x 0

= 2, for x = 0; at x = 0 Solution:

0

limx

f(x) =0

limx

2e 1

5

x

x

=2

0

1 e 1lim 2

5 2

x

x x

=2

0

2 e 1lim

5 2

x

x x

= 2log e

5

[ x 0, 2x 0,0

limx

e 1log e

x

x]

0

limx

f(x) = 2

5 …. (i)


0

limx




f(x) =2e 1

5

x

x, for x 0

= 2

5 , for x = 0 ; at x = 0

v. f(x) =

x

x3

3 2

1, for x 1


1

limx

f(x) =1

limx 3

3 2

1

x

x

=1

limx

3 3

3 2 3 2

1 3 2

x x

x x

=1

limx

2 2

2

3 2

1 1 3 2

x

x x x x

=1

limx 2

3 4

1 1 3 2

x

x x x x

=1

limx

2

1

1 1 3 2

x

x x x x

=1

limx 2

1

1 3 2 x x x

....[x1, x 1 x–1 0]

= 2

1

1 1 1 1 3 2 =

1

3 2 2

1

limx

f(x) = 1

12 …. (i)


1

limx




f(x) = 3

3 2

1

x

x, for x 1

= 1

12 , for x = 1 ; at x = 1

5. If f is continuous at x = 0, then find f(0)

i. f(x) = 2

5 5 2x x

x, x 0

Solution: Given, f is continuous at x = 0

f(0) =0

limx

f(x)

=0

limx 2

5 5 2 x x

x=

0limx 2

15 2

5 x

x

x

=0

limx

2

2

5 2 5 1

5

x x

x x=

0limx

2

2

5 1

5

x

x x

=0

limx

2

0

5 1 1lim

5

x

xxx

= 2 1log 5

5

....[0

limx

a 1log a

x

x]

f(0) = (log 5)2

132


ii. f(x) =

2sin3 1

log 1

x

x x, x 0 [Mar 12]


f(0) =0

limx

f(x)

=0

limx

2sin3 1

log 1

x

x x

=

0limx

2sin 2

2

3 1 sin 1 1log( 1)sin

x x

xx x xx

=

0limx

2 2sin3 1 sinsin

log 1

x xx x

+ x

x

=

2 2sin

0 0

0

3 1 sinlim lim

sin

log(1 + )lim

x

x x

x

xx x

x

x

= 2 2log3 1

1

.... [x0, sinx0, 0

limx

sinx

x=1,

0limx

a 1x

x= loga]

iii. f(x) = 15 3 5 1

tan

x x x

x x

, x 0 [Mar 15]


f(0) =0

limx

f(x)

=0

limx

15 3 5 1

tan

x x x

x x

=0

limx

5 3 3 5 1

tan

x x x

x x

=0

limx

5 3 3 5 1

tan

x x x x

x x

=0

limx

3 5 1 1 5 1

tan

x x x

x x

=0

limx

5 1 3 1

tan

x x

x x

=0

limx

2

2

5 1 3 1

tan

x x

xx x

x

=0

limx

5 1 3 1

tan

x x

x xx

x

=0 0

0

5 1 3 1lim lim

tanlim

x x

x x

x

x xx

x

=

log5 log3

1

….[x 0, x 0,0

limx

a 1x

x

= log a,0

limx

tan x

x= 1]

f(0) = (log 5) (log 3)

iv. f(x) =2

cos3 cosx x

x

, x 0 [Mar 16]


f(0) = 0

limx

f(x)

=0

limx 2

cos3 cosx x

x

=0

limx

3

2

4 cos 3cos cos x x x

x

….[cos3 = 4cos3 – 3cos]

=0

limx

3

2

4 cos 4 cosx x

x

=0

limx

2

2

4cos cos 1x x

x

=0

limx

2

2

4cos 1 cos x x

x

=0

limx

2

2

4cos sin x x

x

= –40

limx

cos x . 0

limx

2sin

x

x

= –4. cos(0) . (1)2

….[0

limx

sin1

x

x]

f (0) = –4

133


6. Find the value of k, if the function

i. g(x) =12 1

1

x

x

, for x 1

= k, for x = 1 is continuous at x = 1 Solution: Given, g is continuous at x = 1 and g(1) = k

g(1) =1

limx

g(x)

k =1

limx

12 1

1

x

x

=1

limx

12 121

1

x

x

=12(1)12–1

.... [a

limx

n nn 1a

n aa

x

x]

k = 12 ii. h(x) = x2 + 1, for x < 0

= 5 2 1x + k, for x 0 is continuous at x = 0 Solution: Given, h is continuous at x = 0

0lim

xh(x) =

0lim

xh(x) = h (0) ….(i)

Now,

0

limx

h(x) =0

limx

x2 + 1

= (0)2 + 1

0lim

xh(x) = 1

and0

limx

h(x) = 0

limx

25 1 kx

= 5 0 1 k

0

limx

h(x) = 5 + k

1 = 5 + k ....[From (i)] k = –4

iii. f(x) =tan 7

2

x

x, for x 0

= k, for x = 0 is continuous at x = 0 [Mar 15] Solution: Given, f is continuous at x = 0 and f (0) = k

f(0) =0

limx

f(x)

k =0

limx

tan7

2

x

x=

1

2 0limx

tan 77

7

x

x

=7

2 0limx

tan7

7

x

x

=7

2(1)

….[x0,7x0, 0

limx

tan x

x= 1]

k = 7

2 iv. h(x) = |x + k|, for x 17 = 20, for x = 17 is continuous at x = 17 Solution: Given h is continuous at x = 17, h(17) = 20

h(17) = 17

limx

h(x)

= 17

limx

| x + k |

= 17

limx

(x + k)

h(17) = (17 + k) 17 + k = 20 or – (17 + k) = 20 k = 20 17 or 17 k = 20 k = 3 or k = 20 17 k = 3 or k = 37.

7. If f(x) = 2

1 sin

2

x

x, for x

2

, is continuous

at x =2

, then find f

2

. [Oct 15]

Solution:

Given, f is continuous at x = 2

f 2

lim f2

x

x = 2

2

1 sinlim

2

x

x

x

Put x = h2

h = x2

as , 0,h 02 2

x x

f2h 0

1 sin h2

lim2

2 h2

= 2h 0

1 cos hlim

2h

.... [sin cos2

]

134


=h 0lim 2

1 cosh 1 cosh

1 cosh2h

=h 0lim

2

2

1 cos h

4h 1 cos h

= h 0lim

2

2

sin h

4h 1 cosh

=1

4 h 0lim

2sin h

h h 0

lim

1

1 cos h

=1

4(1)2

1

1 cos 0….[

0limx

sinx

x= 1]

=1 1

4 1 1

=

1 1

4 2

f 1

2 8

8. If f(x) =2e 1

a

x

x

for x < 0, a 0

= 1 for x = 0

= log 1 7

b

x

x

for x > 0, b 0

is continuous at x = 0, then find a and b. [ Mar 16] Solution: Given, f is continuous at x = 0 and f(0) = 1

0lim

xf(x) =

0lim

xf(x) = f(0)

0

limx

f(x) =0

limx

f(x) = 1 ….(i)

Now,

0

limx

f(x) =0

limx

2e 1

a

x

x

=2

0

1 e 1lim 2

a 2

x

x x

=2

0

2 e 1lim

a 2

x

x x

=2

logea

….[0

a 10, 2 0, lim log a

x

xx x

x]

0

limx

f(x) =2

a ....(ii)

Also,0

limx

f(x) =0

limx

log 1 7

b

x

x

= 0

log 1 71lim 7

b 7

x

x

x

= 0

log 1 77lim

b 7

x

x

x

= 71

b

.... [ 0

log 10,7 0, lim 1

x

xx x

x]

0

limx

f(x) =7

b ....(iii)

Now, 2

1a ….[From (i) and (ii)]

a = 2

and 7

1b ….[From (i) and (iii)]

b = 7

a = 2, b = 7 9. If f is continuous at x = 0 and

f(x) = 23 1x + a, for x < 0

= x3 + a + b, for x 0 and f(1) = 2, then find a, b. Solution: Given, f is continuous at x = 0

0

limx

f(x) =0

limx

f(x) …. (i)

Now,

0

limx

f(x) =0

limx

32 1x + a

= 2 0 1 a

0

limx

f(x) = 2 + a

and0

limx

f(x) =0

limx

x3 + a + b

= 0 + a + b

0

limx

f(x) = a + b

2 + a = a + b ….[From (i)]

b = 2

Also, f(x) = x3 + a + b, for x 0 and f(1) = 2

f(1) = (1)3 + a + b

2 = 1 + a + b

a + b = 1 …. (ii) Substituting b = 2 in (ii) we get a + 2 = 1

a = –1

a = –1, b = 2

135


10. Is the function f(x) = x3 + 2x2 – 5 cos x + 3

continuous at x =2

? Justify.

Solution: Given, f(x) = x3 + 2x2 – 5 cos x + 3

f(x) = (x3 + 2x2 + 3) – 5 cos x Let x3 + 2x2 + 3 = p(x) and 5 cos x = q(x)

f(x) = p(x) – q(x) ....(i) Now, p(x) = x3 + 2x2 + 3

p(x) is a polynomial function

It is continuous at each value of x and q(x) = 5 cos x Here, 5 is constant function and cos x is cosine

function which is continuous.

q(x) is continuous at each value of x

f(x) is a difference of two continuous function which is always continuous

f(x) is continuous at x = 2

Miscellaneous Exercise – 3 1. Discuss the continuity of the function.

f(x) =3

2

64

9 5

x

x, for x 4


4limx

f(x) =4

limx

3

2

64

9 5

x

x

=4

limx

23 3

2 2

9 54

9 5 9 5

xx

x x

=4

limx

2 2

222

4 4 16 9 5

9 5

x x x x

x

=4

limx

2 2

2

4 4 16 9 5

9 25

x x x x

x

=4

limx

2 2

2

4 4 16 9 5

16

x x x x

x

=4

limx

2 24 4 16 9 5

4 4

x x x x

x x

=4

limx

2 24 16 9 5

4

x x x

x

[ x 4, x 4, x – 4 0]

= 2 24 4 4 16 4 9 5

4 4

=

48 10

8

4

limx

f(x) = 60 …. (i)


4

limx


f is discontinuous at x = 4 2. Examine the continuity of the function

f(x) =

23e 1

log 1 3

x

x x, for x 0

= 10, for x = 0; at x = 0. If discontinuous, then state whether the

discontinuity is removable. If so, redefine and make it continuous.

Solution:

0limx

f(x) = 0

limx

23e 1

log 1 3

x

x x=

0limx

23

2

e 1

9 log 1 3

9

x

x x

x

=0

limx

23

2

e 1 1log 1 39

9

x

xx

x

=0

limx

230

0

lim1e 11 log(1 3 )3 lim3 3

xx

x

xxx

=

2 1loge

11

3

....

0

e 10,3 0,lim loge,

x

xx x

x

0

log 1lim 1

x

x

x

= 3(1)2

0

limx

f(x) = 3 …. (i)


136


0

limx




f(x) = 23e 1

log(1 3 )

x

x x

, for x 0

= 3 , for x = 0 ; at x = 0 3. The function f is defined as

f(x) = 7

5

128

32

x

x, for x 2

=1/ 5 1/ 5

1/ 2 1/ 2

2

2

x

x, for x > 2

f(2) = 3 Examine, if f is continuous at x = 2. Solution:

2lim

xf(x) =

2limx

7

5

128

32

x

x

=2

limx

7 7

5 5

2

2

x

x

=2

limx

7 7

5 5

2222

xx

xx

=

7 7

2

5 5

2

2lim

22

lim2

x

x

x

xx

x

=

7 1

5 1

7 2

5 2

....

n nn 1

a

alim na

a

x

x

x

=6

4

7 2

5 2

=27 2

5

2

limx

f(x) = 28

5

and2

limx

f(x) = 2

limx

1 1

5 5

1 1

2 2

2

2

x

x

=2

limx

1 1

5 5

1 1

2 2

2

2

22

x

x

xx

=

1 1

5 5

2

1 1

2 2

2

2lim

2

2lim

2

x

x

x

x

x

x

=

11

5

11

2

12

51

22

....[n n

n 1

a

alim na

a

x

x

x]

=

4

5

1

2

12

51

22

=4 1

5 222

5

2

limx

f(x) =2

5 (2)

3

10

2

limx

f(x) 2

limx

f(x)

f(x) does not exist

f is discontinuous at x = 2 4. The function f defined as

f(x) = 8 8 1

5 7 3

x x

x x, for x 1

is continuous at x = 1. Find f(1) Solution: Given, function is continuous at x = 1

f(1) = 1

limx

f(x)=1

limx

8 8 1

5 7 3

x x

x x

=1

limx

8 8 1

5 7 3

x x

x x

5 7 3

5 7 3

x x

x x

8 8 1

8 8 1

x x

x x

=1

limx

2 2

2 2

8 8 1 5 7 3

5 7 3 8 8 1

x x x x

x x x x

=1

limx

8 8 1 5 7 3

5 7 3 8 8 1

x x x x

x x x x

=1

limx

7 7 5 7 3

8 8 8 8 1

x x x

x x x

=1

limx

7 1 5 7 3

8 1 8 8 1

x x x

x x x

137


=1

limx

7 5 7 3

8 8 8 1

x x

x x

....[ x1, x 1 x 1 0]

=

7 5 1 7 1 3

8 1 8 8 1 1

=

7 2 2

8 3 3

=28

48

f (1) = 7

12

5. Find k if the function given below is

continuous at x =2

f(x) = 3

2cos sin2

2

x x

x, for x

2

= k, for x =2

Solution:

Given, f is continuous at x = 2

f2

=2

lim

x

f(x)

k = 2

lim

x 3

2cos sin 2

2

x x

x

= 2

lim

x 3

2cos 2sin cos

2

x x x

x

[sin2 = 2sin cos]

= 2

lim

x

3

2cos 1 sin

2

x x

x

Put x = h2

h = x – 2

as x 2

, x –

2

0, h 0

h = h 0lim 3

2 cos h 1 sin h2 2

2 h2

= h 0lim

3

2sin h 1 cos h

2h

[cos sin , sin cos2 2

]

= h 0lim

2

3

h2sin h 2sin

28h

= 1

2 h 0lim

2

2

hsinsin h 2hh

44

=

2

h 0 h 0

hsin1 sin h 2lim .lim

h8 h2

= 211 1

8

….h 0

h sin hh 0, 0, lim 1

2 h

6. If the function given below is continuous at

x = 2 as well as at x = 4 , then find the values of a and b.

f(x) = x2 + ax + b, x 2 = 3x + 2, 2 x 4 = 2ax + 5b, 4 x [Oct 14] Solution: Given, f is continuous at x = 2

2lim

xf(x) =

2lim

xf(x) ….(i)

Now,2

limx

f(x) =2

limx

x2 + ax + b

= (2)2 + a(2) + b

2lim

xf(x) = 4 + 2a + b

and2

limx

f(x) =2

limx

(3x + 2)

= 3(2) + 2

2lim

xf(x) = 8

4 + 2a + b = 8 ….[From (i)] 2a + b = 4 ….(ii) Also, f is continuous at x = 4

4lim

xf(x) =

4lim

xf(x) ….(iii)

Now,4

limx

f(x) = 4

limx

3x + 2 = 3(4) + 2

4

limx

f(x) = 14

and4

limx

f(x) = 4

limx

2ax + 5b = 2a(4) + 5b

4

limx

f(x) = 8a + 5b

14 = 8a + 5b ….[From(iii)] 8a + 5b = 14 ….(iv)

138


By eq. (iv) – 5 eq. (iii), we get

8a 5b 14

10a 5b 20

2a 6

a = 3 Substituting a = 3 in eq. (ii), we get 2 3 + b = 4 b = 4 – 6 b = –2 a = 3 or b = 2 7. Find a and b if f is continuous at x = 1,

where

f(x) = sin

1

x

x+ a, x 1

= 2, x = 1

= 2

1 cos

1

x

x+ b, x 1

Solution: Given, f is continuous at x = 1 and f (1) = 2

1lim

xf(x) =

1lim

xf(x) = f (1)

1

limx

f(x) = 1

limx

f(x) = 2 ….(i)

Now, 1

limx

f(x) = 1

limx

sina

1

x

x

Put x = 1 + h h = x – 1 as x1, x 10, h0

1

limx

f(x) = h 0lim

sin 1 ha

1 h 1

= h 0lim

sin ha

h

= h 0lim

sin ha

h

….[sin(+) = –sin]

= h 0lim

sin ha

h

= –h 0 h 0

sin hlim lim a

h

= –(1) + a

….[h0, h0,h 0lim

sin h1

h ]

1

limx

f(x) = – + a ….(ii)

Also 1

limx

f(x) =1

limx 2

1 cosb

1

x

x

Put x = 1 + h h = x – 1 as x1, x – 10, h0

1

limx

f(x) =

2h 0

1 cos 1 hlim b

1 1 h

= h 0lim

2

1 cos hb

1 1 h

= h 0lim 2

1 cos hb

h

….[cos(+) = –cos]

= h 0lim

2

2

h2sin

2 bh

….[1 – cos = 2sin2

2

]

= h 0lim

2

2

h2sin

2 bh

44

= h 0lim

2

2 2

hsin2 2 b

4 h4

=

2

h 0 h 0

hsin

2lim lim bh2

2

= 21 b

2

.... [0

h sinh 0, 0, lim 1

2

x

x

x]

1

limx

f(x) = b2

….(iii)

– + a = 2 ….[From (i) and (ii)] a = 2 + a = 3

and b 22

….[From (i) and (iii)]

139


b = 2 – 2

b = 3

2

a = 3, b = 3

2

8. Find k, if the function f is continuous at

x = 0, where

i. f(x) =

2

e 1 sinx x

x, for x 0

= k, for x = 0 Solution: Given, f is continuous at x = 0 and f(0) = k

f(0) = 0

limx

f(x)

k = 0

limx

2

e 1 sinx x

x

= 0

limx

e 1 sin

x x

x x

= 0

limx

e 1x

x .

0limx

sinx

x

= log e . (1)

….[0

limx

e 1x

x= log e,

0limx

sinx

x= 1]

k = 1

ii. f(x) = 27 3

k 1

x x

x , for x 0

= 2, for x = 0 Solution: Given, f is continuous at x = 0 and f(0) = 2

f(0) = 0

limx

f(x)

2 = 0

limx

27 3

k 1

x x

x=

0limx

3 9 3

k 1

x x

x

= 0

limx

3 9 3

k 1

x x x

x=

0limx

3 9 1

k 1

x x

x

=

0limx

3 9 1

k 1

x x

xx

x

= 0 0

0

9 1lim3 .lim

k 1lim

xx

x x

x

x

x

x

2 = 03 log 9

log k .... [

0limx

a 1log a

x

x]

log k = log9

2

log k = log 129 .... [n log a = log an]

log k = log 3

k = 3

iii. f(x) = log 1 3

5

x

x, for x 0

= k, for x = 0 Solution: Given, f is continuous at x = 0 and f(0) = k

f(0) = 0

limx

f(x)

k = 0

limx

log 1 3

5

x

x

= 0

limx

log 1 313

5 3

x

x

= 3

5 0limx

log 1 3

3

x

x

= 31

5

....[

0

log 10,3 0, lim 1

x

xx x

x]

k = 3

5 9. Find the points of discontinuity, if any, for

the following:

i. f(x) =2

2

cos

1

x x

x

Solution:

Given, f(x) = 2

2

cos

1

x x

x

Let, x2 + cosx = p(x) and x2 + 1 = q(x) Consider, p(x) = x2 + cosx Here, x2 is always continuous for all real

values of x and cosine is a continuous function

p(x) is a continuous function and q(x) = x2 + 1 It is a polynonimal function

It is continuous for all real values of x

f(x) is a continuous function.

140


ii. f(x) =2

5 4

4

x

x

Solution:

Given, f(x) = 2

5 4

4

x

x

f(x) =

5 4

2 2

x

x x

f(x) is a rational function f(x) will be discontinuous if (x + 2) (x – 2) = 0 i.e., x + 2 = 0 or x – 2 = 0 i.e., x = –2 or x = 2 f(x) is discontinuous at x = –2 and x = 2

iii. f(x) =2

2

3 4 9

6 10

x x

x x

Solution:

Given, f(x) = 2

2

3 4 9

6 10

x x

x x

f(x) is a rational function f(x) will be discontinuous if x2 – 6x + 10 = 0

i.e., x = 2

6 6 4 1 10

2

= 6 36 40

2

6 4

2

x

6 2i

2

x

Value of x is a complex number f(x) is continuous for all real values of x

iv. f(x) =2 9

sin 9

x

x

Solution:

Given, f(x) = 2 9

sin 9

x

x

Let x2 – 9 = p(x) and sinx – 9 = q(x) Consider, p(x) = x2 – 9 It is a polynomial function It is continuous function and q(x) = sinx – 9

Here, sine is a continuous function and 9 is a constant function

q(x) is continuous as –1 sinx 1 f(x) is continuous function.

10. If possible, redefine the function to make it continuous.

i. f(x) =1

1 xx , for x 1

= e2, for x = 1; at x = 1. Solution:

1

1

1 1lim f lim

x

x xx x

Put x = 1 + h h = x – 1 as x1, x–10, h0

1

limx

f(x) =

h 0lim

1

1 h 11 h =

h 0lim

1

h1 h

1

limx

f(x) = e

....[ 1

0lim 1 e

xx

x ]....(i)

Also, f(1) = e2 …. (ii) (given)

1

limx




f(x) = 1

1 xx , for x 1

= e, for x = 1 ; at x = 1

ii. f(x) =tan6 1

sin

x

x, for x 0

= log 50, for x = 0; at x = 0. Solution:

0

limx

f(x) =0

limx

tan6 1

sin

x

x

=0

limx

tan6 1sin

coscos

x

xx

x

= 0

limx

tan6 1

tan cos

x

x x

= 0

limx

tan

0

6 1 1.lim

tan cos

x

xx x

= log 6 1

cos 0

.... [x0, tanx0, 0

limx

a 1loga

x

x

]

0

limx

f(x) = log 6

Also, f(0) = log 50 ….(given)

0

limx

f(x) f(0)

f has removable discontinuity at x = 0

141


This discontinuity can be removed by redefining the function as:

f(x) = tan6 1

sin

x

x, for x 0

= log 6, for x = 0 ; at x = 0

iii. f(x) =2

2

sin 5 x

x, for x 0

= 5, for x = 0; at x = 0. [Oct 14]

Solution:

0

limx

f(x) = 0

limx

2

2

sin 5x

x

=0

limx

2

2

sin 525

25

x

x

= 250

limx

2sin5

5

x

x

= 25(1)2

[x0, 5x0, 0

limx

sin1

x

x]

0

limx

f(x) = 25

Also, f(0) = 5 …. (given)

0

limx

f(x) f(0)



f(x) = 2

2

sin 5x

x, for x 0

= 25, for x = 0 ; at x = 0

iv. f(x) =2

3

cos

1 sinx

x, for x <

2

=2

2 1 sin

cos

x

x, for x >

2

= 2

3, for x =

2

; at x =

2

.

Solution:

2

lim

x

f(x) =2

lim

x

2

3

cos

1 sinx

x

=2

lim

x

2

3

1 sin

1 sin

x

x

=2

lim

x

2

1 sin 1 sin

1 sin 1 sin sin

x x

x x x

....[a3 – b3 = (a – b) (a2 + ab + b2)]

=2

lim

x2

1 sin

1 sin sin

x

x x

….[x2

sinxsin2

sinx1, 1–sinx0]

=2

1 sin2

1 sin sin2 2

= 2

1 1

1 1 1

2

lim f ( )x

x

= 2

3

and

2

lim

x

f(x) =2

lim

x2

2 1 sin

cos

x

x

=2

lim

x2

2 1 sin 2 1 sin

cos 2 1 sin

x x

x x

=2

lim

x

2 2

2

2 1 sin

cos 2 1 sin

x

x x

=2

lim

x

2

2 1 sin

1 sin 2 1 sin

x

x x

=2

lim

x

1 sin

1 sin 1 sin 2 1 sin

x

x x x

=2

lim

x 1

1 sin 2 1 sin x x

….[x2

sinxsin2

sinx1, 1–sinx0]

=1

1 sin 2 1 sin2 2

=

1

1 1 2 1 1 =

1

2 2 2

2

lim f ( )x

x

= 1

4 2

2

lim f ( )x

x

2

lim

x

f(x)

limit of the function does not exist

f has irremovable discontinuity at x = 2

142


v. f(x) = 2

2

2

1

x x

x, for x 1

=3 2

2 1

x

x, for x > 1

= 1, for x = 1; at x = 1.

Solution:

1

limx

f(x) =1

limx

2

2

2

1

x x

x

=1

limx

2

2

1 1

1

x x

x

=1

limx

2

2 2

1 1

1 1

x x

x x

=1

limx 2

1 11

1 1

x x

x x

=1

limx

2 2

2 2

11

1 1

x

x x

=1

limx

11

1 1 1

x

x x x

=1

limx

11

1 1

x x

....[x1, x–10]

=1

limx

1 + 1

limx

1

1 1 x x

=1 +

1

1 1 1 1 = 1 + 1

2(2)= 1 +

1

4

1

limx

f(x) = 5

4

and 1

limx

f(x) =1

limx

3 2

2 1

x

x

=1

limx

3 2 3 2 2 1

2 1 3 2 2 1

x x x

x x x

=1

limx

2 2

2 2

3 2 2 1

3 22 1

x x

xx

=1

limx

3 4 2 1

2 1 3 2

x x

x x

=1

limx

1 2 1

1 3 2

x x

x x

=1

limx

1 2 1

1 3 2

x x

x x

=1

limx

2 1

3 2

x

x ….[x1, x 1 x–10]

= 2 1 1 2

41 3 2

1

limx

f(x) = 1

2

1

limx

f(x) 1

limx

f(x)

limit of the function does not exist. f has irremovable discontinuity at x = 1

vi. f(x) = 2 3 4 4

1

x x x x

x, for x 1.

= 5, for x = 1; at x = 1. Solution:

1

limx

f(x) =1

limx

2 3 4 4

1

x x x x

x

=1

limx

2 3 41 1 1 1

1

x x x x

x

=1

limx

2 3 41 1 1 1

1 1 1 1

x x x x

x x x x

=1

limx

1

1

x

x+

1limx

2 21

1

x

x+

1limx

3 31

1

x

x

+1

limx

4 41

1

x

x

= (1) (1)1–1 + (2) (1)2–1 + (3) (1)3–1 + (4) (1)4–1

….[a

limx

n nn 1a

naa

x

x]

= 1 + 2 + 3 + 4

1

limx

f(x) = 10 …. (i)


1

limx

f(x) f(1)

f has removable discontinuity at x = 1

143


This discontinuity can be removed by redefining the function as:

f(x) = 2 3 4 4

1

x x x x

x, for x 1

= 10, for x = 1 ; at x = 1 Additional Problems for Practice Based on Exercise 3.1 1. Are the following functions continuous on the

set of real numbers? Justify your answer. (All functions are defined on R R)

i. f(x) = 8x2 + 9 ii. f(x) = e50

iii. f(x) = log 23

19

iv. f(x) = 8x5 2x4 + 5x3 + 2x2 + x + 9 v. h(x) = cos(9x + 5)

vi. g(x) = 2

4

5

29 2

x x

x x vii. g(x) = 3x + 7x 2. Examine the continuity of the following

funtions at the given point:

i. f(x) = sinx

x + cos x, for x 0

= 2, for x = 0; at x = 0

ii. f(x) = 1

2 sin x2, for x 0

= 0, for x = 0; at x = 0 iii. f(x) = (1 + 2x)1/x, for x 0 = e2, for x = 0; at x = 0

iv. f(x) = 2 6

3

x x

x , for x 3

= 7, for x = 3; at x = 3 v. f(x) = x2 + 6x + 10, for x 4 = x2 x + 38, for x > 4; at x = 4

vi. f(y) = 2

2

e 1 .siny y

y , for y 0

= 4, for y = 0; at y = 0

vii. f(x) =

1

41

5

xx, for x 0

= 4

5e , for x 0 at x = 0

viii. f(x) = 1

2 sin 2

(x + 1), for x 0

= 3

tan sinx x

x , for x > 0; at x = 0

ix. f(x) = 3 2

3 2

2 2 5

3 3 1

x x x

x x x , for x < 1

= 4 3

1 1

1 x x x

, for x 1; at x = 1

x. f(x) = 3

3 2

1

x

x, for x 1

= 1

12, for x = 1; at x = 1

3. Discuss the continuity of the following functions:

i. f(x) = 3 5a ax x

x, for x 0

= log a, for x = 0; at x = 0

ii. f(x) =

1

1a

xx , for x 0

= 1

ae , for x = 0; at x = 0

iii. g(x) =

5log 1

2

x

x, for x 0

=

5

2, for x = 0; at x = 0

iv. f(x) = 5 e

sin 2

x x

x, for x 0

= 1

2(log 5 + 1), for x = 0; at x = 0

v. f(x) = 2

2

sin ax

x, for x 0

= 1, for x = 0; at x = 0

vi. f(x) = x2 sin1

x, for x 0

= 0, for x = 0; at x = 0

vii. f(x) = 1 cos x

x, for x 0

= 0, for x = 0; at x = 0

viii. f(x) = 4 2

3

x

x

, for x ≠ 0

= 1

12, for x = 0; at x = 0

[Mar 16] 4. Discuss the continuity of the following

functions at the points given against them. If the function is discontinuous, determine whether the discontinuity is removable. In that case, redefine the function, so that it becomes continuous.

144


i. f(x) = 1 cos3

tan

x

x x, for x 0

= 9, for x = 0; at x = 0

ii. f(x) = 5 32

2

x

x, for x < 2

= 16x, for x 2; at x = 2

iii. f(x) = sin

5

x

x, for x 0

= 5

, for x = 0; at x = 0

iv. f(x) = 2 4

sin 2

x

x, for x 0

= 8, for x = 0; at x = 0

v. f(x) = 2sin( )x x

x, for x 0

= 2, for x = 0; at x = 0 5. If f is continuous at x = 0, then find f(0).

i. f(x) = 2sin4 1

log (1 2 )

x

x x, x 0

ii. f(x) = log(1 a ) log(1 b ) x x

x

iii. f(x) = log(2 ) log(2 )

tan

x x

x

iv. f(x) = 2 2

2

cos sin 1

1 1

x x

x 6. Find the value of k, if the function i. g(x) = |x 3|, for x 3 = k, for x = 3 is continuous at x = 3

ii. f(x) = 8 2

k 1

x x

x, for x 0

= 2, for x = 0 is continuous at x = 0

iii. f(x) = log(1 k )

sin

x

x, for x 0

= 5, for x = 0 is continuous x = 0 iv. f(x) = x2 + k, for x 0 = x2 k, for x < 0 is continuous at x = 0

v. f(x) = 2

2

3 k

2( 1)

x x

x, for x 1

= 5

4, for x = 1

is continuous at x = 1

7. If f(x) = 2

1 cos[7( )]

5( )

x

x, for x is

continuous at x = , find f()

8. If the function f(x) = k cos

2 x

x, for x

2

= 3, for x = 2

be continuous at x = 2

, then find k

9. Is the function

f(x) = 2x3 + 3x2 + 3x cos x + sin 5x + 3

continuous at x = 4

? Justify

10. If the function f is continuous at x = 1, then

find f(1). Where f(x) = 2 3 2

1

x x

x for x ≠ 1.

[Mar 14] 11. If the function f is continuous at x = 2, then

find ‘k’ where f(x) = 2 5

1

x

x, for 1 < x 2

= kx + 1, for x > 2 [Mar 14]

12. Discuss the continuity of the function f

defined as

f (x) = 3

8 3

1

x

x

, x ≠ 1

= 1

3 , x = 1; at x = 1 [Mar 08]

13. Discuss the continuity of the function f

defined as:

f (x) =

1

51

3

xx

, if x ≠ 0

= 5

3e , if x = 0; at x = 0 [Oct 08] 14. Discuss the continuity of f (x) at x = 2, where

f (x) = 2 4

2

x

x

, for x ≠ 2

= 4, for x = 2 [Mar 09] 15. Discuss the continuity of the function f

defined as,

f(x) = 2x + 3, if 1 ≤ x ≤ 2

= 6x 1, if 2 < x ≤ 3; at x = 2 [Oct 10]

145


16. If f(x) = 4 2x

x

, for x ≠ 0

= 1

4, for x = 0

Discuss the continuity of f(x) at x = 0 [Mar 11, Oct 11 ]

17. Discuss the continuity of the following

function:

1

f 1 3 xx x , for x ≠ 0

= e3, for x = 0; at x = 0 [Oct 12] 18. If f is continuous at x = 0, where f (x) = x2 + a, for x ≥ 0

= 2 2 1 bx , for x < 0

Find a, b given that f (1) = 2. [Mar 08] 19. Find k, if the function f defined as:

f (x) = 2

2 3 cos k x

x, x ≠ 0

= 2, x = 0 is continuous at x = 0 [Oct 08] 20. Find k, if the function

f (x) = 3 64

4

x

x

, for x ≠ 4

= k, for x = 4 is continuous at x = 4 [Oct 09] 21. If f(x) is continuous at x = 0 and it is defined as

a af

x x

xx

, x ≠ 0

= k, x = 0 find k. [Mar 10] 22. The function f defined as

f(x) = sin px

x, if x > 0

= q + 25 16x , if x ≤ 0

is continuous at x = 0. Find the values of p and q, given that f (2) = 3. [Oct 10]

23. If f(x) = tan2

3

x

x+ a, for x < 0

= 1, for x = 0 = x + 4 b, for x > 0 is continuous at x = 0, then find the values of a

and b. [Mar 11]

24. Find f(3) if f(x) = 2 9

3

x

x, x ≠ 3 is contiuous at

x = 3. [Oct 11] 25. If f is continuous at x = 0 where

f(x) = 3e 1

a

x

x, x ≠ 0

= 1 , x = 0 then find a. [Mar 12] 26. If f is continuous at x = 0 where f(x) = x2 + a, x 0

= 22 1 b x , x < 0,

find a and b. Given that f(1) = 2 [Oct 12] Based on Miscellaneous Exercise ‐ 3 1. Examine the continuity of the following

funtions at the given point:

i. f(x) = 10 7 14 5

1 cos

x x x x

x , for x 0

= 10

7 , for x = 0; at x = 0

ii. f(x) = sin3

tan2

x

x , for x < 0

= 3

2, for x = 0

= 2

log(1 3 )

e 1

x

x, for x > 0

iii. f(x) = 23 4 1

( 6)

x

x , for x ≠ 6

= 1

5, for x = 6; at x = 6

iv. f(x) = 1 2 1 2 x x

x, for x < 0

= 2x2 + 3x 2, for x 0; at x = 0

v. f(x) = 3 2

2

16 20

( 2)

x x x

x , for x 2

= 7, for x = 2; at x = 2 2. Discuss the continuity of the following functions:

i. f(x) = 2 5

4 3

x x

x x, for x 0

= log 3

10, for x = 0; at x = 0

ii. f(x) = 2(2 1)

tan .log(1 )

x

x x, for x 0

= log 4, for x = 0

146


3. Discuss the continuity of the following functions at the points given against them. If the function is discontinuous, determine whether the discontinuity is removable. In that case, redefine the function, so that it becomes continuous:

i. f(x) = 4 e

6 1

x x

x, for x 0

= log 2

3

, for x = 0; at x = 0

ii. f(x) = 2

3 3 2 x x

x, for x 0

= 2 log3, for x = 0; at x = 0

iii. f(x) =

6

3

1

6418

x

x, for x

1

2

= 1

3, for x =

1

2; at x =

1

2

iv. f(x) = 2(8 1)

sin log 14

x

xx

, for x 0

= 8 log 8, for x = 0; at x = 0 4. If f is continuous at x = 0, then find f(0).

i. f(x) = 14 2 1

1 cos

x x

x, x 0

ii. f(x) = 5 2e e

sin 3

x x

x

5. Find the value of k, if the function

f(x) =

2

2

3sin

2

x

x, for x 0

= k, for x = 0

is continuous at x = 0

6. If f(x) = sin4

5

x

x+ a, for x > 0

= x + 4 b, for x < 0

= 1, for x = 0

is continuous at x = 0, find a and b.

[Mar 09, 10, Oct 09]

7. If f(x) = 2

1 cos4 x

x, for x < 0

= a, for x = 0

= 16 4

x

x, for x > 0

is continuous at x = 0, then find the value of ‘a’. 8. Discuss the continuity of the function f at

x = 0, where f(x) = 5 5 2

,cos 2 cos 6

x x

x x for x ≠ 0

= 21log5

8, for x = 0

[Mar 14]

Multiple Choice Questions

1. If f(x) = 2 , 0 1

c 2 , 1 2

x

x x is continuous at

x = 1, then c = (A) 2 (B) 4 (C) 0 (D) 1

2. If f(x) =

1 , if 3

a b , if 3 5

7 , if 5

x

x x

x

is continuous,

then the value of a and b is (A) 3, 8 (B) –3, 8 (C) 3, –8 (D) –3, –8 3. The sum of two discontinuous functions (A) is always discontinuous. (B) may be continuous. (C) is always continuous. (D) may be discontinuous. 4. For what value of k the function

f(x) = 5 2 4 4

, if 22

k ,if 2

x xx

xx

is

continuous at x = 2?

(A) 1

4 3

(B) 1

2 3 (C) 1

4 3 (D) 1

2 3

5. The function f(x) = log (1 a ) log (1 b )x x

x

is

not defined at x = 0. The value which should be assigned to f at x = 0 so that it is continuous at x = 0, is

(A) a b (B) a + b

(C) log a + log b (D) log a log b

147


6. In order that the function f(x) = (x + 1) cot x is continuos at x = 0, f(0) must be defined as

(A) f(0) = 1

e (B) f(0) = 0

(C) f(0) = e (D) None of these

7. If f(x) =

sin3, 0

sin

k, 0

xx

x

x

is a continuous

function, then k =

(A) 1 (B) 3

(C) 1

3 (D) 0

8. A function f is continuous at a point x = a in the domain of ‘f’ if

(A) a

limx

f(x) exists (B) a

limx

f(x) = f(a)

(C) a

limx

f(x) f(a) (D) both (A) and (B). 9. Which of the following function is

discontinuous?

(A) f(x) = x2 (B) g(x) = tan x

(C) h(x) = 2

3

1x

x (D) none of these

10. If the function f(x) =

k cos, when

2 2

3, when2

xx

x

x is

continuous at x = 2

, then k =

(A) 3 (B) 6 (C) 12 (D) None of these 11. The points at which the function

f(x) = 2

1

12

x

x x is discontinuous, are

(A) –3,4 (B) 3,–4 (C) –1,–3,4 (D) –1,3,4 12. Which of the following statement is true for

graph f(x) = log x

(A) Graph shows that function is continuous (B) Graph shows that function is

discontinuous

(C) Graph finds for negative and positive values of x

(D) Graph is symmetric along x-axis

13. If f(x) =

2 1, when 1

12, when 1

xx

xx

, then

(A) 1

lim x

f(x) = –2

(B) 1

lim x

f(x) = –2

(C) f(x) is continuous at x = –1 (D) All the above are correct

14. If f(x) =

a, when a

a1, when a

xx

xx

, then

(A) f(x) is continuous at x = a (B) f(x) is discontinuous at x = a

(C) 0limx f(x) = 1

(D) None of these

15. If f(x) =

2

1 cos 4, when 0

a when 0,

, when 016 4

xx <

xx =

xx

x

is continuous at x = 0, then the value of ‘a’ will be (A) 8 (B) –8 (C) 4 (D) None of these

16. If f(x) =

4 16, when 2

216,when 2

xx

xx

, then

(A) f(x) is continuous at x = 2 (B) f(x) is discountinuous at x = 2

(C) 2limx f(x) = 16

(D) None of these 17. The values of A and B such that the function

f(x) =

2sin ,2

Asin B, ,2 2

cos2

x x

x x

x, x

is continuous

everywhere are (A) A = 0, B = 1 (B) A = 1, B = 1 (C) A = –1, B = 1 (D) A = –1, B = 0

148


18. If f(x) = 2

1 k 1 k,for 1 0

2 3 2 ,for 0 1

x xx <

x

x x x

, is

continuous at x = 0, then k = (A) –4 (B) –3 (C) –2 (D) –1 19. The function f(x) = sin |x| is (A) Continuous for all x (B) Continuous only at certain points (C) Differentiable at all points (D) None of these

20. The function f(x) = 1 sin cos

1 sin cos

x x

x x is not

defined at x = . The value of f(), so that f(x) is continuous at x = , is

(A) 1

2 (B)

1

2

(C) –1 (D) 1

21. The function f(x) = 2

3 2

2 7

3 3

x

x x x

is

discontinuous for (A) x = 1 only (B) x = 1 and x = –1 only (C) x = 1, x = –1, x = –3 only (D) x = 1, x = –1, x = –3 and other values of x 22. The function ' f is defined by f(x) = 2x – 1, if

x > 2, f(x) = k if x = 2 and x2 –1, if x < 2 is continuous, then the value of k is equal to

(A) 2 (B) 3 (C) 4 (D) –3

23. Function f(x) = 2

1 cos4

8

x

x , where x 0 and

f(x) = k, where x = 0 is a continous function at x = 0 then the value of k will be?

(A) k = 0 (B) k = 1 (C) k = –1 (D) None of these

24. If f(x) =

, when0 1/ 2

1, when 1/ 2

1 ,when1/ 2 1

x x

x

x x

, then

(A) 1/2lim x f(x) = 2

(B) 1/2lim x f(x) = 2

(C) f(x) is continuous at x = 1

2

(D) f(x) is discontinuous at x = 1

2

25. If f(x) = 2

2

10 25

7 10

x x

x x for x 5 and f is

continuous at x = 5, then f(5) = (A) 0 (B) 5 (C) 10 (D) 25

Answers to Additional Practice Problems Based on Exercise 3.1 1. i. Polynomial function continuous ii. Constant function continuous iii Constant function continuous iv. Polynomial function continuous v. Cosine function continuous vi. Rational function continuous for all x R, except when

x4 + 29x + 2 = 0 vii. Addition of exponential functions continuous 2. i. Continuous ii. Continuous iii. Continuous iv. Discontinuous v. Continuous vi. Discontinuous vii. Continuous viii. Continuous ix. Discontinuous x. Continuous 3. i. Discontinuous ii. Continuous iii. Continuous iv. Discontinuous v. Discontinuous vi. Continuous vii. Discontinuous viii. Discontinuous 4. i. Discontinuous, removable ii. Discontinuous, irremovable iii. Discontinuous, removable iv. Discontinuous, removable v. Discontinuous, removable

5. i. 2(log 4)

2 ii. a + b

iii. 1 iv. 4 6. i. 0 ii. 2 iii. 5 iv. 0 v. 4

7. 49

10

149


8. 6 9. Addition of continuous functions. f(x) is continuous. 10. f(1) = 1 11. k = 4 12. Discontinuous 13. Discontinuous 14. Continuous 15. Discontinuous 16. Continuous 17. Discontinuous 18. a = 1, b = 1 19. k = ± 4 20. k = 48 21. 2 log a 22. p = 1, q = 3

23. a = 1

3, b = 3

24. 6 25. a = 3

26. a = 1, b = 3

4

Based on Miscellaneous Exercise ‐ 3 1. i. Discontinuous ii. Continuous iii. Discontinuous iv. Continuous v. Continuous 2. i. Discontinuous ii. Discontinuous 3. i. Discontinuous, removable ii. Discontinuous, removable iii. Discontinuous, removable iv. Discontinuous, removable 4. i. 2(log 2)2 ii. 1

5. 9

4

6. a = 1

5, b = 3

7. 8 8. Discontinuous

Answers to Multiple Choice Question

1. (B) 2. (C) 3. (B) 4. (C) 5. (B) 6. (C) 7. (B) 8. (D) 9. (B) 10. (B) 11. (B) 12. (A) 13. (D) 14. (B) 15. (A) 16. (B) 17. (C) 18. (C) 19. (A) 20. (C) 21. (C) 22. (B) 23. (B) 24. (D) 25. (A)

335

Board Question Paper : March 2016

BOARD QUESTION PAPER : MARCH 2016 Notes: i. All questions are compulsory. ii. Figures to the right indicate full marks. iii. Answer to every question must be written on a new page. iv. L.P.P. problem should be solved on graph paper. v. Log table will be provided on request. vi. Write answers of Section – I and Section – II in one answer book.

Section I Q.1. Attempt any SIX of the following: [12]

i. If y = (sin x)x, find d

d

y

x. (2)

ii. If A = 1 3

3 1

show that A2 2A is a scalar matrix. (2)

iii. Write the negation of the following statements:

(a) y N, y2 + 3 7

(b) If the lines are parallel then their slopes are equal. (2)

iv. The total revenue R = 720x 3x2 where x is number of items sold. Find x for which total revenue R is increasing. (2)

v. Evaluate: 2

2

secd

tan 4

xx

x (2)

vi. Find d

d

y

x, if y = cos1 (sin 5x) (2)

vii. Discuss the continuity of function f at x = 0

Where f (x) = 4 2

3

x

x

, for x ≠ 0

= 1

12 , for x = 0 (2)

viii. State which of the following sentences are statements. In case of statement, write down the truth value:

(a) Every quadratic equation has only real roots.

(b) 4 is a rational number. (2) Q.2. (A) Attempt any TWO of the following: [6][14] i. Solve the following equations by the inversion method:

2x + 3y = 5 and 3x + y = 3 (3)

ii. Find x and y, if 1

1 2 0 1 5 23 2

0 1 3 3 4 41

x

y

(3)

iii. Evaluate: 1tan dx x . (3)

336


(B) Attempt any TWO of the following: [8] i. (a) Express the truth of each of the following statements using Venn diagram. (1) All teachers are scholars and scholars are teachers. (2) If a quadrilateral is a rhombus then it is a parallelogram. (b) Write converse and inverse of the following statement: “If Ravi is good in logic then Ravi is good in Mathematics.” (4) ii. Find the area of the region bounded by the lines 2y + x = 8, x = 2 and x = 4. (4)

iii. Evaluate: 9 3

3 33

12d

12

xx

x x

(4)

Q.3. (A) Attempt any TWO of the following: [6][14]

i. If f (x) = 2 1

a

xe

x

, for x < 0, a ≠ 0

= 1 , for x = 0

= log (1 7 )

b

x

x

, for x > 0, b ≠ 0

Is continuous at x = 0 then find a and b. (3) ii. If the function f is continuous at x = 0, then find f(0)

where f (x) = 2

cos 3 cosx x

x

, x ≠ 0 (3)

iii. If f(x) = 4x3 3x2 + 2x + k and f(0) = 1, f(1) = 4, find f(x). (3) (B) Attempt any TWO of the following: [8] i. Find MPC (Marginal Propensity to Consume) and APC (Average Propensity to Consume) if

the expenditure Ec of a person with income I is given as Ec = (0.0003) I2 + (0.075) I when I = 1000. (4)

ii. Cost of assembling x wallclocks is 3

2403

xx

and labour charges are 500x. Find the number

of wallclocks to be manufactured for which marginal cost is minimum. (4)

iii. If cos1 2 2

2 2

x y

x y

= 2k,

show that yd

d

y

x= x tan2 k. (4)

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