`
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
Chapter 03: Continuity
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
Std. XII : Commerce (Maths ‐ I)
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
Chapter 03: Continuity
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
Std. XII : Commerce (Maths ‐ I)
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)]
f is continuous at x = 1.
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)]
f is discontinuous at x = 4.
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
Chapter 03: Continuity
Also, f(0) = 0 …. (ii)(given)
0
limx
f(x) = f(0) …. [From (i) and (ii)]
f is continuous at x = 0.
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
Std. XII : Commerce (Maths ‐ I)
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)
f is discontinuous at x = 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
Chapter 03: Continuity
=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
Std. XII : Commerce (Maths ‐ I)
=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(x) ….[From (i) and (ii)]
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
f(x) ….[From (i) and (ii)]
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
f(x) ….[From (i) and (ii)]
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
Chapter 03: Continuity
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)
Also, f(0) = 2 …. (ii)(given)
0
limx
f(x) f(0) ….[From (i) and (ii)]
f has removable discontinuity at x = 0 This discontinuity can be removed by
redefining the function as:
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
= 5, for x = 1; at x = 1 Solution:
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)
Also, f(1) = 5 …. (ii)(given)
1
limx
f(x) f(1) ….[From (i) and (ii)]
f has removable discontinuity at x = 1 This discontinuity can be removed by
redefining the function as:
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
Std. XII : Commerce (Maths ‐ I)
ii. f(x) =
2sin3 1
log 1
x
x x, x 0 [Mar 12]
Solution: Given, f is continuous at x = 0
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]
Solution: Given, f is continuous at x = 0
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]
Solution: Given, f is continuous at x = 0
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
Chapter 03: Continuity
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
Std. XII : Commerce (Maths ‐ I)
=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
Chapter 03: Continuity
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
= 10, for x = 4; at x = 4 Solution:
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)
Also, f(4) = 10 …. (ii)(given)
4
limx
f(x) f(4) ….[From (i) and (ii)]
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)
Also, f(0) = 10 …. (ii)(given)
136
Std. XII : Commerce (Maths ‐ I)
0
limx
f(x) f(0) ….[From (i) and (ii)]
f has removable discontinuity at x = 0 This discontinuity can be removed by
redefining the function as:
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
Chapter 03: Continuity
=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
Std. XII : Commerce (Maths ‐ I)
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
Chapter 03: Continuity
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
Std. XII : Commerce (Maths ‐ I)
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) f(1) ….[From (i) and (ii)]
f has removable discontinuity at x = 1 This discontinuity can be removed by
redefining the function as:
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
Chapter 03: Continuity
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 has removable discontinuity at x = 0 This discontinuity can be removed by
redefining the function as:
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
Std. XII : Commerce (Maths ‐ I)
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)
Also, f(1) = 5 …. (ii)(given)
1
limx
f(x) f(1)
f has removable discontinuity at x = 1
143
Chapter 03: Continuity
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
Std. XII : Commerce (Maths ‐ I)
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
Chapter 03: Continuity
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
Std. XII : Commerce (Maths ‐ I)
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
Chapter 03: Continuity
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
Std. XII : Commerce (Maths ‐ I)
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
Chapter 03: Continuity
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
Std. XII : Commerce (Maths ‐ I)
(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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