Analysis · the analysis problem, which is explicitly useful for those preparing mathematics...

AnalysisNational Board for Higher Mathematics (NBHM)

MA, MSc & PhD Scholarship Tests

Previous Year Questions with Complete Solutions

2005 – 2018

Annamalai N

First Edition

Copyright c� 2019 Annamalai N

SELF PUBLISHED BY ANNAMALAI

HTTPS://ANNAMALAIMATHS.WORDPRESS.COM/

Title of Book: ANALYSIS

ISBN: 978-93-5351-952-0

Rs. 250.00

Licensed under the Creative Commons BY-NC-ND (the “License”). You may not usethis file except in compliance with the License. You may obtain a copy of the License athttp://creativecommons.org/licenses/by-nc/3.0. Unless required by applicablelaw or agreed to in writing, software distributed under the License is distributed on an “ASIS” BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express orimplied. See the License for the specific language governing permissions and limitationsunder the License.First printing, March 2019. Printed in India, Tamil Nadu.

Dedicated To

My

Beloved Parents

PREFACE

The National Board for Higher Mathematics (NBHM) was set up by the Government

of India under the Department of Atomic Energy (DAE), in the year 1983, to foster the

development of higher mathematics in the country, to formulate policies for the development

of mathematics, help in the establishment and development of mathematical centres and give

financial assistance to research projects and masters degree and to doctoral and postdoctoral

scholars in Mathematics.

This book elaborately covers the NBHM previous year questions from real and complex

analysis. It is divided into 3 chapters. Chapter 1 contains previous years’ (2005 - 2018)

solved question papers of M.A./M.Sc scholarships test. Chapter 2 contains previous years’

(2005 - 2018) solved question papers of research scholarships screening Test. Chapter 3

contains some important notes of analysis.

I believed that the content of this book should be of interest to readers who wish to know

the analysis problem, which is explicitly useful for those preparing mathematics oriented

exams. It is hoped that this book will help to the students acquire some basic knowledge

about the solving analysis problems. In addition, I have given some fundamental notes for

reviewing purpose.

Suggestions for the improvement of this book are most welcome. Suggestions are

welcome on this email id: [email protected].

March, 2019 N. ANNAMALAI

About the Author

Dr. N. Annamalai is a Assistant Professor(On Contract) in Department of Mathematics

at Central University of Tamil Nadu, Thiruvarur, Tamil Nadu, India. He Completed Un-

dergraduate at Thiru. A. Govindasamy Arts and Science College, Tindivanam, Villupuram

(D.t), Tamil Nadu, India. He received M. Sc, M. Phil. and Ph. D. degree from Bhrathidasan

University, Tiruchirappalli, Tamil Nadu, India. He got University First Rank in Master

of Science (2009 - 2011). Also, he passed CSIR(NET), GATE examination conducted

by Central Government of India. He received INSPIRE Fellowship (JRF & SRF) from

Department of Science and Technology, India. His main research area is Mathematical

Coding Theory.

More information related to workshops, mathematical activity and previous year ques-

tion papers can be found on https://annamalaimaths.wordpress.com/

ACKNOWLEDGEMENTS

It has been an incredible journey that brought me to the completion of this book. I am

deeply indebted to everyone who shared an important path in my journey to this point. It is

their support and inspiration that made this adventure both pleasant and fruitful.

I thank my Research supervisor Dr. C. Durairajan, Assistant Professor, Department of

Mathematics, School of Mathematical Sciences, Bharathidasan University, Tiruchirappalli,

Tamil Nadu, India for guidance and support. Throughout the years, I have learned a lot

from him on both professional and personal level. Thank You Sir!

I thank Mr. D. Chellapillai, Assistant Professor (On Contract), Department of Mathe-

matics, Central University of Tamil Nadu, Thiruvarur, Tamil Nadu, India for nice discussion

and correction.

I thank Mr. E. Sekar, Research Scholar, Department of Mathematics, Bharathidasan

University, Tiruchirappalli, Tamil Nadu, India for nice discussion.

I thank all my Teacher’s and my family members for supporting.

March, 2019 N. ANNAMALAI

NOMENCLATURE

N The set of all natural numbers

Z The set of all integers

Q The set of all rational numbers

R The set of all real numbers

C The set of all complex numbers

[a,b] {x ∈ R | a ≤ x ≤ b}]a,b[ {x ∈ R | a < x < b}l1 {{xn} | ∑xn < ∞}Rn n-dimensional Euclidean space over R

C[a,b] The space of continuous real-valued functions on an interval [a,b]

C1[a,b] The space of continuously differentiable real-valued functions on [a,b]

f (n)(x) The n-th derivative of the function f (x)

Contents

1 M. A. and M.Sc. Scholarship Test . . . . . . . . . . . . . . . . . . . . 15

1.1 October 22, 2005 151.2 September 23, 2006 231.3 September 22, 2007 301.4 September 20, 2008 381.5 September 19, 2009 441.6 September 25, 2010 481.7 September 24, 2011 531.8 September 22, 2012 591.9 September 21, 2013 671.10 September 20, 2014 731.11 September 19, 2015 791.12 September 17, 2016 851.13 September 23, 2017 901.14 September 22, 2018 96

2 Research Scholarships Screening Test . . . . . . . . . . . . . 103

2.1 February 25, 2006 1032.2 January 27, 2007 1092.3 February 2, 2008 117

2.4 January 24, 2009 1242.5 January 23, 2010 1312.6 January 22, 2011 1382.7 January 28, 2012 1442.8 January 19, 2013 1512.9 January 25, 2014 1572.10 January 24, 2015 1662.11 January 23, 2016 1742.12 January 21, 2017 1812.13 January 20, 2018 190

3 Some Important Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

3.1 Real Analysis 2013.2 Complex Analysis 209

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

1. M. A. and M.Sc. Scholarship Test

1.1 October 22, 2005

1. Let Dn be the open disc of radius n with centre at the point (n,0) ∈ R2. Does there

exist a function f : R2 → R of the form f (x,y) = ax+by such that

∪∞n=1Dn = {(x,y) | f (x,y)> 0}?

If your answer is ‘Yes’, give the values of a and b.

Solution: Yes. For n ∈ N, we have Dn = {(x,y) ∈ R2 | (x− n)2 + y2 < n2}. Then

∪∞n=1Dn = {(x,y) ∈ R2 | x > 0}. From this, we define f : R2 → R by f ((x,y)) = x

such that ∪∞n=1Dn = {(x,y) | f (x,y)> 0}. Therefore, a = 1 and b = 0.

2. Find the sum of the series∞

∑k=1

k2

k!.

Solution: ex = 1+ x1! +

x2

2! +x3

3! + · · ·+ xn

n! + · · · . Consider,

∞

∑k=1

k2

k!=

∞

∑k=1

k(k−1)!

=∞

∑k=1

k−1+1(k−1)!

=∞

∑k=1

k−1(k−1)!

+∞

∑k=1

1(k−1)!

16 Chapter 1. M. A. and M.Sc. Scholarship Test

=∞

∑k=2

k−1(k−2)!(k−1)

+ e =∞

∑k=2

1(k−2)!

+ e = e+ e = 2e.

3. What is the radius of convergence of the power series

∞

∑k=1

�logk xk?

Solution: Let ak =√

logk and let R be the radius of convergence of the given series.

Then

1R= lim

k→∞

��ak+1

ak

��= limk→∞

�log(k+1)√

logk= lim

k→∞

�log(k+1)

logk

= limk→∞

�log(k(1+ 1

k ))

logk= lim

k→∞

�1+

log(1+ 1k )

logk.

Since log(1+ 1k ) → 0 as k → ∞ and

1logk

is bounded,log(1+ 1

k )

logk→ 0 as k → ∞.

Therefore, R = 1.

4. What are the values of α ∈ R for which the following series is convergent?

∞

∑n=1

(−1)n

nα .

Solution: The alternating series∞∑

n=1(−1)nan converges if an ≥ 0,an → 0 as n → ∞

and an+1 ≤ an for all n ∈ N. Let an =1

nα . If α > 0, then an → 0. It is easy to check

that, an ≥ 0 and an+1 ≤ an. Therefore, the given alternating series converges for all

0 < α < ∞.

5. Evaluate:

limn→∞

(�

n2 +n−�

n2 +1).

Solution: Let L = limn→∞(√

n2 +n−√

n2 +1). Then

L = limn→∞

(�

n2 +n−�

n2 +1)×√

n2 +n+√

n2 +1√n2 +n+

√n2 +1

= limn→∞

n2 +n− (n2 +1)√n2 +n+

√n2 +1

= limn→∞

n−1√n2 +n+

√n2 +1

= limn→∞

1− 1n�

1+ 1n +

�1+ 1

n2

=12

since1n→ 0 as n → ∞.


8. Let f (x) = [x]+(x− [x])2 for x ∈R, where [x] denotes the largest integer not exceed-

ing x. What is the set of all values taken by the function f ?

Solution: Given any real number x can be written in a decimal expression, that is,

x = a0 +a1

10+

a2

102 +a3

103 + · · ·= a0 + k, where k = a110 +

a2102 +

a3103 + · · · , 0 ≤ k < 1

and a0 is an integer. Then Z+ [0,1) = R and hence f (R) = { f (x) | x ∈ R} =

{[x]+ (x− [x])2 | x ∈ R}= Z+[0,1) = R.

9. Let n be a positive integer. Let f (x) = xn+2sin( 1x ) if x �= 0 and let f (0) = 0. For what

value of n will f be twice differentiable but with its second derivative discontinuous

at x = 0?

Solution:

For n = 1, f (x) = x3 sin 1x , then

limx→0

f (x)− f (0)x−0

= limx→0

x3 sin 1x

x= 0 = f �(0)

and f �(x) =−xcos( 1x )+3x2 sin( 1

x ). Consider,

limx→0

f �(x)− f �(0)x−0

= limx→0

−xcos( 1x )+3x2 sin( 1

x )

x= lim

x→0−cos

�1x

�+3xsin

�1x

�.

Since cos( 1x ) oscillates between 1 and −1 as x → 0, the second derivative does not

exists for n = 1.

For n = 2, f (x) = x4 sin 1x , then

limx→0

f (x)− f (0)x−0

= limx→0

x4 sin 1x

x= 0 = f �(0)

and f �(x) =−x2 cos( 1x )+4x3 sin( 1

x ). Consider,

limx→0

f �(x)− f �(0)x−0

= limx→0

−x2 cos( 1x )+4x3 sin( 1

x )

x= 0 = f ��(0)

and f ��(x) = sin( 1x )− 2xcos( 1

x )− 4xcos( 1x ) + 12x2 sin( 1

x ). Since sin( 1x ) oscillates

between 1 and −1 as x → 0, f ��(x) discontinuous at x = 0. Therefore, for n = 2, the

function f be twice differentiable but with its second derivative discontinuous at

x = 0.

1.1 October 22, 2005 19

10. What is the coefficient of x8 in the expansion of x2 cos(x2) around x = 0?

Solution: cosy = 1− y2

2! +y4

4! −y6

6! + · · · for all y ∈ R. Then

x2 cos(x2) = x2(1− x4

2!+

x8

4!− x12

6!+ · · ·) = x2 − x6

2!+

x10

4!− x14

6!+ · · · .

Therefore, the coefficient of x8 in the expansion of x2 cos(x2) around x = 0 is 0.

11. Let g : R→ R be a differentiable function such that |g�(x)| ≤ M for all x ∈ R. For

what values of ε will the function f (x) = x+ εg(x) be necessarily one-to-one?

Solution: If f �(x) �= 0, then f is one-one. f �(x) = 1+ εg�(x) for all x ∈ R. Suppose

f �(x) = 0 for all x ∈ R.

1+ εg�(x) = 0 ⇒ g�(x) =−1ε⇒ 1

ε= |g�(x)|≤ M,

i.e.1ε≤ M. Therefore, if

1M

> ε, then f �(x) �= 0, i.e. if ε <1M, then f is one-one.

12. Differentiate:

f (x) =� x2

xet2

dt, x > 1.

Solution: Since

ddx

� β (x)

α(x)F(x, t)dt =

� β (x)

α(x)

∂∂x

F(x, t)dt +F(x,β (x))dβ (x)

dx−F(x,α(x))

dα(x)dx

,

then

f �(x) =ddx

� x2

xet2

dt =� x2

x

∂∂x

et2dt + ex4

2x− ex2= 2xex4 − ex2

.

13. Let f (x) = cosx−1+ x2

2! , x ∈ R. Pick out the true statements:

(a) f (x)≥ 0 for x ≥ 0 and f (x)≤ 0 for x ≤ 0.

(b) f is a decreasing function on the entire real line.

Solution:

(a) False

Let x =−2π < 0. Then f (x) = cos(−2π)−1+ 4π2

2 = cos(2π)−1+2π2 = 2π2 > 0.

(b) False

f (0) = 0 and f (π2 ) = −1+ π2

8 > 0. Since 0 < π2 but f (0) < f (π

2 ). Therefore, f is

non-decreasing function.


14. Let f : [−1,2]→ R be given by f (x) = 2x3 − x4 −10. What is the value of x where

f assumes its minimum value?

Solution: Continuous function on a compact attains its maximum and minimum

values on its boundary. Therefore, f (x) attains minimum on its boundary. Since

f (−1) =−2−1−10 =−13 and f (2) = 16−16−10 =−10, f (x) attains its mini-

mum value at x =−1.

15. Does the integral� 1

0 logxdx exist?. If it exists, give its value.

Solution: Yes.

� 1

0logxdx =

�x logx− x

�1

0= (1log1−1)− (0log0−0) =−1.

16. Let

f (x,y) =

1 if |y|≥ x2,

0 otherwise .

Does the directional derivative of f in the direction of (1,10−6) at the point (0,0)

exist? If yes, give its value.

Solution: Yes. The directional derivative of z = f (x,y) in the direction of unit vector

u = (a,b) is

Du f (x,y) = limh→0

f (x+ah,y+bh)− f (x,y)h

.

Here u = (1,10−6) and (x,y) = (0,0). Therefore,

Du f ((0,0)) = limh→0

f (0+h,0+10−6h)− f (0,0)h

= limh→0

f (h,10−6h)−1h

h is sufficiently small enough

= limh→0

1−1h

= 0.

17. Write down a solution, other than the zero function, of the differential equation:

yy�� − (y�)2 = 0.

Solution: Let y(x) be a non-zero solution of the differential equation yy�� − (y�)2 =

0. Thenyy�� − (y�)2

y2 = 0, i.e.ddx

�y�

y

�= 0. This implies that,

y�

y= c, a constant.

Therefore, y = ecx is a solution, in particular y = ex is a solution.

1.1 October 22, 2005 21

18. Write down the general solution of the system of differential equations:

f �(x) = g(x); g�(x) =− f (x)

Solution: Since f �(x) = g(x), f ��(x) = g�(x) =− f (x). Then f ��(x)+ f (x) = 0. There-

fore, the general solution of this system of equations is f (x) = Acosx+Bsinx.

19. Let u and v be two twice continuously differentiable functions on R2 satisfying the

Cauchy-Riemann equations. Let z = x+ iy. Define f (z) = u(x,y)+ iv(x,y). Express

the complex derivative of f , i.e. f �(z), in terms of the partial derivatives of u and v.

Solution: Since f (z) = u(x,y)+ iv(x,y), then

f �(z) = limΔz→0

f (z+Δz)− f (z)Δz

= limΔx→0

u(x+Δx,y)−u(x,y)Δx

+ i limΔx→0

v(x+Δx,y)− v(x,y)Δx

.

Since the partial derivatives of u and v with respect to x exists, the derivative of f (z)

can be written as f �(z) = ux + ivx.

20. Let (X ,d) be a metric space and let E ⊂ X . For ∈ X , define

d(x,E) = infy∈E

d(x,y).

Pick out the true statements:

(a) |d(x,E)−d(y,E)|≤ d(x,y) for all x and y ∈ X .

(b) d(x,E) = d(x,y0) for some y0 ∈ E.

Solution:

(a) True

Let x,y ∈ X and let z ∈ E. Then d(x,E) ≤ d(x,z) ≤ d(x,y) + d(y,z). Therefore,

d(x,E)−d(x,y)≤ d(y,z) for all z ∈ E. This implies that, d(x,E)−d(x,y)≤ d(y,E),

i.e.

d(x,E)−d(y,E)≤ d(x,y). (1.1)

Also, d(y,E) ≤ d(y,z) ≤ d(y,x)+ d(x,z). Therefore, d(y,E)− d(y,x) ≤ d(x,z) for


all z ∈ E. This implies that, d(y,E)−d(x,y)≤ d(x,E), i.e

d(x,E)−d(y,E)≥−d(x,y). (1.2)

From equations (1.1)&(1.2), we have |d(x,E)−d(y,E)|≤ d(x,y).

(b) False

Let X = R with usual metric and let E = (0,1)⊂ R be an open interval. Let x = 1.

Then d(x,E) = 0. Suppose d(x,E) = d(x,y0) for some y0 ∈ E. Then d(x,y0) = 0,

i.e. x = y0, which is a contradiction to 1 /∈ E.

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