bca1030 smu bca fall 2014 Ist sem assignment

Get fully solved assignment, plz drop a mail with your sub code

computeroperator4@gmail.com

Charges for mba rs 125/subject and rs 700/semester only.

For other rs 125/subject only

if urgent then call us on 08791490301, 08273413412

our website is www.smuassignment.in

PROGRAM BCA

SEMESTER 1

SUBJECT CODE &NAME- BCA1030- MATHEMATICS FOR IT

Q1. : Differentiate x sin x  w.r .t.x Solution: Put y = x sin x
log y = log(x sin x )= sinx log x							(Q log AB			= B log A)
	d	(log y ) =			d	(sin x log x)

Q2: Prove that the set Z4 = {0, 1, 2, 3} is an abelian group w.r.t. addition modulo 4.

Solution: Form the composition table w.r.t. addition modulo 4 as below:

Since 1 + 3 = 4 º 0 (mod 4), 3 + 3 = 6 º 2 (mod 4) 2 + 3 = 5 º 1 (mod 4) etc.

1. Closure law.

Q3.

1.2.3

+

2.4.5

+

3.6.7

+ .........

2 !

3 !

4 !

Solution: The given series can be written as,

¥

n(2n) (2n +1)

¥

4n3 + 2n2

S = å

= å

(n +1)!

(n +1)!

n=1

n=1

Consider 4n³ + 2n²  = a + b(n +1)+ c(n +1)n + d(n +1) n(n -1)

=  a + bn + b + cn² + cn + dn³ - dn

=  dn³ + cn² + (b + c - d )n + (a + b)

\ d = 4, c = 2, b + c - d = 0 Þ b + 2 - 4 = 0 Þ b = 2.

\   a + b = 0 Þ a

Q4. One third of the students in a class are girls and the rest are boys. The probability that a girl gets a first class is 0.4 and that of a boy is 0.3. If a student having first class is selected, find the probability that the student is a girl.

Solution: Let A, B and C denote the event that a student is a boy, a girl and a student having first class. We are given the following

P(A) =	2	,	P(B) =	1	,	P(C / A) =								3			and P(C / B) =			4
	3			3									10						10
So P(A Ç C ) = P(C / A) P(A) =								3	.	2	=		1		.	Similarly P(C Ç B) =						4
							10			3			5									30
P(C) = P(C Ç (A È B)) since A È B = 5

=  P((C Ç A) È (C Ç B)) by Demorgan’s law

=   P(C

Q5. Evaluate  _òx ² sin^-1  x dx

Solution

a)  I = _ò x² sin^-1 x dx

Let u = sin-1 x

dv = x2dx

v =

x3

Q6. The mean and standard deviation of 63 children on an average test are respectively 27.6 and 7.1. To them are added a new group of 26 who have less training and whose mean is 19.2 and standard deviation is 6.2. How will the value of combined group differ from those of the original 63 children as to mean and standard deviation?

Solution:

Given number of children

Mean Mark

S.D. of marks

N1

= 63

X1 = 27.6

s1

= 7.1

N2

= 26

X2  =19.2

s2

= 6.2

=

N1

X1 + N2  X2

\ Combined mean X12

Get fully solved assignment, plz drop a mail with your sub code

computeroperator4@gmail.com

Charges for mba rs 125/subject and rs 700/semester only.

For other rs 125/subject only

if urgent then call us on 08791490301, 08273413412

our website is www.smuassignment.in