smu assignment of Bsc IT 1st sem Mathematics for IT

                                              

                                                     [Fall 2014] ASSIGNMENT

PROGRAM	B.SC-IT
SEMESTER	1
SUBJECT CODE &          NAME	BT0063- MATHEMATICS FOR IT

Q NO.1 Differentiate x power of sin x w.r.t. x

Answer: put y = x power sin x

             log y = log(x power sin x)= sin x log x           (log A power B = B log A )

               d/dx(log y )=d/dx(sin x log x)

             1/y. dy/dx = sin x d/dx(log x)+log x .d/dx(sin x)   (using product rule)

             1/y. dy/dx = sinx.1/x + log x. cos x

         So,     dy/dx = y{1/x sin x + log x .cos x}

           Thus,

                       d/dx(x power sinx) = x power sin x {1/x sinx+ logx.cos x}

Q NO.2 Prove that the set Z4={0, 1, 2, 3,} is an abelian group w.r.t. addition modulo 4.

Answer: From the composition table w.r.t. addition module 4 as below:

  

+4	0	1	2	3
0	0	1	2	3
1	1	2	3	0
2	2	3	0	1
3	3	0	1	2

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

1.     Closure law. From the above composition table for all a, b E G, a + 4b also belongs to Z4.

2.      Associates law. Since a+(b+c) and (a+b) +c leave the same remainder when divided by 4 , we have
         a + 4(b + 4c) = (a +4b) +4c.

3.     Existence of identity elements. From the above table, we observe that                

0 E Z4 satisfies a + 4o = 0 + 4a= a for every a E Z4.

Hence, 0 is the identity element.

4.     Existence of inverse. From the above table, the inverse  of 0, 1, 2, 3 are respectively, 0, 3, 2, 1 because 0+4o=0, 1 +4 3 = 0, 2 +4 2 = 0, and

3+4 1=0.

Hence(z4, +4) is a group

Further, since a+b and b+a leave the same remainder when divided by 4,

a+4b= b+4 a.

hence, (Z4, +4) is an abelian group.

Similarly, we can show that the set of remainders of 5 viz.

Z5= {0, 1, 2, 3, 4} from an abelian group under addition (mod 5).

In general  the set of remainders of a positive integer m.

Zm = {0, 1 ,2, …… (m- 1)} from an abelian group under addition (mod m).

Q NO.3 Find the sum to 1.2.3/2!+2.4.5/3!+3.6.7/4!+………..

Answer:

S

 =

Consider,

4nsquare+2nsquare=a+b(n+1)+c(n+1)n+d(n+1)n(n-1)

= a+bn+b+cnsquare+cn+dnqube-dn

= dncube+cnsquare+(b+c-d)n+(a+b)

Hence:

d=4, c=2, b+c-d=0=>b+2-4=0=>2.

a+b=0=> a+2=0 or a=-2

4ncube+2nsquare=-2+2(n+1)+2(n+1)n+4(n+1)n(n+1)

= -2[e-{1+1/1!}]+2[e-1]+2e+4e

Q NO.4 One third of the student in a class are girls and the rest are boys. The probability that a girl gats 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.

Answer: let A, B and C denotes the event that a student is a boy, a girl and a student having first class. We are given the following

P(A)=2/3,   P(B)=1/3,  P(C/A)=3/10 AND P(C/B)=4/10

So,

   P(A∩C)=P(C/A)P(A)=3/10.2/3=1/5.

Similarly P(C∩B)=4/30

P(C)=P(C∩(AUB)) since A U B=5

       = P((C∩A)U(C∩B))by Demorgan’s law

       = P(C∩A)+P(C∩B)by Addition theorem

       = 3/10+4/30=13/30

We are required to find P(B/C)

   P(B/C)=P(B∩C))/P(C)=4/30 / 13/30 = 4/13

Q NO.5 Evaluate:

ʃxsquare sininverse x dx

                                      

Answer:

I = ʃ xsquare sininverse x dx

Let u= sin inverse x

     dv= xsquare dx 

       v= xcube/3

I=ʃudv=uv - ʃvdu=xcube/3 sininverse x - ʃ xcube/3 . 1/√1-xsquare . dx

I=x cube/3 sininverse x – 1/3ʃxsquare/1- xsquare . xdx

Put 1-xsquare=tsquare in the second term;

xsquare=1-tsquare;

2xdx=-2tdt;

Xdx=-tdt

ʃxsquare/√1-xsquare . xdx=ʃ(1-tsquare)(-tdt)/t

                                       =  ʃ-1dt + ʃtsquare dt = -t +tcube/3 +c

                                   = t/3(tsquare-3)+c

                                    = (1-xsquare)power half/3(-xsquare-2)+c

Hence

  I= xcube/3 sin inverse x -1/3(1-xsquare)power half/3(-xsquare-2)+c

 = xcube/3 sininverse x + 1/9√1-xsquare(2+xsquare)+c

Q NO.6 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 are 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 the standard deviation?

Answer:

Given number of children            mean mark                 S.D. of marks

   N1=63                                       X bar1=27.6                  ϭ1=7.1

    N2=26                                         Xbar2=19.2                  ϭ2=6.2

So,

Combined mean Xbar base12 = N1 Xbar1+N2 Xbar2/N1+N2

                                               = 63 . 27.6 + 26 . 19.2/63+26

                                                = 1738.2 + 499.2/89

                                                = 2238.0/89

                                               = 25.15

    Combined standard deviation:

ϭ12= √N1 ϭ1square+N2 ϭ2square+N1 d1square+N2 d2square/N1+N2

    = √63 . (7.1) square+26 . (6.2) square+63 . (2.25) square +26(-5.95) square

    =  √5473.8925/89

    =√61.5044

    = 7.84