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Ignou Mca Cs-60 Solved Assignment In Excel

Some people cannot do any educational course on a regular basis. For them there is a way as IGNOU provides them with loads of opportunities in the distant learning format. A degree of Bachelor of Computer Applications is provided by IGNOU for those who cannot do it on a regular basis. This course has been recognized by UGC and AICTE. Study materials prepared by efficient faculty are sent home.

To be eligible for the course the candidate must have passed 10+2 with Mathematics as a distinct subject. Those candidates which do not have Mathematics as a distinct subject have to register both for MTE-03 and CIC along with the first semester of BCA. Candidates are required to complete these two courses successfully before registering for the third semester of BCA. The course has a minimum duration of 3 years. There is no provision for extending the counseling and practical 2nd and 3rd year automatically.

The programme structure for BCA second semester:

There are three papers-

1. CS612 PC Software Application Skills

2. CS60 Foundation Course in Mathematics in Computing

3. CS62 ‘C’ Programming and Data Structure

The papers have two sections or parts- Assignment and Term End Examination. The assignments are to be submitted by the students after solving them at their homes. The university conducts the term end exam two times in a year.

Download IGNOU BCA Semester II Papers

 

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Q1. Trace the following curve

y=x/(1+x2)(5 Marks)

Solution:

Given thaty=x/(1+x2)

therefore y(1+x2) = x

therefore y + x2y = x

therefore dy/dx + x2 (dy/dx) + 2xy = 1

therefore dy/dx(1+x2) = 1 – 2xy

Putting the value of y

dy/dx(1+x2) = 1 – 2x[x/(1+x2)]

= 1 – 2x2/(1+x2)= [1 – x2 / 1 + x2]

therefore dy/dx = [1 – x2 / (1 + x2)2]

From eqn(1)y=x/(1+x2)

So if y=0 and x=0

Then f(x,y) = y - x/(1+x2)= 0

Then curve

Y

0

0.5

0.4

0.3

Infinite

x

0

1

2

3

-1

Curve is traced as follow

Q2. Obtain fifth roots of 4+3i.                                        (5 Marks)

Solution:

Given Complex number is       Z = 4 + 3i

Q3. Prove that |4x – 5y| <= 4|x| + 5|y|(5 Marks)

Solution:

L.H.S = |4x – 5y|

= (|4x – 5y|2)(because |x|2 = x for all x belong R)

= (4x – 5y)

Therefor |4x|2 + |5y|2 – 2|4x|.|5y|

= > (|4x| + |5y|)2 - 2|4x|.|5y|

Since |x|>= 0 for all x belong R

So taking square on both side

|4x – 5y| <= 4|x| + 5|y|

Given equation is

5x3 – 8x2 + 7x + 6 = 0

= > x3 – (8/5)x2 + (7/5)x + (6/5) = 0-----------(1)

Let a, b, c are the roots, so

a + b + c = 8/5-------------------(2)

ab + bc + ca = 7/5--------------------(3)

abc = -6/5--------------------(4)

Now Let roots of cubic equation is p, q, r such that

p = a2 + b2 + ba-------------------(5)

q = b2 +c2 + cb---------------------(6)

r = c2 + a2 + ca--------------------(7)

Then cube equation is

(x - p)(x - q)(x - r) = 0

= > (x - p)[x2 – (q + r)x + qr] = 0

= > x3 – x2 (p+q+r) + x(pq+qr+rp) – pqr = 0------------(8)

Now(p+q+r) = 2(a2 + b2 + c2) + ab + bc + ca

= >(p+q+r) = 2[(a + b + c)2 – 2(ab + bc + ca)] + ab + bc + ca

= >(p+q+r) = 2(a + b + c)2 – 3(ab + bc + ca)

= >(p+q+r) = 2(8/5)2 – 3*(7/5)

= >(p+q+r) = 23/25-----------------------(9)

Similarly,pq+qr+rp = (a2 + b2 + ab)(b2 + c2 + bc)

= a2b2 + a2c2 + a2bc + b4 + b2c2 + b3c + ab3 + abc2 + ab2c

We can use,( a2 + b2 + c2)2 = a4 + b4 + c4 + 2(a2b2 + b2c2 + c2a2)

Write equation in term of a2, b2 and c2

Putting the value from (5), (6) and (7) we get

pq+qr+rp = 22/25-------------------(10)

and pqr =-21/25--------------------(11)

putting the value of (9), (10) and (11) in (8)

= > x3 – x2 (23/25) + x(22/25) – (-21/25) = 0

= > 25x3 – 23x2 + 22x – 21 = 0

Q5. Find the perimeter of the cord

Solution:





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