## Solved Assignment CS-60 Foundation Course in Mathematics in Computing

Q1. Trace the following curve

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

Solution:

Given that          y=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:

« Solved Assignment MCS-013 Discrete Mathematics Q-3

Posted on Wednesday, April 22nd, 2009 at 6:12 PM under   Year 2009 | RSS 2.0 Feed

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