## Cs 60 Solved Assignment 2009 Movies

2. Recall any incident in which you had difficulty listening effectively. Describewhich of the factors interfered with your listening effectively.Ans.

One day when I was in market there was a lot of noise of hawkers. It was very crowdedarea suddenly. I met one of my old friends. She was trying to communication with me but I could not listen her properly due to noise in the market. I got distracted from her communication. I had to do many task and there was a lots of other activities were alsotaking place. The main factor that interfered with my listening effectively was gettingdistracted. This can be a problem because it leads to miscommunication, takingattention, and tuning out all together.

3. Given below is a list of ambiguous messages. For each sentence, construct a politequestion you could ask the speaker to help clarify the meaning.a)I need this report right away.Ans.

Please bring this report quickly.

b)This presentation needs to be perfect.Ans.

You should work more on this presentation

c)Whenever I leave to go to a meeting, nothing gets done in this office.Ans.

Please do your work ever when I leave to go to meeting.

d)You did a heck of a job on that proposal.Ans.

You didn’t worked satisfaction on that proposal.

e)I just need a little more time to finish the job.Ans.

I need some more time to finish the job.

4. Write a note on the impact of posture and movement as a non-verbalcommunication.

**Q1. Trace the following curve**

**y=x/(1+x ^{2})(5 Marks)**

Solution:

Given thaty=x/(1+x^{2})

therefore y(1+x^{2}) = x

therefore y + x^{2}y = x

therefore dy/dx + x^{2} (dy/dx) + 2xy = 1

therefore dy/dx(1+x^{2}) = 1 – 2xy

Putting the value of y

dy/dx(1+x^{2}) = 1 – 2x[x/(1+x^{2})]

= 1 – 2x^{2}/(1+x^{2})= [1 – x^{2} / 1 + x^{2}]

therefore dy/dx = [1 – x^{2} / (1 + x^{2})^{2}]

From eq^{n}(1)y=x/(1+x^{2})

So if y=0 and x=0

Then f(x,y) = y - x/(1+x^{2})= 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

5x^{3} – 8x^{2} + 7x + 6 = 0

= > x^{3} – (8/5)x^{2} + (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 = a^{2 }+ b^{2} + ba-------------------(5)

q = b^{2} +c^{2} + cb---------------------(6)

r = c^{2} + a^{2} + ca--------------------(7)

Then cube equation is

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

= > (x - p)[x^{2} – (q + r)x + qr] = 0

= > x^{3} – x^{2} (p+q+r) + x(pq+qr+rp) – pqr = 0------------(8)

Now(p+q+r) = 2(a^{2} + b^{2} + c^{2}) + 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 = (a^{2} + b^{2} + ab)(b^{2 }+ c^{2} + bc)

= a^{2}b^{2} + a^{2}c^{2} + a^{2}bc + b^{4} + b^{2}c^{2} + b^{3}c + ab^{3} + abc^{2} + ab^{2}c

We can use,( a^{2} + b^{2} + c^{2})2 = a^{4} + b^{4} + c^{4} + 2(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})

Write equation in term of a^{2}, b^{2} and c^{2}

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)

= > x^{3} – x^{2} (23/25) + x(22/25) – (-21/25) = 0

= > 25x^{3} – 23x^{2} + 22x – 21 = 0

**Q5. Find the perimeter of the cord**

**Solution:**

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