**Q1. Trace the following curve**

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

Solution:

Given that y=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:**