Detailed Solutions        Mathematics - II         D-23/DC-23       December 2004

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Q1(a)

z₁ = r₁ (Cos q₁ + i Sin q₁)

z₂ = r₂ (Cos q₂ + i Sin q₂)

z₁ z₂ = r₁ r₂ (Cos q₁ + i Sin q₁) (Cos q₂ + i Sin q₂)

= r₁ r₂ [(Cos q₁ Cos q₂ - Sin q₁ Sin q₂) + i (Cos q₁ Sin q₂ +Cos q₂ Sin q₁)]

= r₁ r₂ [Cos (q₁ +q₂ ) + i Sin (q₁ +q₂)]                                                           Ans. B

Q1(b)   If w is cube root of unity then we know that 1+w+w²=0                            Ans. A

Q1(c)   Given  x²-x-12=0  Þ (x-4) (x+3) = 0   Þ x=4,-3                                         Ans. C

Q1(d)  Given

Þ                                  Ans. A

Q1(e) Given

A^-1A = I,  B^-1B = I

Now (AB) (B^-1 A^-1)  = AIA^-1 = AA^-1 = I --......................---------------(1)

Also (B^-1 A^-1) (AB) = B^-1 (A^-1A) B  = B^-1 IB = B^-1 B=I-----------------(2)

from 1 and 2, we get    (AB)^-1 = B^-1 A^-1                                                         Ans. C

Q1(f)    Given   A ( 3,4,5) and   B (6,8,9)

                  Ans. A

Q1(g)   We know that the function ¦(x) = Sin x is periodic and period is 2p            Ans. C

Q1(h)   By definition

=

                                           Ans. B

Q2(a)  

Also |z|² = ----------(2)

From (1) and (2),

Q2(b)   ,

 and

 = 9+16 = 25

from (1) and (3) x²=4, y²=1

from (2) xy is positive so if x=2, y=1 and x=-2, y=-1

Hence

Q3(a)   Given z=Cos q +i Sin qÞ zⁿ = Cos n q + i Sin n q,
z^-n =Cosnq-iSinnqTherefore zⁿ+z^-n= 2 Cos n q.

Q3(b)  Given

.................

Now

        = Cos p + i Sin p = -1

Q4(a)   Since A is invertible matrix, therefore |A| ¹0 Þ|A^T|¹0

Þ A^T is also invertible

Now AA^-1 = I = A^-1A  Þ (AA^-1)^T = I = (A^-1A)^T Þ (A^-1)^T A^T= I = A^T(A^-1)^T

Þ (A^T)^-1 = (A^-1)^T

Q4(b)

  and

,

,

 ,

 , ,

Q.5(a)   = 0, Since

Q5(b) 

C₂ à C₂+C₃

= (x+y+z) 0         [C₁ and C₂ are identical]

= 0

Q6(a)   Let the position vectors of points A and B are  and  respectively. Let P be the point which divides the line joining A and B in the ratio m:n and let be the position vector of P. Then  where O is origin

Given

Now ,

From (i) we get

Q6(b)   ,

, 

Þ sides represented by   are at right angles

Also

\vectors  form the sides of right angled triangle

Q7(a)   Cayley Hamilton Theorem

Every square matrix satisfies its own characteristic equation

Characteristic matrix is

Characteristic equation is

|A-lI| = 0  Þl³-7l²+11l-5 = 0 . By Cayley Hamilton theorem

A³-7A²+11A-5I = 0

Now

 =

A³-7A²+11A-5I

            +

Q7(b)   System of equations is 2x - 3y + z = 0, x +2y – 3z = 0 and 4x – y –2z = 0

This is system of homogeneous equations can be written as

or AX = O, where

A=

Now |A| =

Thus |A| ¹ 0, So, the given system has only the trivial solution given by x=y=z=0

Q8.      The Fourier series of ¦(x) is ¦(x) =

Where ,  ,

Now   

Fourier series is

Q9(a)   Even function:

A function ¦(x) is said to be even function if  ¦(-x) = ¦(x)

Odd Function:

A function ¦(x) is said to be odd function if  ¦(-x) = -¦(x)

Example:

Cos x, x² are even functions and Sin x, x are odd functions

Q9(b)   Let  be the resultant of the three forces,     == 3i + 4j + 5k

Vector moment of  at P about Q

=

=(2i+4j-k)  (3i+4j+5k)

= i [20+4]-j [10+3] +k [8-12]  =24i-13j-4k

Q10(a) , where function f(t) is defined for t ³ 0 and  s>0 is a        parameteparameter.

= =

====

Q10(b) =

=

=  =

=

Q11(a) Differential equation is

Let y = e^mx is the solution of given differential equation.

The auxiliary equation is m²-7m+12 = 0  Þ(m-4) (m-3) = 0 Þm=3,4

solution is  y=C₁e^3x + C₂e^4x

Q11(b) Given

Taking Laplace transform of both sides, we have

Þ

But y(0) = 1, y¢(0) = 0Þ