The Institution of Engineers,India 2003 A.M.I.E.T.E Electronics & Communication Engineering Mathematics - Please find attached the word doc - Question Paper

Detailed Solutions : Mathematics : A-01/C-01/T-01 : JUNE 2003
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1. (a) C Because z = one is a pole for provided function f and it lies outside the circle
|z| = ½ . Therefore, by Cauchy’s Theorem

(b) A Because P (x = 2) = nine P (x = 4) + 90 P (x = 6)
=>
=>
Because m?0, Therefore, 3m2 + m4 – four = 0
=> m = 1

(c) B A vector field is solenoidal if div = 0

(d) D
=> (cos x – sin x)2 (sin x + two cos x) = 0
Its only root which lies in .

(e) A 2y sin x + cos 2x = a
I.F.
Therefore, the solution is provided as

=> 2y sin x + cos 2x = a
(f) A
where u and v are homogeneous functions of order six and 0 respectively. Using Euler’s theorem
= six u + 0 v = six u.

(g) D By Rodrigue’s formula,
Pn(0) = 0 if n is odd.

(h) C
=

2. (a) Let u = a3x2 + b3y2 + c3z2

Let where ? is constant using Lagrange’s multiplier method.
For stationary values,


=> ax = by = cz = k
Then, provide a + b + c = k
Therefore,
Stationary value of u is
.

(b) The region of integration R is bounded by x = 0, y = 0, and lx + my = 1
{projection of lx + my + nz = one on z = 0}


.

3. (a)

This is along normal to the surface and is the maximum directional derivative. Thus is // to line .
Therefore,


Therefore, a = 4?, b= -11?, c = 10? and

Therefore, .

(b) By Divergence theorem,

Now,



Let x = two sin then dx = 2cos , for x = 0, = 0 and for x = 2, =

Now Surface S consists of 3 surfaces, the 1 leaving base S1 (z = 0), 2nd leaving top S2 (z = 3), 3rd the curved surface S3 of cylinder x2 + y2 = four ranging from z = 0, z = 3


On S3 the outer normal is in the direction of . Therefore a unit vector along normal to the curved surface is provided by


, thus



Hence divergence theorem proved.

4. (a) We know that

Putting, , Then

Comparing real and imaginary part

Let

.
Therefore, . Let b > 0, then




Hence,
Putting b = 0, we get

(b) Let U = X(x) Y(y), then formula becomes
.
Let , then
Therefore, X = A cos lx + B sin lx
Y = Cely + De-ly
U (x, y) = (A cos lx + B sin lx) (Cely + De-ly)
(iv) condition provide U (x, y) ? 0 as y? ? => C = 0
? U = (A cos lx + B sin lx) e-ly
(i) provide U (0, y) = 0 => A = 0
Hence, U = B sin lx . e-ly
(iii) provide U (x, 0) = one => one = bl sin x



5. (a) The characteristic formula of A is |A-?I| = 0
=> ?2 – 4? – five = 0
=> A2 – 4A – 5I = 0, by Cayley Hamilton Theorem.
Thus A5 – 4A4 – 7A3 + 11A2 – A – 10I
= (A2 – 4A – 5I) (A3 – 2A + 3I) + A + 5I = A + 5I.

(b) (D2 + 5D + 6) y = e-2x sec2x (1 + two tan x)
A.E. : m2 + 5m + six = 0
m = -2 , -3
C.F. = c1e-2x + c2e-3x




Thus y = c1e-2x + c2 e-3x + e-2x tan x.

6. (a) Let and f(z) = u + iv

since u is an analytic function, thus it must
satisfies C-R equations, thus



Using Milne’s Thomson method, Let x = z , y = 0

, where c is a constant of integration.

(b) Consider the function . It has simple pole at z = 0. where C consists of the part of real axis from –R to -r and from r to R, the small semi circle Cr from –r to r with center at origin and radius r, which is small and large semi-circle CR from R to –R as shown in Fig. f (z) is analytic inside C (z = 0, the only singularity has been deleted by indenting the origin by drawing Cr).

Therefore, by Cauchy’s Theorem,

For CR, we have z = Rei? , 0 ? ? ? ? and Cr , z = rei? , 0 ? ? ? ?


Since reduces from one to as ? increases from 0 to





Putting values in (1) and applying limits r?0, R??, we get


7. (a) Since 31% of items are under 45. Hence 19% of items lies ranging from and 45. Since ,
thus, .Similarly

Solving, we get ? = 10, .

(b) p(A) = Probability of hitting target by A = 3/5
p(B) = Probability of hitting target by B = 2/5
p(C) = Probability of hitting target by C = 3/4
(i) p1 = Chance A, B hit & C fails
=
p2 = Chance B, C hit & A fails = 12/100
p3 = Chance C, A hit & B fails = 27/100
Since all these events are mutually exclusive, therefore,
P(two shots hit the target) = p1 + p2 + p3 = 0.45
(ii) In case atleast 2 shots may hit target, we must also consider case when all hit the target.
p4 = Probability A, B, C hit target = 18/100.
Therefore, P(atleast 2 shot hit the target) = p1 + p2 + p3 + p4 = 63/100 = 0.63.

8. (a) The characteristic formula is |A – ?I| = 0
i.e. ?3 - 7?2 + 36 = 0
=> ? = -2 , 3, 6
These are eigen values of provided matrix A. For eigen vectors we obtain X ? 0, such that (A - ?I) X = 0
For ? = -2
3x + y + 3z = 0
x + 7y + z = 0

Therefore, for ? = -2, eigen vector is (-1, 0, 1)’
Similarly for ? = 3, eigen vector is (1, -1, 1)’
for ? = 6, eigen vector is (1, 2, 1)’
The modal matrix P is provided by


The diagonal matrix D is provided by

(b) We have
i.e. AX = B
(i) (A : B) =
R3?R3 – R1, R2?7R1 – 2R2

If ? = 5, system will have no solution for those values of ?, for which rank A ? rank (A : B). If ? = 5, ? ? 9, then rank (A) = two and rank (A : B) = 3. Hence no solution
(ii) The system admits unique solution iff coefficient matrix is of rank 3
?
Thus for unique solution ? ? five and ? may have any value.
(iii) If ? = 5,? = 9, system of formula have infinitely many solution

9. (a) Let r be base radius and l be slant height of cone.
Total area A = area of base + area of curved surface
= ?r2 + ?rl = ?r (r + l) = ?h2 tan ? (tan ? + sec ?)

?A = 2?h (tan2 ? + tan? sec?) ?h + ?h2 (2tan? sec2? + sec3? + tan2? sec?) ??

The fault in h will be compensated by fault in ?, when
?A = 0
radians = -0.33o .

(b) - (A)
Differentiating (A) partially w.r.t. t, we get

Differentiating (A) partially w.r.t. r, we get



equating (1) and (2), we get n = -3/2.

10. (a) Curl

=
? vector field is irrotational then ? a scalar function ? s.t.

Integrating, we get

Because, field is irrotational



(b)




Therefore, solution is

=> t = one + cx
=> (log z)-1 = one + cx or .
11. (a) We know

For n = 1, two , 3




11. (b)
We have to expand about z = i

Earning:   Approval pending.