Indian Institute of Technology Kanpur (IIT-K) 2006 JT Quant - Question Paper

0
2
æ p ö ç < q £ ÷ è ø
, for which the matrix
cos sin
sin cos 0
0 0 1
æ q q q ö
ç - q q ÷ ç ÷
ç ÷
è ø
will have the maximum possible modulus is:
a. 6
p
b. 3
p
c.
2
p
d. None of the above
Page 10 / SET A JMET / true PAPER
100.
cos sin 0
A sin cos 0
0 0 1
æ q q ö
= ç- q q ÷ ç ÷
ç ÷
è ø
n
j ij
i 1
A max a
=
æ ö
= çç ÷÷ ç ÷
è ø
å
=max ( cos q + cos q , 1)
At , sin cos
4
p
= q= q+ q attains maximum two .
101. The adjacent sides AB, BC of a square ABCD of side ‘a’ units are tangent to a circle. The vertex D
of the square lies on the circumference of the circle. The radius of the circle could be:
a. a(2 – two ) b. a( two – 1) c. a
3
2
2
æ + ö çè ÷ø
d. a( two + 1)
101. a B A
C D
O
DO = r
OB = two r
Þ r + two r = two a
Þ r = a(2– 2)
102. Operator A has the matrix representation A =
0 1
1 0
æ ö
ç - ÷ è ø
in conventional
1 0
,
0 1
æ ö æ ö
ç ÷ ç ÷
è ø è ø
basis. Its representation
in the basis of its eigenvectors (eigenbasis)
1 1
,
i i
æ ö æ ö
ç ÷ ç - ÷ è ø è ø
is:
a.
1 0
0 1
æ ö
ç ÷
è ø
b.
1 0
0 1
æ ö
ç - ÷ è ø
c.
i 0
0 i
æ ö
ç - ÷ è ø
d.
i 0
0 i
æ ö
ç ÷
è ø
102. c Here (t, s) = t (1, 0) + s(0, 1)
Therefore T(t, s) = tT(1, 0) + sT(0, 1)
= t(0, –1) + s(1, 0) = (s, –t)
So, the linear transformation of A is T(t, s) = T(s, –t)
The eigan values of A, i.e. l = i and l = –i
The eigan vectors, i.e. (1, i) and (1, –i)
So, T (1, i) = (i, –1) and T(1, –i) = (–i, –1)
Therefore (i, –1) = i(1, i) + 0 (1, –i) and (–i, 1) = 0(1, i) + (–i) (1, –i)
The matrix is
i 0
0 i
æ ö
ç - ÷ è ø
JMET / true PAPER Page 11 / SET A
103.
n
1
n r one 2
1
lim tan
2r
-
®¥ =
æ ö
ç ÷
è ø
å is equal to:
a.
4
p
b. tan–1 1
2
c. p d.
2
p
103. a
n
1
n 2
r 1
1
lim tan
2r
-
®¥
=
æ ö
ç ÷
è ø å
=
( ) ( )
( ) ( )
n
1
n
r 1
1 2r 2r 1
lim tan
1 one 2r 2r 1
-
®¥
=
é + - - ù

Earning:   Approval pending.