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

x1 = {1, 2} , x2 = f
x1 = {1}, x2 = {1, 2}
x1 = {1, 2} , x2 = {1}
x1 = {2}, x2 = {1, 2}
x1 = {1, 2} , x2 ={2}
x1 = {1} , x2 = {2}
x1 = {2} , x2 = {1}
x1 = {1, 2} , x2 = {1, 2}
9 possibilities
( ) ( ) two 2 m n = two -1 = two -1
Now A = {1, 2}
\ n = 2
Let m = 1
x1 = {1, 2}
Only one possibility
i.e. ( ) ( ) one two m n two -1 = two -1
JMET / true PAPER Page five / SET A
89. If A, B and C are the angles of a triangle and eiA, eiB and eiC are in Arithmetic Progression, then the
triangle is:
a. Right angled but not isosceles b. Isosceles but not right angled
c. Equilateral d. Right angled isosceles
89. c 2eiB = eiA + eiC
Þ 2cosB = cosA + cosCA
A C A C
2cosB 2cos cos
2 2
+ -
=
B A C
cosB sin cos
2 2
-
=
2sinB = sinA +sinC
A C A C
sinB sin cos
2 2
+ -
=
B B B A C
2sin cos cos cos
2 two 2 2
-
=
B A C
2sin cos
2 2
-
=
2A C
2cosB cos
2
-
=
Let A = B = C = 60°
A iC one 3
ei e i
2 2
\ - = + form a
constant AP
90. Let (x) and [x] represent the fractional and integral components of x ÎR . We describe f : R ® R; g :
R ® R by f(x) = (x); g(x) = sin[x]p. The range of gof is:
a. (–1, 1) b. {0} c. [1, 1] d. f
90. b gof = sin éë(x )ùû p
0 £ (x) < 1
[(x)] = 0
gof(x) = sin0p = 0
91. If z ÎC lies on the circle whose formula is z – three i = three two , then the argument of
( )
( )
z – 3
z + three is:
a.
4
p
b. tan–13 c. tan–13 two d.
2
p
Page six / SET A JMET / true PAPER
91. a
–3 –2 –1 one two 3
1
2
3
A B x
y
z
C
( )
z three 1
arg ACB
z three 2
æ - ö ç ÷= Ð ç - - ÷ è ø
1
90
2 4
p
= ´ ° =
92.
0
sinx cosx
p
ò + dx is equal to:
a. 0 b. two c. two 2 d.
1
2
92. c
0
sinx cosx dx
p
ò +
0
2sinx dx
4
p
æ p ö
= ç + ÷
è ø ò
0
2 sinx dx
4
p
æ p ö
= ç + ÷
è ø ò
x y
4
p
+ =
dx = dy
5 / 4
/ 4
2 sinydy
p
p
= ò
5 / 4
/ 4
2 sinydy two sinydy
p p
p p
= ò - ò
5 / 4
/ 4
2 cosy two cosy
p p
p p
= - - -
1 1
2 one two 1
2 2
æ ö æ ö
= ç + ÷ - ç - ÷

Earning:   Approval pending.