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IIT JEE Advanced 2009 Paper 2
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© examsnet.com
Question : 49
Total: 57
Match the statements/expressions in Column I with the values given in Column II:
Column I
Column II
(A) Root(s) of the expression
2
s
i
n
2
θ
+
s
i
n
2
2
θ
= 2
(P)
π
6
(B) Points of discontinuity of the function
f (x) =
[
6
x
π
]
c
o
s
[
3
x
π
]
, where [y] denotes the largest integer less than or equal to y
(Q)
π
4
(C) Volume of the parallelopiped with its edges represented by the vectors
^
i
+
^
j
,
+
^
i
+
2
^
j
and
^
i
+
^
j
+
π
^
k
|x + 1| + |x + 2| = 4k has integer solution(s)(R)
π
3
(D) Angle between vectors
→
a
and
→
b
where
→
a
,
→
b
and
→
c
are unit vectors satisfying
→
a
+
→
b
+
√
3
→
c
=
→
0
(S)
π
2
(T) π
(A)→(Q), (S); (B)→(P), (R), (S), (T); (C)→(T); (D)→(R)
(A)→(Q), (S); (B)→(P), (Q), (S), (T); (C)→(T); (D)→(S)
(A)→(P), (S); (B)→(P), (R), (S), (T); (C)→(S); (D)→(Q)
(A)→(R), (S); (B)→(P), (R), (S), (T); (C)→(T); (D)→(S)
Validate
Solution:
(A) We have
2
s
i
n
2
θ
+
4
s
i
n
2
θ
c
o
s
2
θ
= 2
s
i
n
2
θ
+
2
s
i
n
2
θ
(
1
–
s
i
n
2
θ
)
= 1
3
s
i
n
2
θ
−
2
s
i
n
4
θ
- 1 = 0
⇒ sin θ = ±
1
√
2
, ± 1
⇒ θ =
π
4
,
π
2
(B) Let y =
3
x
π
⇒
1
2
≤ y ≤ 3 ∀ x ∊
[
π
6
,
π
]
Now, f(y) = [2y] cos[y].
The critical points are
y =
1
2
, y = 1 , y =
3
2
and y = 3
⇒ points of discontinuity
{
π
6
,
π
3
,
π
2
,
π
}
(C)
|
1
1
0
1
2
0
1
1
π
|
= π ⇒ volume of parallelepiped = π.
(D) We have
|
→
a
+
→
b
|
=
√
3
⇒
√
2
+
2
c
o
s
α
=
√
3
⇒ 2 + 2 cos α = 3
⇒ α =
π
3
© examsnet.com
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