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IIT JEE Advanced 2009 Paper 2
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Section:
Mathematics
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© examsnet.com
Question : 49 of 57
Marks:
+1
,
-0
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
sin
2
θ
+
sin
2
2
θ
2\sin^2\theta + \sin^2 2\theta
2
sin
2
θ
+
sin
2
2
θ
= 2
(P)
π
6
\frac{\pi}{6}
6
π
(B) Points of discontinuity of the function
f (x) =
[
6
x
π
]
cos
[
3
x
π
]
\left[\frac{6x}{\pi}\right] \cos\left[\frac{3x}{\pi}\right]
[
π
6
x
]
cos
[
π
3
x
]
, where [y] denotes the largest integer less than or equal to y
(Q)
π
4
\frac{\pi}{4}
4
π
(C) Volume of the parallelopiped with its edges represented by the vectors
i
^
+
j
^
\hat{i} + \hat{j}
i
^
+
j
^
,
+
i
^
+
2
j
^
+\hat{i}+\hat{2j}
+
i
^
+
2
j
^
and
i
^
+
j
^
+
π
k
^
\hat{i}+\hat{j}+\pi\hat{k}
i
^
+
j
^
+
π
k
^
|x + 1| + |x + 2| = 4k has integer solution(s)(R)
π
3
\frac{\pi}{3}
3
π
(D) Angle between vectors
a
⃗
\vec{a}
a
and
b
⃗
\vec{b}
b
where
a
⃗
,
b
⃗
\vec{a},\vec{b}
a
,
b
and
c
⃗
\vec{c}
c
are unit vectors satisfying
a
⃗
+
b
⃗
+
3
c
⃗
\vec{a}+\vec{b}+\sqrt{\vec{3c}}
a
+
b
+
3
c
=
0
⃗
\vec{0}
0
(S)
π
2
\frac{\pi}{2}
2
π
(T) π
[JEE Adv 2009 P2]
(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
sin
2
θ
+
4
sin
2
θ
cos
2
θ
2\sin^2\theta + 4\sin^2\theta \cos^2\theta
2
sin
2
θ
+
4
sin
2
θ
cos
2
θ
= 2
sin
2
θ
+
2
sin
2
θ
(
1
−
sin
2
θ
)
\sin^2\theta + 2\sin^2\theta (1 - \sin^2\theta)
sin
2
θ
+
2
sin
2
θ
(
1
−
sin
2
θ
)
= 1
3
sin
2
θ
−
2
sin
4
θ
3\sin^2\theta - 2\sin^4\theta
3
sin
2
θ
−
2
sin
4
θ
- 1 = 0
⇒ sin θ = ±
1
2
\frac{1}{\sqrt{2}}
2
1
, ± 1
⇒ θ =
π
4
,
π
2
\frac{\pi}{4},\frac{\pi}{2}
4
π
,
2
π
(B) Let y =
3
x
π
\frac{3x}{\pi}
π
3
x
⇒
1
2
\frac{1}{2}
2
1
≤ y ≤ 3 ∀ x ∊
[
π
6
,
π
]
\left[\frac{\pi}{6},\pi\right]
[
6
π
,
π
]
Now, f(y) = [2y] cos[y].
The critical points are
y =
1
2
\frac{1}{2}
2
1
, y = 1 , y =
3
2
\frac{3}{2}
2
3
and y = 3
⇒ points of discontinuity
{
π
6
,
π
3
,
π
2
,
π
}
\{\frac{\pi}{6},\frac{\pi}{3},\frac{\pi}{2},\pi\}
{
6
π
,
3
π
,
2
π
,
π
}
(C)
∣
1
1
0
1
2
0
1
1
π
∣
\begin{vmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & \pi \end{vmatrix}
1
1
1
1
2
1
0
0
π
= π ⇒ volume of parallelepiped = π.
(D) We have
∣
a
⃗
+
b
⃗
∣
|\vec{a}+\vec{b}|
∣
a
+
b
∣
=
3
\sqrt{3}
3
⇒
2
+
2
cos
α
\sqrt{2+2\cos\alpha}
2
+
2
cos
α
=
3
\sqrt{3}
3
⇒ 2 + 2 cos α = 3
⇒ α =
π
3
\frac{\pi}{3}
3
π
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