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Section:
Mathematics
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© examsnet.com
Question : 25
Total: 30
Match the statements given in Column I with the intervals/union of intervals given in Column II:
Column I
Column II
(A) The set
{
R
e
(
2
i
z
1
−
z
2
)
:
z
‌
is
‌
a
‌
complex
‌
number
,
|
z
|
=
1
,
z
â‰
±
1
}
is
(P) (−∞,−1)∪(1,∞)
(B) The domain of the function
f (x) =
s
i
n
−
1
(
8
(
3
)
x
−
2
1
−
3
2
(
x
−
1
)
)
is
(Q) (−∞,0)∪(0,∞)
(C) If f (θ) =
|
1
t
a
n
θ
1
−
t
a
n
θ
1
t
a
n
θ
−
1
−
t
a
n
θ
1
|
, then the set
{
f
(
θ
)
:
0
≤
θ
<
Ï€
2
}
is
(R) [2, ∞)
(D) If f (x) =
x
3
2
(3x −10), x ≥ 0, then f (x) is increasing in
(S) (−∞,−1)∪(1,∞)
(T) (−∞,0)∪(2,∞)
[JEE Adv 2011 P2]
(A)→(P), (B)→(Q), (C)→(R), (D)→(R)
(A)→(S), (B)→(T), (C)→(R), (D)→(R)
(A)→(Q), (B)→(T), (C)→(S), (D)→(R)
(A)→(R), (B)→(T), (C)→(P), (D)→(S)
Validate
Solution:
(A) z =
2
i
(
x
+
i
y
)
1
−
(
x
+
i
y
)
2
=
2
i
(
x
+
i
y
)
1
−
(
x
2
−
y
2
+
2
i
x
y
)
Using
1
–
x
2
=
y
2
, we get
Z =
2
i
x
−
2
y
2
y
2
−
2
i
x
y
=
−
1
y
Since - 1 ≤ y ≤ 1 ⇒
−
1
y
or
−
1
y
≥ 1.
(B) For domain:
- 1 ≤
8
(
3
x
−
2
)
1
−
3
2
(
x
−
1
)
≤ 1 ⇒ - 1 ≤
3
x
−
3
x
−
2
1
−
3
2
x
−
2
≤ 1
Case 1 :
3
x
−
3
x
−
2
1
−
3
2
x
−
2
−
≤ 0.
⇒
(
3
x
−
1
)
(
3
x
−
2
−
1
)
(
3
2
x
−
2
−
1
)
≥ 0
⇒ x ∊ (- ∞ , 0] ∪ (1 , ∞)
Case 2:
3
x
−
3
x
−
2
1
−
3
2
x
−
2
+
1
≥ 0
⇒
(
x
x
−
2
−
1
)
(
3
x
+
1
)
(
3
x
.
3
x
−
2
−
1
)
≥ 0
⇒ x ∊ (- ∞ , 1) ∪ [2 , ∞)
So, x ∈(−∞,0] ∪ [2,∞).
(C) R
R
1
→
R
1
+
R
3
:
f (θ) =
|
0
0
2
−
t
a
n
θ
1
t
a
n
θ
−
1
−
t
a
n
θ
1
|
=
2
(
t
a
n
2
θ
+
1
)
=
2
s
e
c
2
θ
(D) f' (x) =
3
2
(
x
)
1
2
(3x - 10) +
(
x
)
3
2
× 3 =
15
2
(
x
)
1
2
(x - 2)
Increasing, when x ≥ 2.
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