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JEE Advanced 2015 Paper 1
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© examsnet.com
Question : 60
Total: 60
Column I
Column II
(A)
In a triangle
△
X
Y
Z
, let
a
,
b
and
c
be the length of the sides opposite to the angles
X
,
Y
and
Z
, respectively. If
2
(
a
2
−
b
2
)
=
c
2
and
λ
=
s
i
n
(
X
−
Y
)
s
i
n
Z
,then possible values of
n
for which
c
o
s
(
n
π
λ
)
=
0
is (are)
(P)
1
(B)
In a triangle
△
X
Y
Z
,let
a
,
b
and
c
be the length of the sides opposite to the angles
X
,
Y
and
Z
,respectively.If
1
+
c
o
s
2
X
−
2
c
o
s
2
Y
=
2
s
i
n
X
,
s
i
n
Y
,then possible value(s) of
a
b
is (are)
(Q)
2
(C)
In
ℝ
3
,let
√
3
∧
i
+
∧
j
,
∧
i
+
√
3
∧
j
and
β
∧
i
+
(
1
−
β
)
∧
j
be the position vectors of
X
,
Y
and
Z
with respect of the origin, respectively. If the distance of Z from the bisector of the acute angle of
→
O
X
with
→
O
Y
is
3
√
2
. then the possible value(s)of |β| is (are)
(R)
3
(D)
Suppose that
F
(
α
)
denotes the area of the region bounded by
x
=
0
,
x
−
2
,
y
2
=
4
x
and
y
=
|
α
x
−
1
|
=
|
α
x
−
2
|
+
α
x
,
where
α
∊
0
,
1
.Then the value(s) of
F
(
α
)
+
8
3
√
2
.where
α
=
0
and
α
=
1
is (are)
(S)
5
(T)
6
(A)
→ (P, R, S),
(B)
→ (P),
(C)
→ (P, Q),
(D)
→ (S, T)
(A)
→ (P, S, T),
(B)
→ (P),
(C)
→ (P, Q),
(D)
→ (S, T)
(A)
→ (P, Q, S),
(B)
→ (P),
(C)
→ (P, Q),
(D)
→ (S, T)
(A)
→ (P, R, S),
(B)
→ (Q),
(C)
→ (P, Q),
(D)
→ (S, T)
Validate
Solution:
(A)→(P,R,S),(B)→(P),(C)→(P,Q),(D)→(S,T)
a
2
−
b
2
=
c
2
3
(given)
4
R
2
(
s
i
n
2
X
−
S
i
n
2
Y
)
=
4
R
2
2
s
i
n
2
(
Z
)
⇒
2
(
s
i
n
(
X
−
Y
)
.
s
i
n
(
X
+
Y
)
)
=
s
i
n
2
(
z
)
⇒
2.
s
i
n
(
X
−
Y
)
.
s
i
n
(
Z
)
=
s
i
n
2
(
Z
)
⇒
s
i
n
(
X
−
Y
)
s
i
n
Z
=
1
2
=
λ
⇒
c
o
s
(
n
π
2
)
=
0
for n=odd integer
(B)
1
+
c
o
s
2
X
−
2
c
o
s
2
Y
=
2
s
i
n
X
s
i
n
Y
s
i
n
2
X
+
s
i
n
X
s
i
n
Y
−
2
s
i
n
2
Y
=
0
(
s
i
n
X
−
s
i
n
Y
)
(
s
i
n
X
+
2
s
i
n
Y
)
=
0
⇒
s
i
n
X
=
s
i
n
Y
⇒
s
i
n
X
s
i
n
Y
=
a
b
=
1
(C) Here,distance of Z from bisectors of
→
O
X
and
→
O
Y
=
3
√
2
⇒
(
β
−
1
2
)
2
+
(
β
−
1
2
)
2
=
9
2
⇒β=2,-1
⇒|β|=2,1
(D)when α=0
Area
=
6
−
∫
0
2
2
√
x
dx
=
6
−
8
√
2
3
when α=1
Area
=
∫
0
1
(
3
−
x
−
2
√
x
)
d
x
+
∫
1
2
(
x
+
1
−
2
√
x
)
d
x
=
|
3
x
−
x
2
2
−
4
3
x
3
2
|
0
1
+
|
x
2
2
+
x
−
4
3
x
3
2
|
0
1
=
5
−
8
3
√
2
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