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JEE Main Math Class 11 Trigonometric Functions Questions
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© examsnet.com
Question : 8
Total: 115
If
2
tan
2
θ
−
5
s
e
c
θ
=
1
has exactly 7 solutions in the interval
[
0
,
n
π
2
]
, for the least value of
n
∈
N
then
n
∑
k
=
1
k
2
k
is equal to :
[27-Jan-2024 Shift 2]
1
2
15
(
2
14
−
14
)
1
2
14
(
2
15
−
15
)
1
−
15
2
13
1
2
13
(
2
14
−
15
)
Validate
Solution:
👈: Video Solution
2
tan
2
θ
−
5
s
e
c
θ
−
1
=
0
⇒
2
s
e
c
2
θ
−
5
s
e
c
θ
−
3
=
0
⇒
(
2
s
e
c
θ
+
1
)
(
s
e
c
θ
−
3
)
=
0
⇒
s
e
c
θ
=
−
1
2
,
3
⇒
cos
θ
=
−
2
,
1
3
⇒
cos
θ
=
1
3
For 7 solutions
n
=
13
So,
13
∑
k
=
1
k
2
k
=
S
(say)
S
=
1
2
+
2
2
2
+
3
2
3
+
.
.
.
+
13
2
13
1
2
S
=
1
2
2
+
1
2
3
+
.
.
.
.
+
12
2
13
+
13
2
14
⇒
S
2
=
1
2
⋅
1
−
1
2
13
1
−
1
2
−
13
2
14
⇒
S
=
2
⋅
(
2
13
−
1
2
13
)
−
13
2
13
© examsnet.com
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