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JEE Main Math Class 11 Relations and Functions Questions 1 Questions
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© examsnet.com
Question : 2
Total: 100
Let
f
:
R
→
R
be a function defined by
f
(
x
)
=
2
e
2
x
e
2
x
+
e
.
Then
f
(
1
100
)
+
f
(
2
100
)
+
f
(
3
100
)
+
.
.
.
+
f
(
99
100
)
is equal to____
[27-Jun-2022-Shift-1]
Your Answer:
Validate
Solution:
Given,
f
(
x
)
=
2
e
2
x
e
2
x
+
e
∴
f
(
1
−
x
)
=
2
e
2
(
1
−
x
)
e
2
(
1
−
s
)
+
e
=
2
⋅
e
2
e
2
x
e
2
e
2
π
+
e
=
2
e
2
e
2
+
e
2
x
⋅
e
=
2
e
2
e
(
e
+
e
2
x
)
=
2
e
e
+
e
2
∆
∴
f
(
x
)
+
f
(
1
−
x
)
=
2
e
2
x
e
2
x
+
e
+
2
e
e
2
x
+
e
=
2
(
e
2
x
+
e
)
e
2
x
+
e
=
2
.
.
.
.
.
.
(1)
Now,
f
(
1
100
)
+
f
(
99
100
)
=
f
(
1
100
)
+
f
(
1
−
1
100
)
=
2
[as
f
(
x
)
+
f
(
1
−
x
)
=
2
]
f
(
2
100
)
+
f
(
1
−
2
100
)
=
2
⁞
f
(
49
100
)
+
f
(
1
−
49
100
)
=
2
∴
Total sum
=
49
×
2
Remaining term
=
f
(
50
100
)
=
f
(
1
2
)
Put
x
=
1
2
in equation (1), we get
f
(
1
2
)
+
f
(
1
−
1
2
)
=
2
⇒
2
f
(
1
2
)
=
2
⇒
f
(
1
2
)
=
1
∴
Sum
=
49
×
2
+
1
=
99
© examsnet.com
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