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JEE Main Math Class 11 Relations and Functions Part 2 Questions
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© examsnet.com
Question : 95
Total: 95
Let
f
(
x
)
=
x
−
1
x
+
1
,
x
∈
R
−
{
0
,
−
1
,
1
}
. If
f
n
+
1
(
x
)
=
f
(
f
n
(
x
)
)
for all
n
∈
N
, then
f
6
(
6
)
+
f
7
(
7
)
is equal to :
[26-Jun-2022-Shift-1]
7
6
−
3
2
7
12
−
11
12
Validate
Solution:
👈: Video Solution
Given,
f
(
x
)
=
x
−
1
x
+
1
Also given,
f
n
+
1
(
x
)
=
f
(
f
n
(
x
)
)
.
.
.
.
(
1
)
∴
For
n
=
1
f
1
+
1
(
x
)
=
f
(
f
1
(
x
)
)
⇒
f
2
(
x
)
=
f
(
f
(
x
)
)
=
f
(
x
−
1
x
+
1
)
=
x
−
1
x
+
1
−
1
x
−
1
x
+
1
+
1
=
x
−
1
−
x
−
1
x
+
1
x
−
1
+
x
+
1
x
+
1
=
−
2
2
x
=
−
1
x
From equation (1), when
n
=
2
f
2
+
1
(
x
)
=
f
(
f
2
(
x
)
)
⇒
f
3
(
x
)
=
f
(
f
2
(
x
)
)
=
f
(
−
1
x
)
=
−
1
x
−
1
−
1
x
+
1
=
−
1
−
x
−
1
−
1
+
x
x
=
−
1
−
x
−
1
+
x
=
−
(
x
+
1
)
x
−
1
Similarly,
f
4
(
x
)
=
f
(
f
3
(
x
)
)
=
f
(
−
x
+
1
x
−
1
)
=
−
(
x
+
1
)
x
−
1
−
1
−
(
x
+
1
)
x
−
1
+
1
=
x
−
1
−
x
+
1
x
−
1
−
x
−
1
+
x
−
1
x
−
1
=
−
2
x
−
2
=
x
∴
f
5
(
x
)
=
f
(
f
4
(
x
)
)
=
f
(
x
)
=
x
−
1
x
+
1
f
6
(
x
)
=
f
(
f
5
(
x
)
)
=
f
(
x
−
1
x
+
1
)
=
−
1
x
(Already calculated earlier)
f
7
(
x
)
=
f
(
f
6
(
x
)
)
=
f
(
−
1
x
)
=
−
1
x
−
1
−
1
x
+
1
=
−
(
x
+
1
)
x
−
1
∴
f
6
(
6
)
=
−
1
6
and
f
7
(
7
)
=
−
(
7
+
1
)
7
−
1
=
−
8
6
So,
f
6
(
6
)
+
f
7
(
7
)
=
−
1
6
−
8
6
=
−
3
2
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