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JEE Main Math Class 11 Relations and Functions Part 2 Questions
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© examsnet.com
Question : 94
Total: 95
Let
f
:
R
→
R
be a function defined by
f
(
x
)
=
(
2
(
1
−
x
25
2
)
(
2
+
x
25
)
)
1
50
. If the function
g
(
x
)
=
f
(
f
(
f
(
x
)
)
+
f
(
f
(
x
)
)
, then the greatest integer less than or equal to
g
(
1
)
is___
[25-Jun-2022-Shift-1]
Your Answer:
Validate
Solution:
👈: Video Solution
Given,
f
(
x
)
=
(
2
(
1
−
x
25
2
)
(
2
+
x
25
)
)
1
50
and
g
(
x
)
=
f
(
f
(
f
(
x
)
)
)
+
f
(
f
(
x
)
)
∴
g
(
1
)
=
f
(
f
(
f
(
1
)
)
)
+
f
(
f
(
1
)
)
Now,
f
(
1
)
=
(
2
(
1
−
1
25
2
)
(
2
+
1
25
)
)
1
50
=
(
2
(
1
−
1
2
)
(
2
+
1
)
)
1
50
=
(
3
)
1
50
∴
f
(
f
(
1
)
)
=
f
(
3
1
50
)
=
(
2
(
1
−
(
3
1
50
)
25
2
)
(
2
+
(
3
1
50
)
25
)
)
1
50
=
(
2
(
1
−
3
1
2
2
)
(
2
+
3
1
2
)
)
1
50
=
(
2
×
(
2
−
√
3
2
)
(
2
+
√
3
)
)
1
50
=
[
(
2
−
√
3
)
(
2
+
√
3
)
]
1
50
=
(
4
−
3
)
1
50
=
1
1
50
=
1
Now,
f
(
f
(
f
(
1
)
)
)
=
f
(
1
)
=
3
1
50
∴
g
(
1
)
=
f
(
f
(
f
(
1
)
)
)
+
f
(
f
(
1
)
)
=
3
1
50
+
1
Now, greatest integer less than or equal to
g
(
1
)
=
[
g
(
1
)
]
=
[
3
1
50
+
1
]
=
[
3
1
50
]
+
[
1
]
=
[
1.02
]
+
1
=
1
+
1
=
2
© examsnet.com
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