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JEE Advanced 2020 Concept Recapitulation Test 1 Paper 2
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© examsnet.com
Question : 48
Total: 54
Consider the functions
f
(
x
)
and
g
(
x
)
, both defined from
R
→
R
and are defined as
f
(
x
)
=
2
x
−
x
2
and
g
(
x
)
=
x
n
where
n
∈
N
. If the area between
f
(
x
)
and
g
(
x
)
in first quadrant is
1
2
then
n
is a divisor of
12
15
20
30
Validate
Solution:
Solving
f
(
x
)
=
2
x
−
x
2
and
g
(
x
)
=
x
n
We have
2
x
−
x
2
=
x
n
⇒
x
=
0
and
x
=
1
A
=
1
∫
0
(
2
x
−
x
2
−
x
n
)
d
x
=
x
2
−
x
3
3
−
x
n
+
1
n
+
1
]
0
1
=
1
−
1
3
−
1
n
+
1
=
2
3
−
1
n
+
1
Hence,
2
3
−
1
n
+
1
=
1
2
⇒
2
3
−
1
2
=
1
n
+
1
⇒
4
−
3
6
=
1
n
+
1
⇒
n
+
1
=
6
⇒
n
=
5
Hence
n
is a divisor of
15
,
20
,
30
© examsnet.com
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