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JEE Main 29 June 2022 Shift 2 Solved Paper
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© examsnet.com
Question : 7
Total: 90
Let
f
be a real valued continuous function on
[
0
,
1
]
and
f
(
x
)
=
x
+
1
∫
0
(
x
−
t
)
f
(
t
)
d
t
Then, which of the following points
(
x
,
y
)
lies on the curve
y
=
f
(
x
)
?
[29-Jun-2022-Shift-2]
(
2
,
4
)
(
1
,
2
)
(
4
,
17
)
(
6
,
8
)
Validate
Solution:
👈: Video Solution
f
(
x
)
=
(
1
+
1
∫
0
f
(
t
)
dt
)
x
−
1
∫
0
tf
(
t
)
dt
f
(
x
)
=
Ax
−
B
A
=
1
+
1
∫
0
f
(
t
)
dt
=
1
+
1
∫
0
(
At
−
B
)
dt
⇒
A
=
2
(
1
−
B
)
Also
B
=
1
∫
0
tf
(
t
)
dt
=
1
∫
0
(
At
t
2
−
Bt
)
dt
A
=
9
2
B
From
(
2
)
, (3)
A
=
18
13
,
B
=
4
13
so
f
(
6
)
=
8
© examsnet.com
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