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JEE Mains 24-Jan-2023 Shift 2 Solved Paper
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© examsnet.com
Question : 65
Total: 90
If
f
(
x
)
=
x
3
−
x
2
f
′
(
1
)
+
x
f
′
′
(
2
)
−
f
′
′
′
(
3
)
,
x
∈
R
, then
[24-Jan-2023 Shift 2]
3
f
(
1
)
+
f
(
2
)
=
f
(
3
)
f
(
3
)
−
f
(
2
)
=
f
(
1
)
2
f
(
0
)
−
f
(
1
)
+
f
(
3
)
=
f
(
2
)
f
(
1
)
+
f
(
2
)
+
f
(
3
)
=
f
(
0
)
Validate
Solution:
f
(
x
)
=
x
3
−
x
2
f
′
(
1
)
+
x
f
′
′
(
2
)
−
f
′
′
′
(
3
)
,
x
∈
R
Let
f
′
(
1
)
=
a
,
f
′
′
(
2
)
=
b
,
f
′
′
′
(
3
)
=
c
f
(
x
)
=
x
3
−
x
2
+
b
x
−
c
f
′
(
x
)
=
3
x
2
−
2
a
x
+
b
f
′
′
(
x
)
=
6
x
−
2
a
f
′
′
′
(
x
)
=
6
c
=
6
,
a
=
3
,
b
=
6
f
(
x
)
=
x
3
−
3
x
2
+
6
x
−
6
f
(
1
)
=
−
2
,
f
(
2
)
=
2
,
f
(
3
)
=
12
,
f
(
0
)
=
−
6
2
f
(
0
)
−
f
(
1
)
+
f
(
3
)
=
2
=
f
(
2
)
© examsnet.com
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