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Test Index
JEE Mains 28-July-2022 Shift 1 Solved Paper
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© examsnet.com
Question : 20
Total: 90
The minimum value of the twice differentiable function
f
(
x
)
=
∫
0
x
e
x
−
t
f
′
(
t
)
dt
−
(
x
2
−
x
+
1
)
e
x
,
x
∈
R
, is
:
[28-Jul-2022-Shift-1]
−
2
√
e
−
2
√
e
−
√
e
2
√
e
Validate
Solution:
f
(
x
)
=
x
−
t
∫
0
x
e
f
′
(
t
)
d
t
−
(
x
2
−
x
+
1
)
e
x
f
(
x
)
=
e
x
−
t
∫
0
x
e
f
′
(
t
)
d
t
−
(
x
2
−
x
+
1
)
e
x
e
−
x
f
(
x
)
=
−
t
∫
0
x
e
f
′
(
t
)
d
t
−
(
x
2
−
x
+
1
)
Differentiate on both side
e
−
x
f
′
(
x
)
+
(
−
f
(
x
)
e
−
x
)
=
e
−
x
f
′
(
x
)
−
2
x
+
1
f
(
x
)
=
e
x
(
2
x
−
1
)
f
′
(
x
)
=
e
x
(
2
)
+
e
x
(
2
x
−
1
)
=
e
x
(
2
x
+
1
)
x
=
−
1
2
f
′
′
(
x
)
=
e
x
(
2
)
+
(
2
x
+
1
)
e
x
=
e
x
(
2
x
+
3
)
For
x
=
−
1
2
f
′
′
(
x
)
>
0
⇒
Maxima
∴
Max.
=
e
−
1
2
(
−
1
−
1
)
∴
−
2
√
e
© examsnet.com
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