Tardigrade
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Question
Mathematics
Suppose that f'' (x) is continuous for all x and f(0) = f' (1) = 1 If ∫ limits01 tfn (t)dt=0, then the value of f(1) is
Q. Suppose that
f
′′
(
x
)
is continuous for all
x
and
f
(
0
)
=
f
′
(
1
)
=
1
If
0
∫
1
tf
n
(
t
)
d
t
=
0
,
then the value of f(1) is
1460
216
Integrals
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A
2
33%
B
3
28%
C
4
2
1
26%
D
none of these
13%
Solution:
Since
0
∫
1
t
f
′
(
t
)
d
t
=
0
∴
∣
t
f
′
(
t
)
∣
0
1
−
0
∫
1
1
f
′
(
t
)
d
t
=
0
⇒
f
′
(
t
)
−
0
∫
1
f
′
(
t
)
d
t
=
0
⇒
f
′
(
1
)
−
∣
f
(
t
)
∣
0
1
=
0
⇒
f
′
(
1
)
−
[
f
(
1
)
−
f
(
0
)
]
=
0
⇒
f
′
(
1
)
−
f
(
1
)
+
f
(
0
)
=
0
⇒
f
(
1
)
=
f
′
(
1
)
+
f
(
0
)
=
1
+
1
=
2