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KEAM 2019 Math Paper
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© examsnet.com
Question : 47
Total: 121
Let
f
:
R
→
R
be a differentiable function and
f
(
l
)
=
4
, then
lim
x
→
1
f
(
x
)
∫
4
2
t
x
−
1
d
t
=
?
8
f
′
(
1
)
4
f
′
(
1
)
3
f
′
(
1
)
2
f
′
(
1
)
f
′
(
1
)
2
Validate
Solution:
lim
x
→
1
[
t
2
]
4
f
(
x
)
x
−
1
=
lim
x
→
1
{
f
(
x
)
}
2
−
16
(
x
−
1
)
=
lim
x
→
1
2
f
(
x
)
f
′
(
x
)
−
0
1
=
2
f
(
1
)
f
′
(
1
)
=
2
×
4
×
f
′
(
1
)
=
8
f
′
(
1
)
© examsnet.com
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