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Question
Mathematics
Let f(1)=-2and f'(x) ge4.2 for 1 le x le6The smallest possible value of f(6) is
Q. Let
f
(
1
)
=
−
2
and
f
′
(
x
)
≥
4.2
for
1
≤
x
≤
6
The smallest possible value of
f
(
6
)
is
1475
198
Application of Derivatives
Report Error
A
9
B
12
C
15
D
19
Solution:
Using Lagrange’s mean value theorem, for some
c
∈
(
1
,
6
)
f
′
(
c
)
=
5
f
(
6
)
−
f
(
1
)
=
5
f
(
6
)
+
2
≥
4.2
or
f
(
6
)
+
2
≥
21
or
f
(
6
)
≥
19