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Problem
Problem
If
f
(
x
)
f(x)
f
(
x
)
is twice differentiable function such that
f
(
a
)
=
0
,
f
(
b
)
=
2
,
f
(
c
)
=
−
1
,
f
(
d
)
=
2
,
f
(
e
)
=
0
f(a) = 0, f(b) = 2, f(c) = -1, f(d) = 2, f(e) = 0
f
(
a
)
=
0
,
f
(
b
)
=
2
,
f
(
c
)
=
−
1
,
f
(
d
)
=
2
,
f
(
e
)
=
0
where,
a
<
b
<
c
<
d
<
e
a < b < c < d < e
a
<
b
<
c
<
d
<
e
, then the minimum number of zeros of
g
(
x
)
=
(
f
′
(
x
)
)
2
+
f
′
′
(
x
)
f
(
x
)
g(x) = (f'(x))^2 + f''(x) f(x)
g
(
x
)
=
(
f
′
(
x
)
)
2
+
f
′′
(
x
)
f
(
x
)
in the interval
[
a
,
e
]
[a, e]
[
a
,
e
]
is
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