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Question
Mathematics
If f1(x), f2(x), f3(x) are three polynomials of degree 2 on R and g(x)=|f1(x) f2(x) f3(x) f1 prime(x) f2 prime(x) f3 prime(x) f1 prime prime(x) f2 prime prime(x) f3 prime prime(x)|, then the value of g(2)-g(1), is
Q. If
f
1
(
x
)
,
f
2
(
x
)
,
f
3
(
x
)
are three polynomials of degree
2
on
R
and
g
(
x
)
=
∣
∣
f
1
(
x
)
f
1
′
(
x
)
f
1
′′
(
x
)
f
2
(
x
)
f
2
′
(
x
)
f
2
′′
(
x
)
f
3
(
x
)
f
3
′
(
x
)
f
3
′′
(
x
)
∣
∣
,
then the value of
g
(
2
)
−
g
(
1
)
, is
404
168
Report Error
A
0
B
1
C
2
D
3
Solution:
∵
g
(
x
)
=
∣
∣
f
1
(
x
)
f
1
′
(
x
)
f
1
′′
(
x
)
f
2
(
x
)
f
2
′
(
x
)
f
2
′′
(
x
)
f
3
(
x
)
f
3
′
(
x
)
f
3
′′
(
x
)
∣
∣
⇒
g
′
(
x
)
=
∣
∣
f
1
′
(
x
)
f
1
′
(
x
)
f
1
′′
(
x
)
f
2
′
(
x
)
f
2
′
(
x
)
f
2
′′
(
x
)
f
3
′
(
x
)
f
3
′
(
x
)
f
3
′′
(
x
)
∣
∣
+
∣
∣
f
1
(
x
)
f
1
′′
(
x
)
f
1
′′
(
x
)
f
2
(
x
)
f
2
′′
(
x
)
f
2
′′
(
x
)
f
3
(
x
)
f
3
′′
(
x
)
f
3
′′
(
x
)
∣
∣
+
∣
∣
f
1
(
x
)
f
1
′
(
x
)
f
1
′′′
(
x
)
f
2
(
x
)
f
2
′
(
x
)
f
2
′′′
(
x
)
f
3
(
x
)
f
3
′
(
x
)
f
3
′′′
(
x
)
∣
∣
=
0
+
0
+
∣
∣
f
1
(
x
)
f
1
′
(
x
)
0
f
2
(
x
)
f
2
′
(
x
)
0
f
3
(
x
)
f
3
′
(
x
)
0
∣
∣
=
0
⇒
g
(
x
)
=
constant
∴
g
(
2
)
−
g
(
1
)
=
0