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Problems
Let
h
(
x
)
=
{
ln
(
1
+
2
x
)
−
2
x
x
2
,
x
<
0
,
m
x
+
n
,
x
≥
0
,
h(x)= \begin{cases} \dfrac{\ln(1+2x)-2x}{x^{2}}, & x<0,\\ mx+n, & x\ge 0, \end{cases}
h
(
x
)
=
⎩
⎨
⎧
x
2
ln
(
1
+
2
x
)
−
2
x
,
m
x
+
n
,
x
<
0
,
x
≥
0
,
…
Mathematics
Problem
Let
h
(
x
)
=
{
ln
(
1
+
2
x
)
−
2
x
x
2
,
x
<
0
,
m
x
+
n
,
x
≥
0
,
h(x)= \begin{cases} \dfrac{\ln(1+2x)-2x}{x^{2}}, & x<0,\\ mx+n, & x\ge 0, \end{cases}
h
(
x
)
=
⎩
⎨
⎧
x
2
ln
(
1
+
2
x
)
−
2
x
,
m
x
+
n
,
x
<
0
,
x
≥
0
,
where
m
m
m
and
n
n
n
are real constants. If
h
h
h
is continuous at
x
=
0
x=0
x
=
0
, what is the value of
n
n
n
? a)
−
1
-1
−
1
b)
−
2
-2
−
2
c)
0
0
0
d)
2
2
2
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