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Mathematics
Mathematics
Problem
Let
f
(
x
)
=
{
x
−
3
,
x
<
0
x
2
−
3
x
+
2
,
x
≥
0
f(x)=\begin{cases}x-3,&x<0\\x^2-3x+2,&x\geq 0\end{cases}
f
(
x
)
=
{
x
−
3
,
x
2
−
3
x
+
2
,
x
<
0
x
≥
0
and
g
(
x
)
=
f
(
∣
x
∣
)
+
∣
f
(
x
)
∣
g(x)=f(|x|)+|f(x)|
g
(
x
)
=
f
(
∣
x
∣
)
+
∣
f
(
x
)
∣
. The value of
lim
x
→
π
2
g
(
x
)
\lim_{x\to\frac{\pi}{2}} g(x)
lim
x
→
2
π
g
(
x
)
is equal to (a)
π
2
+
1
\frac{\pi}{2}+1
2
π
+
1
(b)
π
2
2
−
3
π
+
4
\frac{\pi^2}{2}-3\pi+4
2
π
2
−
3
π
+
4
(c)
π
2
2
+
2
π
−
3
\frac{\pi^2}{2}+2\pi-3
2
π
2
+
2
π
−
3
(d) 0
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