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Mathematics
Mathematics
Problem
If
f
(
x
)
=
f(x) =
f
(
x
)
=
b
(
[
x
]
2
+
[
x
]
)
+
1
b([x]^2 + [x]) + 1
b
([
x
]
2
+
[
x
])
+
1
for
x
≥
−
1
\text{for } x \ge -1
for
x
≥
−
1
sin
(
π
(
x
+
a
)
)
\sin(\pi(x+a))
sin
(
π
(
x
+
a
))
for
x
<
−
1
\text{for } x < -1
for
x
<
−
1
, where
[
x
]
[x]
[
x
]
denotes the integral part of
x
x
x
, then for what values of
a
,
b
a, b
a
,
b
the function is continuous at
x
=
−
1
x = -1
x
=
−
1
? a.
a
=
2
n
+
(
3
/
2
)
;
b
∈
R
;
n
∈
I
a = 2n + (3/2); b \in \mathbb{R}; n \in \mathbb{I}
a
=
2
n
+
(
3/2
)
;
b
∈
R
;
n
∈
I
b.
a
=
4
n
+
2
;
b
∈
R
;
n
∈
I
a = 4n + 2; b \in \mathbb{R}; n \in \mathbb{I}
a
=
4
n
+
2
;
b
∈
R
;
n
∈
I
c.
a
=
4
n
+
(
3
/
2
)
;
b
∈
R
+
;
n
∈
I
a = 4n + (3/2); b \in \mathbb{R}^+; n \in \mathbb{I}
a
=
4
n
+
(
3/2
)
;
b
∈
R
+
;
n
∈
I
d.
a
=
4
n
+
1
;
b
∈
R
+
;
n
∈
I
a = 4n + 1; b \in \mathbb{R}^+; n \in \mathbb{I}
a
=
4
n
+
1
;
b
∈
R
+
;
n
∈
I
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