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Problems
Let
f
(
x
)
=
cos
(
a
1
+
x
)
+
1
2
cos
(
a
2
+
x
)
+
1
2
2
cos
(
a
3
+
x
)
+
⋯
+
1
2
n
−
1
cos
(
a
n
+
x
)
f(x) = \cos(a_1 + x) + \frac{1}{2} \cos(a_2 + x) + \frac{1}{2^2} \cos(a_3 + x) + \dots + \frac{1}{2^{n-1}} \cos(a_n + x)
f
(
x
)
=
cos
(
a
1
+
x
)
+
2
1
cos
(
a
2
+
x
)
+
2
2
1
cos
(
a
3
+
x
)
+
⋯
+
2
n
−
1
1
cos
(
a
n
+
x
)
…
Problem
Let
f
(
x
)
=
cos
(
a
1
+
x
)
+
1
2
cos
(
a
2
+
x
)
+
1
2
2
cos
(
a
3
+
x
)
+
⋯
+
1
2
n
−
1
cos
(
a
n
+
x
)
f(x) = \cos(a_1 + x) + \frac{1}{2} \cos(a_2 + x) + \frac{1}{2^2} \cos(a_3 + x) + \dots + \frac{1}{2^{n-1}} \cos(a_n + x)
f
(
x
)
=
cos
(
a
1
+
x
)
+
2
1
cos
(
a
2
+
x
)
+
2
2
1
cos
(
a
3
+
x
)
+
⋯
+
2
n
−
1
1
cos
(
a
n
+
x
)
where
a
1
,
a
2
,
…
,
a
n
∈
R
a_1, a_2, \dots, a_n \in \mathbb{R}
a
1
,
a
2
,
…
,
a
n
∈
R
. If
f
(
x
1
)
=
f
(
x
2
)
=
0
f(x_1) = f(x_2) = 0
f
(
x
1
)
=
f
(
x
2
)
=
0
, then
∣
x
2
−
x
1
∣
|x_2 - x_1|
∣
x
2
−
x
1
∣
may be equal to a.
π
\pi
π
b.
2
π
2\pi
2
π
c.
3
π
3\pi
3
π
d.
π
2
\frac{\pi}{2}
2
π
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