Tardigrade
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Tardigrade
Question
Mathematics
If f(x)=| cos (x+α) cos (x+β) cos (x+γ) sin (x+α) sin (x+β) sin (x+γ) sin (β-γ) sin (γ-α) sin (α-β)| and f(0)=-2 then displaystyle∑ r =130|f(r)| equals
Q. If
f
(
x
)
=
∣
∣
cos
(
x
+
α
)
sin
(
x
+
α
)
sin
(
β
−
γ
)
cos
(
x
+
β
)
sin
(
x
+
β
)
sin
(
γ
−
α
)
cos
(
x
+
γ
)
sin
(
x
+
γ
)
sin
(
α
−
β
)
∣
∣
and
f
(
0
)
=
−
2
then
r
=
1
∑
30
∣
f
(
r
)
∣
equals
362
86
Determinants
Report Error
A
2
B
30
C
60
D
120
Solution:
f
′
(
x
)
=
0
⇒
f
(
x
)
is constant
∴
∣
f
(
1
)
∣
+
∣
f
(
2
)
∣
+
…
..
+
∣
f
(
30
)
∣
=
60
.