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Byju's Answer
Standard XII
Mathematics
Differentiation of a Determinant
If Δx=|[ xn...
Question
If
Δ
(
x
)
=
∣
∣ ∣ ∣
∣
x
n
s
i
n
x
c
o
s
x
n
!
s
i
n
n
π
2
c
o
s
n
π
2
a
a
2
a
3
∣
∣ ∣ ∣
∣
, then the value of
d
n
d
x
n
[
Δ
(
x
)
]
a
t
x
=
0
is
A
-1
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B
1
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C
Dependent of a
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Solution
d
n
d
x
n
[
Δ
(
x
)
]
=
∣
∣ ∣ ∣ ∣
∣
d
n
d
x
n
x
n
d
n
d
x
n
s
i
n
x
d
n
d
x
n
c
o
s
x
n
!
s
i
n
(
n
π
2
)
c
o
s
(
n
π
2
)
a
a
2
a
3
∣
∣ ∣ ∣ ∣
∣
=
∣
∣ ∣ ∣ ∣
∣
n
!
s
i
n
(
x
+
n
π
2
)
c
o
s
(
x
+
n
π
2
)
n
!
s
i
n
(
n
π
2
)
c
o
s
(
n
π
2
)
a
a
2
a
3
∣
∣ ∣ ∣ ∣
∣
⇒
[
Δ
n
(
x
)
]
x
=
0
=
∣
∣ ∣ ∣ ∣
∣
n
!
s
i
n
(
0
+
n
π
2
)
c
o
s
(
0
+
n
π
2
)
n
!
s
i
n
(
n
π
2
)
c
o
s
(
n
π
2
)
a
a
2
a
3
∣
∣ ∣ ∣ ∣
∣
=
0
{
S
i
n
c
e
R
1
≡
R
2
}
.
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0
Similar questions
Q.
If
Δ
(
x
)
=
∣
∣ ∣ ∣
∣
x
n
s
i
n
x
c
o
s
x
n
!
s
i
n
n
π
2
c
o
s
n
π
2
a
a
2
a
3
∣
∣ ∣ ∣
∣
, then the value of
d
n
d
x
n
[
Δ
(
x
)
]
a
t
x
=
0
is
Q.
If
f
(
x
)
=
∣
∣ ∣ ∣
∣
x
n
s
i
n
x
c
o
s
x
n
!
s
i
n
(
n
π
2
)
c
o
s
(
n
π
2
)
a
a
2
a
3
∣
∣ ∣ ∣
∣
then the value of
d
n
f
d
x
n
at
x
=
0
, is equal to
Q.
If
f
(
x
)
=
∣
∣ ∣ ∣
∣
x
n
sin
x
cos
x
n
!
sin
n
π
2
cos
n
π
2
a
a
2
a
3
∣
∣ ∣ ∣
∣
, then the value of
d
n
d
x
n
(
f
(
x
)
)
at
x
=
0
for
n
=
2
m
+
1
is
Q.
If
f
(
x
)
=
∣
∣ ∣ ∣
∣
x
n
sin
x
cos
x
n
!
sin
n
π
2
cos
n
π
2
a
a
2
a
3
∣
∣ ∣ ∣
∣
, then the value of
d
n
d
x
n
(
f
(
x
)
)
at
x
=
0
for
n
=
2
m
+
1
is
Q.
If
f
(
x
)
=
∣
∣ ∣ ∣ ∣
∣
x
n
sin
x
−
cos
x
n
!
sin
(
n
π
2
)
cos
(
n
π
2
)
a
a
2
a
3
∣
∣ ∣ ∣ ∣
∣
, then the value of
d
n
d
x
n
(
f
(
x
)
)
at
x
=
0
for
n
=
2
m
+
1
is
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