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Byju's Answer
Standard XII
Mathematics
Adjoint of a Matrix
If fx=n x n...
Question
If
f
(
x
)
=
n
∣
∣ ∣ ∣ ∣ ∣
∣
x
n
n
!
2
cos
x
cos
n
π
2
4
sin
x
sin
n
π
2
8
∣
∣ ∣ ∣ ∣ ∣
∣
then find the value of
(
d
n
d
x
n
f
(
x
)
)
x
=
0
f
o
r
a
l
l
(
n
∈
Z
)
.
Open in App
Solution
Given
f
(
x
)
=
n
∣
∣ ∣ ∣ ∣ ∣
∣
x
n
n
!
2
cos
x
cos
n
π
2
4
sin
x
sin
n
π
2
8
∣
∣ ∣ ∣ ∣ ∣
∣
d
n
d
x
n
[
f
(
x
)
]
=
n
∣
∣ ∣ ∣ ∣ ∣
∣
n
!
n
!
2
cos
(
x
+
n
π
2
)
cos
n
π
2
4
sin
(
x
+
n
π
2
)
sin
n
π
2
8
∣
∣ ∣ ∣ ∣ ∣
∣
(
d
n
d
x
n
[
f
(
x
)
]
)
x
=
0
=
n
∣
∣ ∣ ∣ ∣ ∣
∣
n
!
n
!
2
cos
n
π
2
cos
n
π
2
4
sin
n
π
2
sin
n
π
2
8
∣
∣ ∣ ∣ ∣ ∣
∣
Since,
C
1
and
C
2
are identical
∴
Δ
=
0
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0
Similar questions
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
Q.
Let
f
(
x
)
=
∣
∣ ∣ ∣ ∣ ∣
∣
π
n
x
sin
π
x
cos
π
x
(
−
1
)
n
n
!
−
sin
(
n
π
2
)
−
cos
(
n
π
2
)
−
1
1
√
2
√
3
2
∣
∣ ∣ ∣ ∣ ∣
∣
then value of
d
n
d
x
n
[
f
(
x
)
]
a
t
x
=
1
is
Q.
If
f
(
x
)
=
e
x
−
e
−
x
−
2
sin
x
−
2
3
x
3
,
then the least value of
n
for which
d
n
d
x
n
f
(
x
)
at
x
=
0
is non-zero is ?
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