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Byju's Answer
Standard XII
Mathematics
Transpose of a Matrix
Let f x =1+s...
Question
Let
f
(
x
)
=
∣
∣ ∣ ∣
∣
1
+
sin
2
x
cos
2
x
4
sin
2
x
sin
2
x
1
+
cos
2
x
4
sin
2
x
sin
2
x
cos
2
x
1
+
4
sin
2
x
∣
∣ ∣ ∣
∣
find the maximum value of
f
(
x
)
Open in App
Solution
f
(
x
)
=
∣
∣ ∣ ∣
∣
1
+
sin
2
x
cos
2
x
4
sin
2
x
sin
2
x
1
+
cos
2
x
4
sin
2
x
sin
2
x
cos
2
x
1
+
4
sin
2
x
∣
∣ ∣ ∣
∣
a
p
p
l
y
i
n
g
C
1
→
C
1
+
C
2
w
e
g
e
t
f
(
x
)
=
∣
∣ ∣ ∣
∣
2
cos
2
x
4
sin
2
x
2
1
+
cos
2
x
4
sin
2
x
1
cos
2
x
1
+
4
sin
2
x
∣
∣ ∣ ∣
∣
a
p
p
l
y
i
n
g
R
2
→
R
2
−
R
1
a
n
d
R
3
→
R
3
−
R
1
f
(
x
)
=
∣
∣ ∣ ∣
∣
2
cos
2
x
4
sin
2
x
2
−
2
1
+
cos
2
−
c
o
s
2
x
4
sin
2
x
−
4
s
i
n
2
x
1
−
2
cos
2
x
−
c
o
s
2
x
1
+
4
sin
2
x
−
4
s
i
n
2
x
∣
∣ ∣ ∣
∣
f
(
x
)
=
∣
∣ ∣
∣
2
cos
2
x
4
sin
2
x
0
1
0
−
1
0
1
∣
∣ ∣
∣
=
2
+
4
sin
2
x
S
i
n
c
e
t
h
e
m
a
x
i
m
u
m
v
a
l
u
e
o
f
sin
2
x
i
s
1
∴
t
h
e
m
a
x
i
m
u
m
v
a
l
u
e
o
f
f
(
x
)
i
s
6
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0
Similar questions
Q.
Let
f
(
x
)
=
∣
∣ ∣ ∣
∣
1
+
sin
2
x
cos
2
x
4
sin
2
x
sin
2
x
1
+
cos
2
x
4
sin
2
x
sin
2
x
cos
2
x
1
+
4
sin
2
x
∣
∣ ∣ ∣
∣
, then the maximum value of
f
(
x
)
=
.
Q.
Let
f
(
x
)
=
∣
∣ ∣ ∣
∣
1
+
s
i
n
2
x
c
o
s
2
x
4
s
i
n
2
x
s
i
n
2
x
1
+
c
o
s
2
x
4
s
i
n
2
x
s
i
n
2
x
c
o
s
2
x
1
+
4
s
i
n
2
x
∣
∣ ∣ ∣
∣
, then the maximum value of
f
(
x
)
=
Q.
Let
f
(
x
)
=
∣
∣ ∣ ∣
∣
sin
2
x
−
2
+
cos
2
x
cos
2
x
2
+
sin
2
x
cos
2
x
cos
2
x
sin
2
x
cos
2
x
1
+
cos
2
x
∣
∣ ∣ ∣
∣
,
x
∈
[
0
,
π
]
.
Then the maximum value of
f
(
x
)
is equal to
Q.
Let
m
and
M
be respectively the minimum and maximum values of
∣
∣ ∣ ∣
∣
cos
2
x
1
+
sin
2
x
sin
2
x
1
+
cos
2
x
sin
2
x
sin
2
x
cos
2
x
sin
2
x
1
+
sin
2
x
∣
∣ ∣ ∣
∣
Then the ordered pair
(
m
,
M
)
is equal to:
Q.
If
sin
2
x
=
α
−
1
and
cos
2
x
=
β
−
1
, then the value of
sec
2
x
[
(
cos
2
x
−
sin
2
x
)
−
2
sin
x
cos
x
]
1
+
sin
2
x
is equal to
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