can I differentiate the function 10 cos^(2)x -6sinx.cosx+2sin^(2) to get max and min value
Answers
Answered by
1
Answer:
Answer
Correct option is
D
(a−b)
2
Given u=
a
2
cos
2
θ+b
2
sin
2
θ
+
a
2
sin
2
θ+b
2
cos
2
θ
∴u
2
=(a
2
+b
2
)cos
2
θ+(a
2
+b
2
)sin
2
θ+2
a
2
cos
2
θ+b
2
sin
2
θ
a
2
sin
2
θ+b
2
cos
2
θ
u
2
=a
2
+b
2
+2
a
4
cos
2
θsin
2
θ+b
4
cos
2
θsin
2
θ+a
2
b
2
(cos
4
θ+sin
4
θ)
=a
2
+b
2
+2
(a
4
+b
4
)cos
2
θsin
2
θ+a
2
b
2
(1−2cos
2
θsin
2
θ)
u
2
=a
2
+b
2
+2
4
a
4
+b
4
(sin2θ)
2
+a
2
b
2
(1−
2
sin
2
2θ
)
u
2
=a
2
+b
2
+2
4
(a
2
−b
2
)
2
(sin2θ)
2
+a
2
b
2
For max. value of u
2
, sin
2
2θ=1
u
2
=a
2
+b
2
+2
2
(a
2
+b
2
)
2
=2(a
2
+b
2
)
And for min. u
2
,sin2θ=0
u
2
=a
2
+b
2
+2ab=(a+b)
2
∴ Difference between max. and min. value of u
2
is 2(a
2
+b
2
)−(a+b)
2
=(a−b)
2
.
Answered by
1
Answer
Correct option is
D
(a−b)
2
Given u=
a
2
cos
2
θ+b
2
sin
2
θ
+
a
2
sin
2
θ+b
2
cos
2
θ
∴u
2
=(a
2
+b
2
)cos
2
θ+(a
2
+b
2
)sin
2
θ+2
a
2
cos
2
θ+b
2
sin
2
θ
a
2
sin
2
θ+b
2
cos
2
θ
u
2
=a
2
+b
2
+2
a
4
cos
2
θsin
2
θ+b
4
cos
2
θsin
2
θ+a
2
b
2
(cos
4
θ+sin
4
θ)
=a
2
+b
2
+2
(a
4
+b
4
)cos
2
θsin
2
θ+a
2
b
2
(1−2cos
2
θsin
2
θ)
u
2
=a
2
+b
2
+2
4
a
4
+b
4
(sin2θ)
2
+a
2
b
2
(1−
2
sin
2
2θ
)
u
2
=a
2
+b
2
+2
4
(a
2
−b
2
)
2
(sin2θ)
2
+a
2
b
2
For max. value of u
2
, sin
2
2θ=1
u
2
=a
2
+b
2
+2
2
(a
2
+b
2
)
2
=2(a
2
+b
2
)
And for min. u
2
,sin2θ=0
u
2
=a
2
+b
2
+2ab=(a+b)
2
∴ Difference between max. and min. value of u
2
is 2(a
2
+b
2
)−(a+b)
2
=(a−b)
2
.
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