To find the maximum and minimum values of
f(θ), we first need to calculate the determinant:
f(θ)=|| 1 | cos‌θ | −1 |
| −sin‌θ | 1 | −cos‌θ |
| −1 | sin‌θ | 1 |
| We can expand the determinant along the first row:
Now, calculate each
2×2 determinant:
Substitute these back into the expression for
f(θ) :
f(θ)=1(1+sin‌θ‌cos‌θ)−cos‌θ(−sin‌θ−cos‌θ)−1(cos2θ)
f(θ)=1+sin‌θ‌cos‌θ+sin‌θ‌cos‌θ+cos2θ−cos2θ
f(θ)=1+2sin‌θ‌cos‌θ We know the trigonometric identity
2sin‌θ‌cos‌θ=sin‌(2θ).So,
f(θ)=1+sin‌(2θ).Now we need to find the maximum and minimum values of
f(θ).The range of the sine function
sin‌(x) is
[−1,1]. Therefore, the range of
sin‌(2θ) is als
[−1,1].
The minimum value of
f(θ) occurs when
sin‌(2θ) is at its minimum, which is -1 .
Minimum value
B=1+(−1)=0.
The maximum value of
f(θ) occurs when
sin‌(2θ) is at its maximum, which is 1 .
Maximum value
A=1+(1)=2.
So,
A=2 and
B=0. The pair
(A,B) is
(2,0).
The calculated pair
(2,0) matches Option B.
The final answer is
(2,0)