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If $\mathrm{f}(\theta)=\left|\begin{array}{ccc}1 & \cos \theta & 1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1\end{array}\right|$ and
$\mathrm{A}$ and $\mathrm{B}$ are respectively the maximum and the minimum values of $f(\theta)$, then $(A, B)$ is equal to:
Options:
$\mathrm{A}$ and $\mathrm{B}$ are respectively the maximum and the minimum values of $f(\theta)$, then $(A, B)$ is equal to:
Solution:
1828 Upvotes
Verified Answer
The correct answer is:
$(2+\sqrt{2}, 2-\sqrt{2})$
$(2+\sqrt{2}, 2-\sqrt{2})$
$$
\begin{aligned}
&\text { Let } f(\theta)=\left|\begin{array}{ccc}
1 & \cos \theta & 1 \\
-\sin \theta & 1 & -\cos \theta \\
-1 & \sin \theta & 1
\end{array}\right| \\
&=(1+\sin \theta \cos \theta)-\cos \theta(-\sin \theta-\cos \theta) \\
&+1\left(-\sin ^2 \theta+1\right) \\
&=1+\sin \theta \cos \theta+\sin \theta \cos \theta+ \\
&=2+2 \sin \theta \cos \theta+\cos 2 \theta \\
&=2+\sin 2 \theta+\cos 2 \theta
\end{aligned}
$$
Now, maximum value of (1)
is $2+\sqrt{1^2+1^2}=2+\sqrt{2}$ and minimum value of (1) is
$$
2-\sqrt{1^2+1^2}=2-\sqrt{2} \text {. }
$$
\begin{aligned}
&\text { Let } f(\theta)=\left|\begin{array}{ccc}
1 & \cos \theta & 1 \\
-\sin \theta & 1 & -\cos \theta \\
-1 & \sin \theta & 1
\end{array}\right| \\
&=(1+\sin \theta \cos \theta)-\cos \theta(-\sin \theta-\cos \theta) \\
&+1\left(-\sin ^2 \theta+1\right) \\
&=1+\sin \theta \cos \theta+\sin \theta \cos \theta+ \\
&=2+2 \sin \theta \cos \theta+\cos 2 \theta \\
&=2+\sin 2 \theta+\cos 2 \theta
\end{aligned}
$$
Now, maximum value of (1)
is $2+\sqrt{1^2+1^2}=2+\sqrt{2}$ and minimum value of (1) is
$$
2-\sqrt{1^2+1^2}=2-\sqrt{2} \text {. }
$$
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