Question

If $f\left(\theta \right)=\left|\begin{array}{ccc}1& cos\theta & 1\\ -sin\theta & 1& -cos\theta \\ -1& sin\theta & 1\end{array}\right|$ and A and B are respectively the maximum and the minimum values of $f\left(\theta \right)$, then (A, B) is equal to :

• A. $(3,-1)$
• B. $(4,2-\sqrt{2})$
• C. $(2+\sqrt{2},2-\sqrt{2})$
• D. $(2+\sqrt{2}-1)$

Answer 1

Answer: (C)

$(2+\sqrt{2},2-\sqrt{2})$

$f\left(\theta \right)=\left|\begin{array}{ccc}1& cos\theta & 1\\ -sin\theta & 1& -cos\theta \\ -1& sin\theta & 1\end{array}\right|$
By solving the determinant, we get
$f\left(\theta \right)=(1+sin\theta cos\theta )-cos\theta (-sin\theta -cos\theta )+(-si{n}^2\theta +1)$
By simplifying this, we get
$f\left(\theta \right)=2+sin2\theta +cos2\theta$
Maximum and minimum value of $sin2\theta +cos2\theta$ will be $\sqrt{2},-\sqrt{2}$ respectively
So, the final maximum and minimum value becomes $2+\sqrt{2},2-\sqrt{2}$ respectively