Question

Let $f(\theta )=\left|\begin{array}{ccc}1& \cos \theta & -1\\ -\sin \theta & 1& -\cos \theta \\ -1& \sin \theta & 1\end{array}\right|$
Suppose $A$ and $B$ are respectively maximum and minimum value of $f(\theta )$. Then $(A,B)$ is equal to:

• A. $(2,1)$
• B. $(2,0)$
• C. $(\sqrt{2},1)$
• D. $\left(2,\frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (B)

Using $C_1\to C_1+C_3$, we get
$f(\theta )=\left|\begin{array}{ccc}0& \cos \theta & -1\\ -(\sin \theta +\cos \theta )& 1& -\cos \theta \\ 0& \sin \theta & 1\end{array}\right|$
Evaluating along $C_1$, we get
$f(\theta )=(\sin \theta +\cos \theta )\left|\begin{array}{cc}\cos \theta & -1\\ \sin \theta & 1\end{array}\right|$
$=(\sin \theta +\cos \theta {)}^2$
$={\left[\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta \right)\right]}^2$
$=2{\sin }^2\left(\frac{\pi }{4}+\theta \right)$
As $0\le {\sin }^2\left(\frac{\pi }{4}+\theta \right)\le 1,\text{ we get }$
$0\le f(\theta )\le 2$
$A=2\text{ for }\theta =\frac{\pi }{4}$
$B=0\text{ for }\theta =-\frac{\pi }{4}$
As $0\le {\sin }^2\left(\frac{\pi }{4}+\theta \right)\le 1$, we get
and $B=0$ for $\theta =-\frac{\pi }{4}$
Thus, $(A,B)=(2,0)$