Question

The maximum value of $\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+sin\theta & 1\\ 1+cos\theta & 1& 1\end{array}\right|$ is ($\theta$ is real number)

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\sqrt{2}$
• D. $\frac{2\sqrt{3}}{4}$

Answer 1

Answer: (A)

$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+sin\theta & 1\\ 1+cos\theta & 1& 1\end{array}\right|$
Applying $R_1\to R_1-R_{3}and{R}_2\to R_2-R_3$, we get
$\Delta =\left|\begin{array}{ccc}-cos\theta & 0& 0\\ -cos\theta & sin\theta & 0\\ 1+cos\theta & 1& 1\end{array}\right|$
Expanding along $C_3$, we get
$\Delta =1\left(-cos\theta sin\theta -0\right)=-cos\theta sin\theta =\frac{-1}{2}sin2\theta$
We know, $-1\le sin2\theta \le 1$
$\Rightarrow -\frac{1}{2}\le \frac{1}{2}sin2\theta \le \frac{1}{2}\Rightarrow \frac{1}{2}\ge -\frac{1}{2}sin2\theta \ge \frac{-1}{2}$
i.e., $-\frac{1}{2}\le -\frac{1}{2}sin2\theta \le \frac{1}{2}\Rightarrow \frac{-1}{2}\le \Delta \le \frac{1}{2}$
Hence, maximum value of $\Delta$ is $\frac{1}{2}$.