Question

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

(a) $\frac{1}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\sqrt{2}$
(d) $-\frac{\sqrt{3}}{2}$

Answer 1

$$\begin{gathered} ∆=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+\sin \theta & 1\\ 1+\cos \theta & 1& 1\end{array}\right| \\ =\left|\begin{array}{ccc}1& 1& 1\\ 0& \sin \theta & 0\\ \cos \theta & 0& 0\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_1\mathrm{and}{R}_3\to R_3-R_1\right] \\ =-\sin \theta \cos \theta \\ =-\frac{\sin 2\theta }{2} \\ \mathrm{Now},\mathrm{Maximum}\mathrm{and}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{sin\theta }\mathrm{is}1\mathrm{and}-1. \\ \mathrm{So},\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}-\sin \theta \mathrm{is}1. \\ \mathrm{So},\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}-\sin 2\theta \mathrm{is}1. \\ \mathrm{Therefore},\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}-\frac{\sin 2\theta }{2}\mathrm{is}\frac{1}{2}. \end{gathered}$$

Hence, the correct option is (a).