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  • . The maximum value of delta is (θ is a real number) A. 1/2

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|_{\text {is }(\theta \text { is a real number })}\\

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

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A)

\\ \begin{aligned} &\text { We have: }\\ &\Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1+\cos \theta & 1 & 1 \end{array}\right|\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3} \text { and } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}\\ &\Rightarrow \Delta=\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & \sin \theta & 1 \\ \cos \theta & 0 & 1 \end{array}\right| \end{aligned}

\vspace{\baselineskip} = 0 - 0 + 1 (\sin \theta \cdot \cos \theta)\\ \vspace{\baselineskip} \text{Multiply and divide by } 2,\\ \vspace{\baselineskip} = \frac{1}{2} (2\sin \theta \cos \theta)\\ \vspace{\baselineskip} \text{We already know that } \underline{2 \sin \theta \cos \theta = \sin 2\theta}\\ = \frac{1}{2} (\sin 2\theta)\\ \vspace{\baselineskip}

The maximum value of :\sin 2\theta \text{ is 1, } \theta = 45^{\circ}.


\therefore \Delta = \frac{1}{2} (\sin 2(45^{\circ}))\\ \vspace{\baselineskip} = \frac{1}{2} \sin 90^{\circ}\\ \vspace{\baselineskip} = \frac{1}{2} (1)\\

\therefore \Delta = \frac{1}{2}

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