Question:
The maximum value of \(\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1+\cos\theta & 1 & 1 \end{vmatrix}\) is (\(\theta\) is real number).
\(\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1+\cos\theta & 1 & 1 \end{vmatrix}\) का महत्तम मान क्या है? (\(\theta\) एक वास्तविक संख्या है)
Options:
- \(\dfrac{2\sqrt{3}}{4}\)
- \(\dfrac{\sqrt{3}}{2}\)
- \(\sqrt{2}\)
- \(\dfrac{1}{2}\)
- \(\dfrac{2\sqrt{3}}{4}\)
- \(\dfrac{\sqrt{3}}{2}\)
- \(\sqrt{2}\)
- \(\dfrac{1}{2}\)
Correct Answer: d)
Explanation:
Applying row operations \(R_2 \to R_2 – R_1\) and \(R_3 \to R_3 – R_1\):
\(\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & \sin\theta & 0 \\ \cos\theta & 0 & 0 \end{vmatrix} = 1(0) – 1(0) + 1(0 – \sin\theta\cos\theta) = -\sin\theta\cos\theta = -\dfrac{1}{2}\sin 2\theta\).
Maximum value occurs when \(\sin 2\theta = -1\), giving \(\Delta_{\max} = \dfrac{1}{2}\).
संक्रियाएँ \(R_2 \to R_2 – R_1\) तथा \(R_3 \to R_3 – R_1\) लगाने पर:
\(\Delta = -\sin\theta\cos\theta = -\dfrac{1}{2}\sin 2\theta\)।
इसका महत्तम मान \(\dfrac{1}{2}\) होगा।