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प्रश्न
The maximum value of \[∆ = \begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1\end{vmatrix}\] is (θ is real)
पर्याय
`1/2`
`sqrt3/2`
`sqrt2`
`-sqrt3/2`
उत्तर
\[∆ = \begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1\end{vmatrix}\]
\[ = \begin{vmatrix}1 & 1 & 1 \\ 0 & \sin\theta & 0 \\ \cos\theta & 0 & 0\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1\text{ and }R_3 \to R_3 - R_1 \right]\]
\[ = - \sin\theta\cos\theta\]
\[ = - \frac{\sin2\theta}{2}\]
Now, Maximum and minimum value of \[sin\theta\] is 1 and - 1.
\[\text{ So, the maximum value of} - \sin\theta \text{ is 1 .} \]
\[\text{ So, the maximum value of } - \sin2\theta\text{ is }1 . \]
\[\text{ Therefore, the maximum value of} - \frac{\sin2\theta}{2}\text{ is }\frac{1}{2} .\]
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