Question

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`

Answer 1

\[∆ = \begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1\end{vmatrix}\]

\[ = \begin{vmatrix}1 & 1 & 1 \\ 0 & \sin\theta & 0 \\ \cos\theta & 0 & 0\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1\text{ and }R_3 \to R_3 - R_1 \right]\]

\[ = - \sin\theta\cos\theta\]

\[ = - \frac{\sin2\theta}{2}\]

Now, Maximum and minimum value of \[sin\theta\] is 1 and - 1. 

\[\text{ So, the maximum value of} - \sin\theta \text{ is 1 .} \]

\[\text{ So, the maximum value of } - \sin2\theta\text{ is }1 . \]

\[\text{ Therefore, the maximum value of}  - \frac{\sin2\theta}{2}\text{ is }\frac{1}{2} .\]