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DETERMINANTS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDETERMINANTS1 Mark
Question
Choose the correct answer from given four options in each of the Exercise:
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):
  1. $\frac{1}{2}$
  2. $\frac{\sqrt{3}}{2}$
  3. $\sqrt{2}$
  4. $\frac{2\sqrt{3}}{4}$
✓

Answer

  1. $\frac{1}{2}$
Solution:
Since,
$\Delta=\begin{vmatrix}1 & 1&1 \\1 &1+\sin\theta&1\\1+\cos\theta &1&1\end{vmatrix}$
$=\begin{vmatrix}0 &0&0 \\0 &\sin\theta&1\\\cos\theta &0&1\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_2-\text{C}_3\text{ and C}_2\rightarrow\text{C}_2-\text{C}_3\big]$
$=-(\sin\theta.\cos\theta)$
$=\frac{1}{2}.2\sin\cos\theta=\frac{1}{2}\sin2\theta$
Since, the maximum value of $\sin2\theta$ is 1. So, for maximum value of $\theta$ should be 45°
$\therefore\ \Delta-\frac{1}{2}\sin2.45^\circ$
$=\frac{1}{2}\sin90^\circ=\frac{1}{2}.1=\frac{1}{2}$

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