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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
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$):$
  • ✓
    $\frac{1}{2}$
  • B
    $\frac{\sqrt{3}}{2}$
  • C
    $\sqrt{2}$
  • D
    $\frac{2\sqrt{3}}{4}$
✓

Answer

Correct option: A.
$\frac{1}{2}$
Since,
$\Delta=\begin{vmatrix}1 & 1&1 \\1 &1+\sin\theta&1\\1+\cos\theta &1&1\end{vmatrix}$
$=\begin{vmatrix}0 &0&0 \\0 &\sin\theta&1\\\cos\theta &0&1\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_2-\text{C}_3$ 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 $45^\circ$
$\therefore\ \Delta-\frac{1}{2}\sin2.45^\circ$
$=\frac{1}{2}\sin90^\circ=\frac{1}{2}.1=\frac{1}{2}$

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