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Home  >>  CBSE XII  >>  Math  >>  Determinants
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The maximum value of $\begin{vmatrix}1 & 1 & 1\\1 & 1-sin\theta & 1\\1-\cos\theta & 1 & 1\end{vmatrix}\;(is\;\theta\; is\; real\;number)$

\[\begin{array}{1 1}(A)\quad \frac{1}{2} & (B)\quad \frac{\sqrt 3}{2}\\(C)\quad \sqrt 2 & (D)\quad \frac{2\sqrt 3}{4}\end{array}\]
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  • If $A=\begin{vmatrix}a_{11} &a_{21} & a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} &a_{33}\end{vmatrix}$
  • $|A|=a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
  • $sin 2\theta=2sin\theta cos\theta$
Let $\Delta=\begin{vmatrix}1 & 1 & 1\\1 & 1-sin\theta & 1\\1-cos\theta & 1 & 1\end{vmatrix}$
Apply $R_2\rightarrow R_3-R_2$ and $R_3\rightarrow R_3-R_1$
$\Delta=\begin{vmatrix}1 & 0 & 0\\1 & sin\theta & 0\\1-cos \theta &0 & cos \theta\end{vmatrix}$
Expanding along $R_1$
$\Delta=1(sin\theta cos\theta)$
$\Delta=sin\theta cos\theta$
But $sin2\theta=2sin\theta cos\theta$
$\frac{1}{2}sin 2\theta=sin\theta cos\theta$
Substituting this we get,
If $\theta=0$ then $sin2\theta=0$
If $\theta=\frac{\pi}{2}$ then $sin2\frac{\pi}{2}=0$ ($sin\pi=0$)
If $\theta=\frac{\pi}{3}$ then $sin2\frac{\pi}{3}=\frac{\sqrt 3}{2}$
If $\theta=\frac{\pi}{6}$ then $sin2\frac{\pi}{6}=\frac{\sqrt 3}{2}$
If $\theta=\frac{\pi}{4}$ then $sin2\frac{\pi}{4}=sin\frac{\pi}{2}=1.$
Therefore $\Delta=\frac{1}{2}\times 1=\frac{1}{2}$
Hence $\Delta$ will have maximum value of $\frac{1}{2}$
Hence A is the correct answer.
answered Mar 26, 2013 by sreemathi.v

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