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The number of distinct real roots of $\begin{vmatrix}\sin x & \cos x & \cos x\\\cos x & \sin x & \cos x\\\cos x & \cos x &\sin x\end{vmatrix}=0\;$ in the interval $\large \frac{-\pi}{4}$$ \leq x \leq\large \frac {\pi}{4}$ is

\[\begin{array}{1 1}(A)\quad 0 & (B)\quad 2\\(C)\quad 1 & (D)\quad 3\end{array}\]

1 Answer

  • If $A=\begin{vmatrix}a_{11} &a_{21} & a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} &a_{33}\end{vmatrix}$
  • $|A|=a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
  • $\sin\frac{\pi}{4}=\frac{1}{\sqrt 2}$
Let $\Delta=\begin{vmatrix}sin x & cos x & cos x\\cos x & sin x & cos x\\cos x & cos x & sin x\end{vmatrix}$
Apply $R_1\rightarrow R_1-R_2$ and $R_2\rightarrow R_2-R_3$
$\Delta=\begin{vmatrix}sin x-cos x& cos x-sin x& 0\\0 & sin x-cos x& cos x-sin x\\cos x& cos x&sin x\end{vmatrix}$
Let us take (sin x-cos x) as a common factor from $R_1$ and $R_2$
$\Delta=(sin x-cos x)\begin{vmatrix}1 & -1 & 0\\0 & 1 & -1\\cos x & cos x & sin x\end{vmatrix}$
Now expanding along $R_1$ we get,
$\Delta=(sin x-cos x)^2[1(sin x+cos x)-1(0+cos x)]$
$\quad=(sin x-cos x)^2[sin x+cos x-cos x]$
But it is given $\Delta=0$
($sin x-cos x)^2[sin x]=0.$
If $(sin x-cos x)^2=0.$
$\Rightarrow sin x=cos x=\frac{1}{\sqrt 2}$
This is possible if $x=\frac{\pi}{4}$
and sin x=0.if x=0.
We know $\frac{1}{\sqrt 2}$ is irrational.
Hence sin x=0 is the only distinct real root,between the interval $\frac{-\pi}{4}\leq x\leq \frac{\pi}{4}$
Hence C is the correct answer.
answered Mar 26, 2013 by sreemathi.v

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