Matrices and Determinants 2 Question 15
17. 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 $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is
(a) 0
(b) 2
(c) 1
(d) 3
(2001, 1M)
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Answer:
Correct Answer: 17. (c)
Solution:
- Given, $\begin{vmatrix}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{vmatrix}=0$
Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$
$ \begin{vmatrix} & \sin x+2 \cos x & \cos x & \cos x \\ & \sin x+2 \cos x & \sin x & \cos x \\ & \sin x+2 \cos x & \cos x & \sin x \\ \end{vmatrix} $
$ =(2 \cos x+\sin x) $ $ \begin{vmatrix} 1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x \end{vmatrix}=0 $
Applying $R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1}$
$(2 \cos x+\sin x)$ $ \begin{vmatrix} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 0 & 0 & \sin x-\cos x \end{vmatrix}$ =0
$\Rightarrow (2 \cos x+\sin x)(\sin x-\cos x)^{2}=0 $
$\Rightarrow 2 \cos x+\sin x=0 \text { or } \sin x-\cos x=0 $
$\Rightarrow 2 \cos x=-\sin x \text { or } \sin x=\cos x$
$\Rightarrow \cot x=-1 / 2$ gives no solution in $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$
and $\sin x=\cos x \Rightarrow \tan x=1 \quad \Rightarrow \quad x=\pi / 4$
