Question

The number of the distinct real roots of the equation $\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} \le x \le \frac{\pi}{4}$ is

• A. 4
• B. 3
• C. 1
• D. 2

Answer 1

The Correct Option is C

We have, $\begin{vmatrix} \sin\, x &\cos\, x & \cos\, x\\ \cos\, x & \sin\, x &\cos\,x\\ \cos\, x &\cos\,x &\sin\, x\end{vmatrix}=0$
Applying $C_{1} \to C_{1}+C_{2}+C_{3}$, we get
$\begin{vmatrix}2\, \cos\, x + \sin\, x &\cos\, x &\cos\, x\\ 2\, \cos\, x + \sin\, x &\sin\, x & \cos\, x\\ 2\, \cos\, x +\sin\, x &\cos\, x &\sin\,x\end{vmatrix}=0$
Taking $(2\, \cos\, x + \sin\, x)$ common from $C_1$, we get
$\left(2\, \cos\, x + \sin\,x\right)\begin{vmatrix}1& \cos\, x &\cos\,x\\ 1&\sin\, x & \cos\,x\\ 1&\cos\, x &\sin\,x\end{vmatrix}=0$
Applying $R_{2} \to R_{2} -R_{1}$ and $R_{3} \to R_{1}$, we get
$\left(2\, \cos\, x +\sin\,x\right)\begin{vmatrix}1&\cos\, x &\cos\,x\\ 0&\sin\,x-\cos\,x&0\\ 0&0&\sin\,x-\cos\,x\end{vmatrix}=0$
Expanding along $C_{1}$, we get
$\left(2\, \cos \,x + \sin \,x\right) \left[1 \left(\sin\, x - \cos \,x\right)^{2}\right] = 0$
$\Rightarrow \left(2 \,\cos \,x + \sin \,x\right) \left(\sin \,x - \cos \,x\right)^{2}=0$
Now, if $2\cos\,x + \sin\,x = 0$, then $2\,\cos \,x = - \sin\, x$
$\Rightarrow \tan\,x=-2$
But here $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$, we have
$-1 \le \tan\,x \le 1$, so no solution possible.
or if $\left(\sin\, x-\cos\,x\right)^{2}=0$, then $\sin\,x =\cos\,x$
$\Rightarrow \tan\,x=1=\tan \frac{\pi}{4}$
$\Rightarrow x=\frac{\pi}{4}$
So, only one distinct real root exist.