Question

The number of distinct real root of $\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0$ in the interval $\left[-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right],$ is
(a) 0
(b) 2
(c) 1
(d) 3

Answer 1

Given: $\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0$



$$\begin{gathered} \left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right| \\ \mathrm{Applying}{R}_2\to R_2-R_1 \\ =\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x-\sin x& \sin x-\cos x& \cos x-\cos x\\ \cos x& \cos x& \sin x\end{array}\right| \\ =\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ -\left(\sin x-\cos x\right)& \sin x-\cos x& 0\\ \cos x& \cos x& \sin x\end{array}\right| \\ \mathrm{Taking}\left(\sin x-\cos x\right)\mathrm{common}\mathrm{from}{R}_2 \\ =\left(\sin x-\cos x\right)\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ -1& 1& 0\\ \cos x& \cos x& \sin x\end{array}\right| \\ \mathrm{Applying}{C}_2\to C_2+C_1 \\ =\left(\sin x-\cos x\right)\left|\begin{array}{ccc}\sin x& \cos x+\sin x& \cos x\\ -1& 1-1& 0\\ \cos x& \cos x+\cos x& \sin x\end{array}\right| \\ =\left(\sin x-\cos x\right)\left|\begin{array}{ccc}\sin x& \cos x+\sin x& \cos x\\ -1& 0& 0\\ \cos x& 2\cos x& \sin x\end{array}\right| \\ \mathrm{Expanding}\mathrm{through}{R}_2 \\ =\left(\sin x-\cos x\right)\left[-\left(-1\right)\left\{\left(\cos x+\sin x\right)\left(\sin x\right)-2\cos x\left(\cos x\right)\right\}\right] \\ =\left(\sin x-\cos x\right)\left[1\left\{\cos x\sin x+{\sin }^{2}x-2{\cos }^{2}x\right\}\right] \\ =\left(\sin x-\cos x\right)\left(\cos x\sin x+{\sin }^{2}x-2{\cos }^{2}x\right) \\ =\left(\sin x-\cos x\right)\left({\sin }^{2}x+2\cos x\sin x-\cos x\sin x-2{\cos }^{2}x\right) \\ =\left(\sin x-\cos x\right)\left(\sin x\left(\sin x+2\cos x\right)-\cos x\left(\sin x+2\cos x\right)\right) \\ =\left(\sin x-\cos x\right)\left(\sin x-\cos x\right)\left(\sin x+2\cos x\right) \\ ={\left(\sin x-\cos x\right)}^2\left(\sin x+2\cos x\right) \\ \mathrm{Thus},\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|={\left(\sin x-\cos x\right)}^2\left(\sin x+2\cos x\right) \\ \mathrm{But}\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that},\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0 \\ \Rightarrow {\left(\sin x-\cos x\right)}^2\left(\sin x+2\cos x\right)=0 \\ \Rightarrow {\left(\sin x-\cos x\right)}^2=0\mathrm{or}\sin x+2\cos x=0 \\ \Rightarrow \sin x-\cos x=0\mathrm{or}\sin x=-2\cos x \\ \Rightarrow \sin x=\cos x\mathrm{or}\tan x=-2 \\ \mathrm{Since}x\in \left[-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right],\mathrm{Thus}\tan x\ne -2. \\ \Rightarrow \sin x=\cos x\mathrm{at}x=\frac{\mathrm{\pi }}{4} \\ \mathrm{Therefore},\mathrm{we}\mathrm{have}1\mathrm{distinct}\mathrm{real}\mathrm{root}. \end{gathered}$$

Hence, the correct option is (c).