A) \[0\]
B) 1
C) \[2\]
D) \[4\]
Correct Answer: B
Solution :
Given, \[\left| \begin{matrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \\ \end{matrix} \right|=0\] On applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{2}},\] \[\Rightarrow \] \[\left| \begin{matrix} \sin x & \cos x-\sin x & 0 \\ \cos x & \sin x-\cos x & \cos x-\sin x \\ \cos x & 0 & \sin x-\cos x \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\sin x[{{(\sin x-\cos x)}^{2}}]-(cosx-sinx)\times \] \[[\cos x\,(\sin x-\cos x)-\cos x(\cos x-\sin x)]=0\] \[\Rightarrow \] \[\sin x{{(\sin x-\cos x)}^{2}}-2\cos x(\cos x-sinx)\] \[(\sin x-\cos x)=0\] \[\Rightarrow \] \[{{(\operatorname{sinx}-cos\,x)}^{2}}\,(\sin x+2\cos x)=0\] \[\Rightarrow \] \[\sin x=\cos x\] or \[\sin x=-2\cos x\] \[\Rightarrow \] \[\tan \,x=1\] or \[\tan x=2\] But \[\frac{-\pi }{4}\le x\le \frac{\pi }{4}\] \[\Rightarrow \] \[\tan \left( \frac{-\pi }{4} \right)\le \tan \,x\,\le \tan \left( \frac{\pi }{4} \right)\] \[\Rightarrow \] \[-1\le \tan \,x\,\le 1\] \[\therefore \] \[\tan \,x=1\] \[\Rightarrow \] \[x=\frac{\pi }{4}\] Hence, only one real root exist.
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