A) \[\left[ 0,\,\sqrt{2} \right]\]
B) \[\left[ -\sqrt{2},0 \right]\]
C) \[\left[ -\sqrt{2},\sqrt{2} \right]\]
D) None of these
Correct Answer: C
Solution :
[c] Since the system has a non-trivial solution, therefore \[\left| \begin{matrix} \lambda & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ -1 & \sin \alpha & -\cos \alpha \\ \end{matrix} \right|=0\] \[\Rightarrow \,\,\,\,\lambda (-{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha )\] \[-(-\sin \alpha \cos \alpha -\sin \alpha \cos \alpha )\] \[-({{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha )=0\] \[\Rightarrow \,\,\,-\lambda +\sin 2\alpha +\cos 2\alpha =0\] \[\Rightarrow \,\,\,\,\lambda =\sin 2\alpha +\cos 2\alpha \] \[\Rightarrow \,\,\,\,\lambda =\sqrt{2}\,\cos \left( 2\alpha -\frac{\pi }{4} \right).\] Since \[-1\le \cos \left( 2\alpha -\frac{\pi }{4} \right)\le 1\forall \in R\] \[\therefore \,\,\,-\sqrt{2}\le \lambda \le \sqrt{2}\] i.e. \[\lambda \in \left[ -\sqrt{2},\sqrt{2} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
