A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{2}\]
Correct Answer: D
Solution :
Given equation is \[{{x}^{2}}-(\sin \alpha -2)x-(1+\sin \alpha )=0\] Let\[{{x}_{1}}\] and \[{{x}_{2}}\] be two roots of quadratic equation. \[\therefore \]\[{{x}_{1}}+{{x}_{2}}=\sin \alpha -2\]and\[{{x}_{1}}{{x}_{2}}=-(1+\sin \alpha )\] \[{{({{x}_{1}}+{{x}_{2}})}^{2}}={{(\sin \alpha -2)}^{2}}={{\sin }^{2}}\alpha +4-4\sin \alpha \] \[\Rightarrow \]\[x_{1}^{2}+x_{2}^{2}={{\sin }^{2}}-\alpha +4-4\sin \alpha -2{{x}_{1}}{{x}_{2}}\] \[={{\sin }^{2}}\alpha +4-4\sin \alpha +2(1+\sin \alpha )\] \[={{\sin }^{2}}\alpha -2\sin \alpha +6\] ?(A) Now, By putting \[\alpha =\frac{\pi }{6},\alpha =\frac{\pi }{4}\alpha =\frac{\pi }{3}\]and\[\alpha =\frac{\pi }{2}\]in (A) one by one We get least value of \[x_{1}^{2}+x_{2}^{2}\]at\[\frac{\pi }{2}\] Hence, \[\alpha =\frac{\pi }{2}\]
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