A) 4
B) 5
C) \[\frac{5}{2}\]
D) 6
E) \[\frac{15}{2}\]
Correct Answer: A
Solution :
\[{{\sin }^{-1}}\left( {{x}^{2}}-3x+\frac{5}{2} \right)=\frac{\pi }{6}\] \[\Rightarrow \] \[{{x}^{2}}-3x+\frac{5}{2}=\sin \left( \frac{\pi }{6} \right)=\frac{1}{2}\] \[\Rightarrow \] \[2{{x}^{2}}-6x+5=0\] Since,\[{{x}_{1}}\,{{x}_{2}}\], are the solution set, then \[{{x}_{1}}+\,{{x}_{2}}=\frac{6}{2}=3\] and \[{{x}_{1}}\,{{x}_{2}}=\frac{5}{2}\] \[\therefore \]\[x_{1}^{2}+x_{2}^{2}={{({{x}_{1}}+{{x}_{2}})}^{2}}-2{{x}_{1}}{{x}_{2}}=9-5=4\]
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