Question

Let $ {{x}_{1}} $ and $ {{x}_{2}} $ be solutions of the equation $ {{\sin }^{-1}}\left( {{x}^{2}}-3x+\frac{5}{2} \right)=\frac{\pi }{6} $ .Then, the value of $ x_{1}^{2}+x_{2}^{2} $ is:

• A. 4
• B. 5
• C. $ \frac{5}{2} $
• D. 6
• E. $ \frac{15}{2} $

Answer 1

Answer: (A)

$ {{\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 $