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BCECE Engineering BCECE Engineering Solved Paper-2013

  • question_answer
    The value of \[a\] for which \[a{{x}^{2}}+{{\sin }^{-1}}({{x}^{2}}-2x+2)+{{\cos }^{-1}}\]\[({{x}^{2}}-2x+2)=0\] has a real solution, is

    A) \[-\frac{2}{\pi }\]                             

    B)         \[\frac{2}{\pi }\]                             

    C)  \[-\frac{\pi }{2}\]                            

    D)         \[\frac{\pi }{2}\]

    Correct Answer: C

    Solution :

    Here, \[{{x}^{2}}-2x+2={{(x-1)}^{2}}+1\ge 1\] But \[-1\le ({{x}^{2}}-2x+2)\le 1\] which is possible only when \[{{x}^{2}}-2x+2=1\]                 \[\therefore \]  \[x=1\] Then,   \[a{{(1)}^{2}}+{{\sin }^{-1}}(1)+co{{s}^{-1}}(1)=0\] \[\Rightarrow \]               \[a+\frac{\pi }{2}+0=0\] \[\therefore \]  \[a=-\frac{\pi }{2}\]


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