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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    If\[4{{\sin }^{-1}}x+{{\cos }^{-1}}x=\pi ,\]then\[x\]is equal to:

    A)  \[\frac{1}{2}\]                  

    B)         \[2\]

    C)  \[1\]                    

    D)         \[\frac{1}{3}\]

    E)  \[\frac{1}{5}\]

    Correct Answer: A

    Solution :

    \[\therefore \] \[4{{\sin }^{-1}}(x)+{{\cos }^{-1}}x=\pi \] \[\Rightarrow \] \[4{{\sin }^{-1}}x+\frac{\pi }{2}-{{\sin }^{-1}}x=\pi \] \[\Rightarrow \]               \[3{{\sin }^{-1}}x=\frac{\pi }{2}\] \[\Rightarrow \]               \[x=\sin \left( \frac{\pi }{6} \right)=\frac{1}{2}\]


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