Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    \[y=\int_{1/8}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}dt+\int_{1/8}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}dt}}\],\[0\le x\le x/2\]

    A)  Is the equation of a straight line parallel to the x-axis

    B)  Is the equation of a straight line which is the bisector of first quadrant

    C)  Is the equation of a straight line which is the bisector of second quadrant

    D)  None of the above

    Correct Answer: A

    Solution :

    Here, we have to prove that, \[y=\]constant or derivative of\[y\] with respect to\[x\]is zero. \[y=\int_{1/8}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}dt}+\int_{1/8}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}dt}\]       ... (i) \[\frac{dy}{dx}={{\sin }^{-1}}\sqrt{{{\sin }^{2}}x}\cdot 2\sin x\cdot \cos x+{{\cos }^{-1}}\sqrt{{{\cos }^{2}}x}\] \[(-2\sin x\cdot \cos x)=2x\sin x\cdot \cos x-2x\cdot \sin x\cdot \cos x\]\[=0\]for all\[x\].


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