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Continuity And Differentiability

Question
CBSEENMA12035448
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If space straight y equals left parenthesis cos to the power of negative 1 end exponent space straight x right parenthesis squared comma space prove space that space left parenthesis 1 minus straight x squared right parenthesis space straight y subscript 2 minus straight x space straight y subscript 1 minus 2 equals 0.

Solution
space space space space space space space space space space space space space straight y equals left parenthesis cos to the power of negative 1 end exponent space straight x right parenthesis squared space space space space space space space space space space space space space space space space space rightwards double arrow space space straight y subscript 1 equals 2 space cos to the power of negative 1 end exponent straight x. fraction numerator negative 1 over denominator square root of 1 minus straight x squared end root end fraction therefore space space square root of 1 minus straight x squared end root space straight y subscript 1 equals negative 2 space cos to the power of negative 1 end exponent straight x space space space space space rightwards double arrow space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 1 squared equals 4 left parenthesis cos to the power of negative 1 end exponent straight x right parenthesis squared therefore space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 1 squared equals 4 space straight y Differentiating space both space sides space straight w. straight r. straight t. straight x comma space we space get comma left parenthesis 1 minus straight x squared right parenthesis. space 2 space straight y subscript 1 space straight y subscript 2 space plus space straight y subscript 1 squared left parenthesis negative 2 space straight x right parenthesis equals 4 space straight y subscript 1 Dividing space both space sides space by space 2 space straight y 1 comma space we space get comma space space left parenthesis 1 minus straight x squared right parenthesis space straight y subscript 2 minus straight x space straight y subscript 1 equals 2 space or space left parenthesis 1 minus space straight x squared right parenthesis straight y subscript 2 minus space straight x space straight y subscript 1 minus 2 equals 0

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