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Continuity And Differentiability

Question
CBSEENMA12035453
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If space space straight y equals sin left parenthesis straight m space sin to the power of negative 1 end exponent straight x right parenthesis comma space prove space that space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 plus straight m squared space straight y equals 0.

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

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