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Relations And Functions

Question
CBSEENMA12036220

Suppose f(x) is differentiable x = 1 and limit as straight h rightwards arrow 0 of 1 over straight h space straight f left parenthesis 1 plus straight h right parenthesis space equals space 5 space comma then space straight f apostrophe left parenthesis 1 right parenthesis space equals

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Solution

C.

5

straight f apostrophe left parenthesis 1 right parenthesis space equals space limit as straight h rightwards arrow 0 of space fraction numerator straight f left parenthesis 1 plus straight h right parenthesis minus straight f left parenthesis 1 right parenthesis over denominator straight h end fraction semicolon As the function is differentiable so it is continuous as it is given that limit as straight h rightwards arrow 0 of space fraction numerator straight f left parenthesis 1 plus straight h right parenthesis over denominator straight h end fraction space equals space 5 and hence f(1) = 0 Hence  limit as straight h rightwards arrow 0 of space fraction numerator straight f left parenthesis 1 plus straight h right parenthesis over denominator straight h end fraction space equals space 5