JEE Main 16 March 2021 Shift 2 Solved Paper
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Question : 76 of 90
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Let f : S → S , where S = ( 0 , ∞ ) be a twice differentiable function, such that f ( x + 1 ) = x f ( x ) . If g : S → R be defined as g ( x ) = log e f ( x ) , then the value of g ′ ′ ( 5 ) − g ′ ′ ( 1 ) | is equal to
[16 Mar 2021 Shift 2]
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