QJEEM21Mafc86: JEE Main 2021 Mathematics - Class 12 JEEM21Mafc86 Solution: Let f : S S where S = (0, ) be a twice differentiable function…

Question

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Let f : S

\to

S where S = (0,

\infty

) be a twice differentiable function such that f(x + 1) = xf(x). If g : S

\to

R be defined as g(x) = log_e f(x), then the value of |g''(5)

-

g''(1)| is equal to :

Accepted Answer

f(x + 1) = xf(x)

\ln (f(x + 1)) = \ln x + \ln f(x)

g(x + 1) = \ln x + g(x)

g(x + 1) - g(x) = \ln x

..... (i)

g'(x + 1) - g'(x) = {1 \over x}

g''(x + 1) - g''(x) = {{ - 1} \over {{x^2}}}

g''(2) - g'(1) = {{ - 1} \over 1}

.... (ii)

g''(3) - g''(2) = {{ - 1} \over 4}

.... (iii)

g''(4) - g''(3) = {{ - 1} \over 9}

..... (iv)

g''(5) - g''(4) = {{ - 1} \over {16}}

....(v)

Adding (ii), (iii), (iv) & (v)

g''(5) - g''(1) = - \left( {{1 \over 1} + {1 \over 4} + {1 \over 9} + {1 \over {16}}} \right) = {{ - 205} \over {144}}

|g''(5) - g''(1)|\, = {{205} \over {144}}

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