JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 6)
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) = loge f(x), then the value of |g''(5) $$-$$ g''(1)| is equal to :
1
$${{187} \over {144}}$$
$${{197} \over {144}}$$
$${{205} \over {144}}$$
Explanation
$$f(x + 1) = xf(x)$$
$$\ln (f(x + 1)) = \ln x + \ln f(x)$$
$$g(x + 1) = \ln x + g(x)$$
$$g(x + 1) - g(x) = \ln x$$ ..... (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)
$$g''(5) - g''(1) = - \left( {{1 \over 1} + {1 \over 4} + {1 \over 9} + {1 \over {16}}} \right) = {{ - 205} \over {144}}$$
$$|g''(5) - g''(1)|\, = {{205} \over {144}}$$
$$\ln (f(x + 1)) = \ln x + \ln f(x)$$
$$g(x + 1) = \ln x + g(x)$$
$$g(x + 1) - g(x) = \ln x$$ ..... (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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