Let $f: S \rightarrow S$ where $S=(0, \infty)$ be a twice differentiable function such that $f(x+1)=x f(x)$. If $g: S \rightarrow R$ be defined as $g(x)=\log _{e} f(x)$, then the value of $\left|g^{\prime \prime}(5)-g^{\prime \prime}(1)\right|$ is equal to :

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Continuity and Differentiability

Let

f : S \to S

where

S = (0, \infty)

be a twice differentiable function such that

f (x + 1) = x f (x)

. If

g : S \to R

be defined as

g (x) = lo g_{e} f (x)

, then the value of

∣ g^{''} (5) - g^{''} (1) ∣

is equal to :

Answer: (A)

Taking logarithms in

f (x + 1) = x f (x)

gives

g (x + 1) = lo g_{e} x + g (x)

. Differentiating twice,

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

Adding this relation for

x = 1, 2, 3, 4

g^{''} (5) - g^{''} (1) = - (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16}) = - \frac{144 + 36 + 16 + 9}{144} = - \frac{205}{144}

∣ g^{''} (5) - g^{''} (1) ∣ = \frac{205}{144}

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option