Question

Let $f$ be real function defined on $R$ (the set of real numbers) such that $f^{\prime}(x)=100(x-1)(x-2)^2(x-3)^3 \ldots .(x-100)^{100}$, for all $x \in R$. If $g$ is a function defined on $R$ such that $\int\limits_a^x e^{f(t)} d t=\int\limits_0^x g(x-t) d t+2 x+3$, if sum of the all the values of $x$ for which $g(x)$ has a local extremum be $\lambda$ then find $\frac{\lambda}{500}$.

Answer 1

$ \because \int_a^{\mathrm{x}} \mathrm{e}^{\mathrm{f}(\mathrm{t})} \mathrm{dt}=\int_0^{\mathrm{x}} \mathrm{g}(\mathrm{x}-1) \mathrm{dt}+2 \mathrm{x}+3 $
$ \Rightarrow \int_{\mathrm{a}}^{\mathrm{x}} \mathrm{e}^{\mathrm{f}(\mathrm{t})} \mathrm{dt}=\int_0^{\mathrm{x}} \mathrm{g}(\mathrm{t}) \mathrm{dt}+2 \mathrm{x}+3 $
(Using King Property)
Differential both sides, we get
$ \mathrm{e}^{f(x)}=g(x)+2$
$ \Rightarrow g(x)=\mathrm{e}^{f(x)}-2$
$ \Rightarrow g^{\prime}(x)=\mathrm{e}^{f(x)} \cdot \mathrm{f}^{\prime}(x) $
$ \because \mathrm{e}^{f(x)} \text { is always greater than zero. }$

Clearly, local extremum (maximum or minimum) will occur at
$x=99,97,95, \ldots, 3,1$
$\therefore$ Sum of all the values $=1+3+5+\ldots .+99$
$=\frac{50}{2}[2 \times 1+(50-1) \times 2]=2500$