Question

Let $f$ be real function defined on $R$ (the set of real numbers) such that $f^{\prime }(x)=100(x-1)(x-2{)}^2(x-3{)}^3\dots .(x-100{)}^{100}$, for all $x\in R$. If $g$ is a function defined on $R$ such that $\underset{a}{\overset{x}{\int }}{e}^{f(t)}dt=\underset{0}{\overset{x}{\int }}g(x-t)dt+2x+3$, 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

Answer: 5.00

$\because {\int }_a^{\mathrm{x}}{\mathrm{e}}^{\mathrm{f}(\mathrm{t})}\mathrm{dt}={\int }_0^{\mathrm{x}}\mathrm{g}(\mathrm{x}-1)\mathrm{dt}+2\mathrm{x}+3$
$\Rightarrow {\int }_{\mathrm{a}}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{f}(\mathrm{t})}\mathrm{dt}={\int }_0^{\mathrm{x}}\mathrm{g}(\mathrm{t})\mathrm{dt}+2\mathrm{x}+3$
(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
$x=99,97,95,\dots ,3,1$
$\therefore$ Sum of all the values $=1+3+5+\dots .+99$
$=\frac{50}{2}[2\times 1+(50-1)\times 2]=2500$