Let f be real function defined on R (the set of real numbers) such that  $f^{\text{'}}\left(x\right)=100(x-1){(x-2)}^2{(x-3)}^3\mathrm{...}{(x-100)}^{100}$  for all  $x\in R$. If g is a function defined on R such that  $\underset{a}{\overset{x}{\int }}{e}^{f\left(t\right)}dt=\underset{0}{\overset{x}{\int }}g(x-t)dt+2x+3$, if sum of the all the values of x for which g(x) has a local extremum be $\lambda$  then find  $\frac{\lambda }{500}$.

Given Information:

1. f(x) Function:

f'(x) = 100(x-1)(x-2)^2(x-3)^3 ... (x-100)^100
This is a derivative of the function f(x), where each term involves a factor of (x - k)^{k} for k = 1, 2, ..., 100.

2. Integral Condition:

∫_a^x e^f(t) dt = ∫₀^x g(x-t) dt + 2x + 3 We have two integrals here, one involving the function e^f(t) and the other involving g(x - t).

3. Extrema of g(x): The function g(x) has a local extremum for some values of x, and you are asked to find λ/500, where λ is the sum of all values of x for which g(x) has a local extremum.

Solution:

1. Function Analysis:

We can analyze f(x) by looking at the roots of the equation f'(x) = 0. The factors in the expression for f'(x) suggest that the function will have zeros at x = 1, 2, ..., 100, each with multiplicity depending on the power of the corresponding factor.

2. Local Extrema of g(x):

The integral condition gives us information about the behavior of the function g(x), but we need to identify the values of x where g(x) has local extrema. To do this, we look for points where the derivative of g(x) with respect to x is zero.

3. Sum of Values:

The values of x that lead to local extrema for g(x) are tied to the roots of the function f(x), so they correspond to the points where f(x) changes behavior, which would be x = 1, 2, ..., 100.

4. Final Answer:

The sum λ is the sum of all these values, which is:

λ = 1 + 2 + 3 + ... + 100

The sum of the first 100 natural numbers is given by the formula:

λ = (100(100+1))/2 = 5050

Finally, you are asked to find λ/500:

λ/500 = 5050/500 = 10.1

Answer:

λ/500 = 10.1