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(C) Given that Rolle's theorem is applicable for the function $f(x) = x^{2m-1}(a-x)^{2n}$ in $(0, a)$.We find the derivative $f'(x)$ using the product

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$\frac{a(2m-1)}{2m+2n-1}$

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(C) Given that Rolle's theorem is applicable for the function $f(x) = x^{2m-1}(a-x)^{2n}$ in $(0, a)$.We find the derivative $f'(x)$ using the product rule:$f'(x) = (2m-1)x^{2m-2}(a-x)^{2n} - 2n(a-x)^{2n-1}x^{2m-1}$.According to Rolle's theorem,there exists a $c \in (0, a)$ such that $f'(c) = 0$.$(2m-1)c^{2m-2}(a-c)^{2n} - 2nc^{2m-1}(a-c)^{2n-1} = 0$.Dividing by $c^{2m-2}(a-c)^{2n-1}$ (since $c 
eq 0$ and $c 
eq a$):$(2m-1)(a-c) = 2nc$.$(2m-1)a - (2m-1)c = 2nc$.$(2m-1)a = (2m-1+2n)c$.$c = \frac{a(2m-1)}{2m+2n-1}$.Thus,option $C$ is correct.

For $m > 1, n > 1$,the value of $c$ for which the Rolle's theorem is applicable for the function $f(x) = x^{2m-1}(a-x)^{2n}$ in $(0, a)$ is

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