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(B) Given $f(x) = \int_{0}^{x^{2}} \frac{t^{2}-5t+4}{2+e^{t}} dt$.Using the Leibniz rule for differentiation under the integral sign,we have:$f'(x) =

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$(2, 3)$

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(B) Given $f(x) = \int_{0}^{x^{2}} \frac{t^{2}-5t+4}{2+e^{t}} dt$.Using the Leibniz rule for differentiation under the integral sign,we have:$f'(x) = \frac{(x^{2})^{2}-5(x^{2})+4}{2+e^{x^{2}}} \cdot \frac{d}{dx}(x^{2})$$f'(x) = \frac{x^{4}-5x^{2}+4}{2+e^{x^{2}}} \cdot (2x)$$f'(x) = \frac{2x(x^{2}-1)(x^{2}-4)}{2+e^{x^{2}}}$$f'(x) = \frac{2x(x-1)(x+1)(x-2)(x+2)}{2+e^{x^{2}}}$The critical points are $x = -2, -1, 0, 1, 2$.We analyze the sign of $f'(x)$ around these points:For $x For $-2  0$.For $-1 For $0  0$.For $1 For $x > 2$,$f'(x) > 0$.Local minima occur where $f'(x)$ changes from negative to positive: at $x = -2, 0, 2$. Thus,$n = 3$.Local maxima occur where $f'(x)$ changes from positive to negative: at $x = -1, 1$. Thus,$m = 2$.Therefore,the ordered pair $(m, n)$ is $(2, 3)$.

If $m$ and $n$ respectively are the number of local maximum and local minimum points of the function $f(x) = \int_{0}^{x^{2}} \frac{t^{2}-5t+4}{2+e^{t}} dt$,then the ordered pair $(m, n)$ is equal to

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