1. Key Concepts and Formulas

• Leibniz's Rule for Differentiation of Integrals: If $f(x) = \int_{a(x)}^{b(x)} g(t) dt$, then $f'(x) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$. This is a direct application of the Fundamental Theorem of Calculus for integrals with variable limits.
• First Derivative Test for Local Extrema: To find local maxima and minima, we analyze the sign changes of the first derivative, $f'(x)$.
  • If $f'(x)$ changes from positive to negative at a critical point $c$, then $f(c)$ is a local maximum.
  • If $f'(x)$ changes from negative to positive at a critical point $c$, then $f(c)$ is a local minimum.
• Critical Points: These are points where $f'(x) = 0$ or $f'(x)$ is undefined. For functions defined by integrals, the denominator of $f'(x)$ (if any) must also be analyzed for points where it could be zero.

1. Step-by-Step Solution

Step 1: Differentiate the function $f(x)$ using Leibniz's Rule. The given function is $f(x) = \int\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt}$. Here, $g(t) = \frac{t^2 - 5t + 4}{2 + e^t}$, the upper limit is $b(x) = x^2$, and the lower limit is $a(x) = 0$. We need to find the derivatives of the limits: $b'(x) = \frac{d}{dx}(x^2) = 2x$, and $a'(x) = \frac{d}{dx}(0) = 0$.

Applying Leibniz's Rule: $f'(x) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$ $f'(x) = \left( \frac{(x^2)^2 - 5(x^2) + 4}{2 + e^{x^2}} \right) \cdot (2x) - \left( \frac{0^2 - 5(0) + 4}{2 + e^0} \right) \cdot (0)$ The second term is zero because $a'(x)=0$. $f'(x) = 2x \left( \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \right)$

Step 2: Find the critical points by setting $f'(x) = 0$. Critical points are candidates for local extrema. We set $f'(x) = 0$: $2x \left( \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \right) = 0$ The denominator $2 + e^{x^2}$ is always positive for all real $x$ since $e^{x^2} > 0$. Thus, the denominator is never zero or undefined. Therefore, $f'(x)$ is defined for all real $x$. For $f'(x)$ to be zero, the numerator must be zero: $2x (x^4 - 5x^2 + 4) = 0$ This equation holds if $2x = 0$ or $x^4 - 5x^2 + 4 = 0$.

Case 1: $2x = 0 \implies x = 0$.

Case 2: $x^4 - 5x^2 + 4 = 0$. This is a quadratic equation in $x^2$. Let $y = x^2$. $y^2 - 5y + 4 = 0$ Factoring the quadratic: $(y - 1)(y - 4) = 0$ Substituting back $y = x^2$: $(x^2 - 1)(x^2 - 4) = 0$ This yields two possibilities:

• $x^2 - 1 = 0 \implies x^2 = 1 \implies x = \pm 1$.
• $x^2 - 4 = 0 \implies x^2 = 4 \implies x = \pm 2$.

Combining all values, the critical points are $x = -2, -1, 0, 1, 2$.

Step 3: Analyze the sign of $f'(x)$ around the critical points. The expression for $f'(x)$ is: $f'(x) = \frac{2x(x-1)(x+1)(x-2)(x+2)}{2 + e^{x^2}}$ Since the denominator $2 + e^{x^2}$ is always positive, the sign of $f'(x)$ is determined by the sign of the numerator: $N(x) = 2x(x-1)(x+1)(x-2)(x+2)$. We arrange the critical points in increasing order: $-2, -1, 0, 1, 2$. These points divide the number line into six intervals. We analyze the sign of $N(x)$ in each interval.

Step 4: Identify the number of local maximum (m) and local minimum (n) points.

• Local Maximum Points (m): occur where $f'(x)$ changes from positive to negative.

  • At $x = -1$, $f'(x)$ changes from $+$ to $-$. This is a local maximum.
  • At $x = 1$, $f'(x)$ changes from $+$ to $-$. This is a local maximum. So, $m = 2$.

• Local Minimum Points (n): occur where $f'(x)$ changes from negative to positive.

  • At $x = -2$, $f'(x)$ changes from $-$ to $+$. This is a local minimum.
  • At $x = 0$, $f'(x)$ changes from $-$ to $+$. This is a local minimum.
  • At $x = 2$, $f'(x)$ changes from $-$ to $+$. This is a local minimum. So, $n = 3$.

Step 5: Form the ordered pair (m, n). We found $m=2$ and $n=3$. The ordered pair is $(2, 3)$.

1. Common Mistakes & Tips

• Incorrect application of Leibniz's Rule: Ensure the derivative of the variable limit ($b'(x)$ and $a'(x)$) is multiplied correctly. Forgetting $b'(x)$ is a common error.
• Sign analysis errors: Carefully check the sign of each factor and the overall product in each interval. A sign error will lead to misclassification of critical points.
• Ignoring the denominator: While the denominator $2+e^{x^2}$ is always positive here, in other problems, a denominator could be zero or change sign, introducing additional critical points or affecting sign changes. Always check the denominator.
• Confusing Maxima and Minima: Remember the sign changes: positive to negative for a maximum, negative to positive for a minimum.

1. Summary

To determine the number of local maximum and minimum points of the given function, we first found its derivative $f'(x)$ using Leibniz's Rule for differentiating integrals. We then identified the critical points by setting $f'(x)=0$. The critical points were found by solving $2x(x^4 - 5x^2 + 4) = 0$, leading to $x \in \{-2, -1, 0, 1, 2\}$. Finally, we analyzed the sign changes of $f'(x)$ around these critical points. A change from positive to negative in $f'(x)$ indicates a local maximum, and a change from negative to positive indicates a local minimum. This analysis revealed 2 local maximum points ($m=2$) and 3 local minimum points ($n=3$). Therefore, the ordered pair $(m, n)$ is $(2, 3)$.

The final answer is $\boxed{(2, 3)}$, which corresponds to option (B).