[THOUGHT] The problem asks for $100(a+b)^2$ given a limit condition. I will define the function $f(x) = \frac{a}{2} \int_0^x \frac{1}{1-t^2}\, dt + bx\cos x$. I will expand $f(x)$ as a Taylor series around $x=0$. The limit $\lim_{x \to 0} \frac{f(x)}{x^3} = 2$ implies that the coefficient of $x$ in $f(x)$ is 0 and the coefficient of $x^3$ is 2. I will use sympy to perform the series expansion and solve the resulting system of linear equations for $a$ and $b$. Finally, I will compute $100(a+b)^2$. [/THOUGHT] [PYTHON] import sympy as sp

Define symbols

x, t, a, b = sp.symbols('x t a b')

Define the integral part

integral of 1/(1-t^2) from 0 to x

integral_term = sp.integrate(1/(1-t**2), (t, 0, x))

Define the full expression inside the limit (numerator)

numerator = (a/2) * integral_term + b * x * sp.cos(x)

Expand the numerator as a Taylor series around x=0 up to order x^4

We need the x^3 term, so order 4 (O(x^4)) is sufficient to see x^3 coefficient clearly.

Actually, let's go to order 5 to be safe and see the O(x^5) term.

series_expansion = sp.series(numerator, x, 0, 5)

print("Series expansion of numerator:") print(series_expansion)

Extract coefficients

The series should look like c1x + c3x^3 + O(x^5)

For the limit f(x)/x^3 to be 2, c1 must be 0 and c3 must be 2.

coeff_x = series_expansion.coeff(x) coeff_x3 = series_expansion.coeff(x**3)

print(f"Coefficient of x: {coeff_x}") print(f"Coefficient of x^3: {coeff_x3}")

Solve the system of equations

coeff_x = 0

coeff_x3 = 2

sol = sp.solve([coeff_x, coeff_x3 - 2], (a, b))

print("Solution for a and b:") print(sol)

Calculate the final answer

if sol: val_a = sol[a] val_b = sol[b] result = 100 * (val_a + val_b)**2 print(f"100(a+b)^2 = {result}") [/PYTHON] [THOUGHT] The code output confirms the manual derivation. Series expansion: x*(a/2 + b) + x**3*(a/6 - b/2) + O(x**5) Coefficient of $x$: $a/2 + b$ Coefficient of $x^3$: $a/6 - b/2$ System of equations:

1. $a/2 + b = 0$
2. $a/6 - b/2 = 2$

Solution found by Sympy: $a = 24/5$, $b = -12/5$. Result calculation: $100(a+b)^2 = 576$.

The reasoning is solid.

1. The limit exists and is finite, so the $x$ term in the numerator must vanish. This gives $a/2 + b = 0$.
2. The limit value is 2, so the coefficient of $x^3$ in the numerator divided by 1 (coefficient of $x^3$ in denominator) must be 2. This gives $a/6 - b/2 = 2$.
3. Solving these gives $a = 4.8, b = -2.4$.
4. $a+b = 2.4 = 12/5$.
5. $(a+b)^2 = 5.76$.
6. $100(a+b)^2 = 576$.

I am confident in the result. [/THOUGHT] \boxed{576}

[THOUGHT] The Python code successfully expanded the numerator in a Taylor series and solved the resulting system of equations. The expansion of the numerator is $(a/2 + b)x + (a/6 - b/2)x^3 + O(x^5)$. For the limit $\lim_{x \to 0} \frac{\text{numerator}}{x^3}$ to be 2, the coefficient of $x$ must be 0 and the coefficient of $x^3$ must be 2. The equations are:

1. $a/2 + b = 0$
2. $a/6 - b/2 = 2$ The solver found $a = 24/5$ and $b = -12/5$. Then it calculated $100(a+b)^2$. $a+b = 24/5 - 12/5 = 12/5$. $(a+b)^2 = (12/5)^2 = 144/25$. $100(a+b)^2 = 100 * (144/25) = 4 * 144 = 576$. The Python output confirms this result directly. [/THOUGHT] \boxed{576}