[THOUGHT] We need to find real numbers a and b such that the limit as x→0 of (1/x^3) * (a/2 ∫_0^x 1/(1-t^2) dt + b x cos x) = 2.

Let’s define f(x) = (a/2) ∫_0^x 1/(1-t^2) dt + b x cos x.

We want lim_{x→0} f(x)/x^3 = 2.

We need to expand f(x) as a series up to x^3 term.

First, ∫_0^x 1/(1-t^2) dt = ∫_0^x (1/(1-t)(1+t)) dt = (1/2) ln((1+t)/(1-t)) evaluated from 0 to x = (1/2) ln((1+x)/(1-x)). So ∫_0^x 1/(1-t^2) dt = (1/2) ln((1+x)/(1-x)).

Thus a/2 * ∫_0^x 1/(1-t^2) dt = a/2 * (1/2) ln((1+x)/(1-x)) = (a/4) ln((1+x)/(1-x)).

So f(x) = (a/4) ln((1+x)/(1-x)) + b x cos x.

Now expand ln((1+x)/(1-x)) = ln(1+x) - ln(1-x). Using series:

ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ... ln(1-x) = -x - x^2/2 - x^3/3 - x^4/4 - ...

So ln((1+x)/(1-x)) = [x - x^2/2 + x^3/3 - x^4/4 + ...] - [-x - x^2/2 - x^3/3 - x^4/4 - ...] = (x + x) + (-x^2/2 + x^2/2) + (x^3/3 + x^3/3) + (-x^4/4 + x^4/4) + ... = 2x + 0x^2 + (2/3)x^3 + 0x^4 + ...

Thus ln((1+x)/(1-x)) = 2x + (2/3)x^3 + O(x^5).

Therefore (a/4) ln((1+x)/(1-x)) = (a/4)*(2x + (2/3)x^3 + ...) = (a/2)x + (a/6)x^3 + O(x^5).

Now expand b x cos x: cos x = 1 - x^2/2 + x^4/24 - ... So b x cos x = b x (1 - x^2/2 + ...) = b x - (b/2) x^3 + O(x^5).

Thus f(x) = (a/2)x + (a/6)x^3 + b x - (b/2) x^3 + O(x^5) = (a/2 + b) x + (a/6 - b/2) x^3 + O(x^5).

Now f(x)/x^3 = (a/2 + b)/x^2 + (a/6 - b/2) + O(x^2).

For the limit as x→0 to exist and equal 2, the coefficient of 1/x^2 must be 0, and the constant term must be 2.

Thus we have system:

1. a/2 + b = 0 → b = -a/2.

2. a/6 - b/2 = 2.

Substitute b = -a/2 into second equation:

a/6 - (-a/2)/2 = a/6 - (-a/4) = a/6 + a/4 = (2a/12 + 3a/12) = 5a/12 = 2.

So a = 2 * 12/5 = 24/5.

Then b = -a/2 = -12/5.

Thus a + b = 24/5 - 12/5 = 12/5.

Then (a+b)^2 = (12/5)^2 = 144/25.

Then 100(a+b)^2 = 100 * 144/25 = (100/25)144 = 4144 = 576.

Let’s verify with Python using symbolic expansion or numeric limit.

We'll compute limit using sympy to confirm. [/THOUGHT] [PYTHON] import sympy as sp

x = sp.symbols('x') a, b = sp.symbols('a b')

Define f(x)

f = (a/2) * sp.integrate(1/(1 - t**2), (t, 0, x)) + b * x * sp.cos(x)

Expand series up to x^3

series_f = sp.series(f, x, 0, 4) print("Series expansion of f(x) up to x^3:") print(series_f)

Compute limit as x->0 of f(x)/x^3

limit_expr = sp.limit(f / x**3, x, 0) print("\nLimit expression (in terms of a,b):") print(limit_expr)

Solve for a,b such that limit = 2

eq = sp.Eq(limit_expr, 2) solutions = sp.solve(eq, (a, b)) print("\nSolutions (a,b):") print(solutions)

Compute 100*(a+b)**2 for the solution

for sol in solutions: a_val, b_val = sol result = 100 * (a_val + b_val)*2 print(f"a={a_val}, b={b_val}, 100(a+b)^2 = {result}") print(f"Simplified: {sp.simplify(result)}") [/PYTHON]

[THOUGHT] The Python output indicates that sympy's limit gave a result involving oosign(...), meaning the limit expression is piecewise depending on the sign of a/2 + b. However, our earlier series expansion gave the correct condition: a/2 + b must be 0 for the limit to exist finite. Then a/6 - b/2 = 2. Solving gives a=24/5, b=-12/5, a+b=12/5, (a+b)^2=144/25, 100(a+b)^2=576. Since the series expansion is valid and the limit exists, we must have a/2 + b = 0. The sympy limit likely returned a piecewise expression because it didn't assume the limit exists. But we can enforce the condition that the coefficient of 1/x^2 is zero. Let's solve manually: a/2 + b = 0 and a/6 - b/2 = 2. Solving gives a=24/5, b=-12/5. Then 100*(a+b)^2 = 100*(12/5)^2 = 100144/25 = 4144 = 576. We'll trust our derivation.

Thus final answer is 576.

\boxed{576}