[THOUGHT] To solve this, I will use SymPy to compute the Taylor series expansions of the integral and $x \cos x$ up to sufficient order, form the expression inside the limit, extract the coefficients of $x$ and $x^3$, set the coefficient of $x$ to zero (for the limit to exist), set the coefficient of $x^3$ to 2, solve for $a$ and $b$, compute $a + b$, then $100(a + b)^2$. [/THOUGHT] [PYTHON] from sympy import *

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

Compute the integral

integral = integrate(1/(1 - t**2), (t, 0, x)) print('Integral series:') print(series(integral, x, 0, 6))

x cos x series

xcos = x * cos(x) print('x cos x series:') print(series(xcos, x, 0, 6))

The expression inside

expr = (a/2) * integral + b * xcos

Series of expr

series_expr = series(expr, x, 0, 6) print('Series of expr:') print(series_expr)

Extract coefficients

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

print('Coeff of x:', coeff_x) print('Coeff of x^3:', coeff_x3)

Solve system: coeff_x = 0, coeff_x3 = 2

solutions = solve([coeff_x, coeff_x3 - 2], [a, b]) print('Solutions:', solutions)

Compute a + b

if solutions: sol_a, sol_b = solutions[0] a_plus_b = sol_a + sol_b result = 100 * (a_plus_b)**2 print('100(a + b)^2 =', result) print('Numerical value:', float(result)) [/PYTHON]

[THOUGHT] The KeyError occurs because coeff() on a series with O() term fails for low orders. Remove the O term using .removeO() before extracting coefficients. [/THOUGHT] [PYTHON] from sympy import *

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

Integral

integral = integrate(1/(1 - t**2), (t, 0, x))

x cos x

xcos = x * cos(x)

Expression

expr = (a/2) * integral + b * xcos

Series up to x^5 (order 6)

series_expr = series(expr, x, 0, 6).removeO()

print('Series expr:', series_expr)

Coefficients

coeff_x1 = series_expr.coeff(x) coeff_x3 = series_expr.coeff(x**3)

print('Coeff x^1:', coeff_x1) print('Coeff x^3:', coeff_x3)

Solve coeff_x1 == 0, coeff_x3 == 2

eq1 = coeff_x1 eq2 = coeff_x3 - 2 sols = solve([eq1, eq2], [a, b]) print('Solutions:', sols)

if sols: a_sol, b_sol = sols[a], sols[b] ab_sum = a_sol + b_sol result = 100 * ab_sum**2 print('a + b =', ab_sum) print('100(a+b)^2 =', result) print('Numerical:', N(result)) [/PYTHON]

[THOUGHT] The Python output confirms $100(a + b)^2 = 576$. [/THOUGHT] $\boxed{576}$

Explanation

The reasoning follows the limit condition by expanding the expression in Taylor series around x=0 up to order 5, ensuring the limit as x approaches 0 of the given fraction equals 2. This requires the series of the numerator, $\frac{a}{2} \int_0^x \frac{1}{1-t^2} \, dt + b x \cos x$, to have its constant, x, and x^2 terms vanish, with the x^3 coefficient set to 2 (since divided by x^3).

SymPy computes the integral $\int_0^x \frac{1}{1-t^2} \, dt$, which expands to $x + \frac{x^3}{3} + \frac{x^5}{5} + O(x^7)$. The term $x \cos x$ expands to $x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7)$.

The full expression's series is then $\frac{a}{2} (x + \frac{x^3}{3} + \frac{x^5}{5}) + b (x - \frac{x^3}{6} + \frac{x^5}{120}) + O(x^7)$, simplifying to:

• Coefficient of x: $\frac{a}{2} + b$
• Coefficient of x^3: $\frac{a}{6} - \frac{b}{2}$

Set $\frac{a}{2} + b = 0$ and $\frac{a}{6} - \frac{b}{2} = 2$. Solving yields $a = \frac{24}{5}$, $b = -\frac{12}{5}$, so $a + b = \frac{12}{5}$. Thus, $100(a + b)^2 = 100 \left(\frac{12}{5}\right)^2 = 100 \cdot \frac{144}{25} = 576$, as verified by the Python output.