Problem Analysis #17

[THOUGHT] To solve for $a$ and $b$, I need to evaluate the given limit:

$$\lim_{x \to 0} \frac{1}{x^3} \left( \frac{a}{2} \int_0^x \frac{1}{1-t^2}\, dt + bx\cos x \right) = 2.$$

First, consider the integral $\int_0^x \frac{1}{1-t^2}\, dt$. We can use the series expansion for $\frac{1}{1-t^2}$ around $t = 0$:

$$\frac{1}{1-t^2} = 1 + t^2 + t^4 + \cdots \quad \text{for} \ |t| < 1.$$

Integrating term by term from 0 to $x$:

$$\int_0^x \frac{1}{1-t^2}\, dt = \int_0^x 1\, dt + \int_0^x t^2\, dt + \int_0^x t^4\, dt + \cdots = x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots.$$

Substitute this expansion into the given limit:

$$\frac{1}{x^3} \left( \frac{a}{2} \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots \right) + bx\cos x \right).$$

Next, expand $bx \cos x$ around $x = 0$:

$$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots,$$

so

$$bx \cos x = bx \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \right) = bx - \frac{bx^3}{2} + \frac{bx^5}{24} - \cdots.$$

Combine the expansions:

$$\frac{1}{x^3} \left( \frac{a}{2} x + \frac{a}{2} \frac{x^3}{3} + \frac{a}{2} \frac{x^5}{5} + \cdots + bx - \frac{bx^3}{2} + \frac{bx^5}{24} - \cdots \right).$$ $$= \frac{1}{x^3} \left( \left( \frac{a}{2} + b \right) x + \left( \frac{a}{6} - \frac{b}{2} \right) x^3 + \left( \frac{a}{10} + \frac{b}{24} \right) x^5 + \cdots \right).$$ $$= \frac{a}{2} + b + \left( \frac{a}{6} - \frac{b}{2} \right) + \left( \frac{a}{10} + \frac{b}{24} \right) x^2 + \cdots.$$

For the limit to exist and equal 2 as $x \to 0$, the coefficient of $x$ must be zero, and the coefficient of $x^3$ must be 2:

$$\frac{a}{2} + b = 0,$$ $$\frac{a}{6} - \frac{b}{2} = 2.$$

Solve these equations simultaneously: From the first equation:

$$b = -\frac{a}{2}.$$

Substitute into the second equation:

$$\frac{a}{6} - \frac{-\frac{a}{2}}{2} = 2,$$ $$\frac{a}{6} + \frac{a}{4} = 2,$$ $$\frac{2a + 3a}{12} = 2,$$ $$\frac{5a}{12} = 2,$$ $$a = \frac{24}{5}.$$

Substitute $a = \frac{24}{5}$ back into $b = -\frac{a}{2}$:

$$b = -\frac{24}{5} \cdot \frac{1}{2} = -\frac{12}{5}.$$

Now, calculate $100(a + b)^2$:

$$a + b = \frac{24}{5} - \frac{12}{5} = \frac{12}{5},$$ $$(a + b)^2 = \left( \frac{12}{5} \right)^2 = \frac{144}{25},$$ $$100(a + b)^2 = 100 \cdot \frac{144}{25} = 576.$$

The final answer is:

$$\boxed{576}$$

[/THOUGHT]