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]