[THOUGHT] To solve this problem, we need to analyze the function ( f(x) ) and its inverse ( f^{-1}(x) ) based on the given conditions. The goal is to find the values ( a ) and ( b ) where the number of intersection points ( g(m) ) is discontinuous, and then compute the specified expression.

1. Understand ( f^{-1}(x) ):

   • For ( |x| \leq 1 ), ( 4(f^{-1}(x))^2 = x^2(x^2 - 5)^2 ).
   • For ( |x| > 1 ), ( |f^{-1}(x)| = e^{|x| - 1} + 1 ).

2. Find ( f(x) ):

   • Since ( f ) is increasing and continuous, ( f^{-1} ) is also increasing and continuous.
   • For ( |x| \leq 1 ), solve for ( f^{-1}(x) ): [ f^{-1}(x) = \pm \frac{x(x^2 - 5)}{2} ] But since ( f^{-1} ) is increasing, we take the positive root: [ f^{-1}(x) = \frac{x(5 - x^2)}{2} ]
   • For ( |x| > 1 ), ( f^{-1}(x) = e^{x - 1} + 1 ) (since ( f^{-1} ) is increasing).

3. Find ( f(x) ):

   • For ( x \leq 1 ), ( f(x) ) is the inverse of ( f^{-1}(x) = \frac{x(5 - x^2)}{2} ).
   • For ( x > 1 ), ( f(x) ) is the inverse of ( f^{-1}(x) = e^{x - 1} + 1 ).

4. Find ( g(m) ):

   • ( g(m) ) counts the number of intersection points between the line ( y = m(x - 1) ) and ( y = f(x) ).
   • Discontinuities in ( g(m) ) occur when the line is tangent to ( f(x) ).

5. Find ( a ) and ( b ):

   • Solve for ( m ) where the line is tangent to ( f(x) ).

6. Compute the expression:

   • Evaluate ( g(a) \times \left(\lim_{m \to a^+} g(m)\right) + g(b) \times \left( \frac{\ln b}{b} \right)^2 ).

[/THOUGHT] [PYTHON] import sympy as sp

Define the variables

x, m = sp.symbols('x m')

Define the inverse function f^{-1}(x)

def f_inv(x_val): if abs(x_val) <= 1: return (x_val * (5 - x_val**2)) / 2 else: return sp.exp(abs(x_val) - 1) + 1

Define the line equation: y = m(x - 1)

line = m * (x - 1)

Find the points of intersection between y = f(x) and the line

We need to find x such that f(x) = m(x - 1)

Since f(x) is the inverse of f^{-1}(x), we can write:

f(x) = y where f^{-1}(y) = x

So, x = f^{-1}(y) and y = m(x - 1)

Substitute y into the inverse function:

x = f^{-1}(m(x - 1))

For |m(x - 1)| <= 1:

x = (m(x - 1) * (5 - (m(x - 1))**2)) / 2

For |m(x - 1)| > 1:

x = exp(|m(x - 1)| - 1) + 1

To find the number of solutions, we can analyze the equation numerically or graphically.

However, for the purpose of finding discontinuities in g(m), we look for tangency conditions.

For |x| <= 1, the equation is:

x = (m(x - 1) * (5 - (m(x - 1))**2)) / 2

Let's expand and solve for m:

2x = m(x - 1)(5 - m^2(x - 1)^2)

This is a nonlinear equation in x and m.

For |x| > 1, the equation is:

x = exp(|m(x - 1)| - 1) + 1

This is also nonlinear.

To find the critical m values where g(m) is discontinuous, we need to find m where the line is tangent to f(x).

This requires solving for m where the system has a double root.

Let's consider the case where |x| <= 1:

The equation is 2x = m(x - 1)(5 - m^2(x - 1)^2)

Let u = x - 1, then x = u + 1

2(u + 1) = m u (5 - m^2 u^2)

2u + 2 = 5m u - m^3 u^3

m^3 u^3 - 5m u + 2u + 2 = 0

u(m^3 u^2 - 5m + 2) + 2 = 0

This is complex to solve analytically, so we might need to find m numerically.

Alternatively, we can find the derivative condition for tangency.

The slope of f(x) must equal m at the point of tangency.

The derivative of f(x) is 1 / (derivative of f^{-1}(y)) where y = f(x).

For |y| <= 1:

f^{-1}(y) = y(5 - y^2)/2

derivative of f^{-1}(y) = (5 - y^2)/2 + y(-2y)/2 = (5 - y^2 - 2y^2)/2 = (5 - 3y^2)/2

So, f'(x) = 1 / ((5 - 3y^2)/2) = 2 / (5 - 3y^2)

At the point of tangency, f'(x) = m, so:

m = 2 / (5 - 3y^2)

Also, y = m(x - 1) and x = f^{-1}(y) = y(5 - y^2)/2

Substitute x into y = m(x - 1):

y = m(y(5 - y^2)/2 - 1)

y = m(y(5 - y^2) - 2)/2

2y = m(y(5 - y^2) - 2)

2y = 5m y - m y^3 - 2m

m y^3 - 5m y + 2y + 2m = 0

y(m y^2 - 5m + 2) + 2m = 0

This is the same as before.

Let's solve for y when m = 2 / (5 - 3y^2):

Substitute m into the equation:

y( (2 / (5 - 3y^2)) y^2 - 5(2 / (5 - 3y^2)) + 2 ) + 2(2 / (5 - 3y^2)) = 0

Multiply through by (5 - 3y^2):

y(2y^2 - 10 + 2(5 - 3y^2)) + 4 = 0

y(2y^2 - 10 + 10 - 6y^2) + 4 = 0

y(-4y^2) + 4 = 0

-4y^3 + 4 = 0

y^3 = 1

y = 1

Then, m = 2 / (5 - 3(1)^2) = 2 / 2 = 1

So, one critical point is at m = 1.

For |y| > 1:

f^{-1}(y) = exp(y - 1) + 1

derivative of f^{-1}(y) = exp(y - 1)

So, f'(x) = 1 / exp(y - 1) = exp(1 - y)

At the point of tangency, f'(x) = m, so:

m = exp(1 - y)

Also, y = m(x - 1) and x = f^{-1}(y) = exp(y - 1) + 1

Substitute x into y = m(x - 1):

y = m(exp(y - 1) + 1 - 1) = m exp(y - 1)

But m = exp(1 - y), so:

y = exp(1 - y) exp(y - 1) = exp(0) = 1

But |y| > 1, so no solution here.

Thus, the only critical point is at m = 1.

However, the problem states there are two discontinuities at m = a and m = b with a < b.

This suggests we might have missed another case.

Let's consider the transition point where |x| = 1.

At x = 1, f^{-1}(1) = 1(5 - 1)/2 = 2

So, f(2) = 1

The line y = m(x - 1) passes through (1,0).

At x = 2, y = m(2 - 1) = m

For the line to be tangent to f(x) at x = 2, f(2) = m and f'(2) = m.

f(2) = 1, so m = 1.

f'(2) = 1 / f^{-1}'(1) = 1 / ((5 - 3(1)^2)/2) = 2/2 = 1

So, m = 1 is indeed a critical point.

For the other critical point, consider the behavior as m increases.

For large m, the line y = m(x - 1) will intersect the exponential part of f(x) at one point.

The transition occurs when the line is tangent to the exponential part.

Let's solve for m where the line is tangent to f(x) for x > 1.

f(x) is the inverse of f^{-1}(y) = exp(y - 1) + 1 for y > 1.

So, f(x) = ln(x - 1) + 1 for x > 2 (since f^{-1}(1) = 2).

The line y = m(x - 1) must be tangent to y = ln(x - 1) + 1.

Set ln(x - 1) + 1 = m(x - 1)

And the derivative condition: 1/(x - 1) = m

Substitute m = 1/(x - 1) into the first equation:

ln(x - 1) + 1 = (x - 1)/(x - 1) = 1

So, ln(x - 1) = 0 => x - 1 = 1 => x = 2

But x = 2 is the transition point, not in the x > 2 region.

Thus, no other tangency points exist in the x > 2 region.

This suggests that the only discontinuity is at m = 1, but the problem states two discontinuities.

Perhaps the other discontinuity is at m = 0.

For m = 0, the line is y = 0.

The number of intersections with f(x) is the number of solutions to f(x) = 0.

For x <= 1, f(x) is the inverse of f^{-1}(y) = y(5 - y^2)/2.

Solve f(x) = 0 => x = f^{-1}(0) = 0.

For x > 1, f(x) = ln(x - 1) + 1 > 1 > 0.

So, only one intersection at x = 0.

For m slightly positive, the line y = m(x - 1) will intersect f(x) at two points (one for x < 1 and one for x > 1).

Thus, g(m) jumps from 1 to 2 at m = 0.

Therefore, the discontinuities are at m = 0 and m = 1.

Now, compute the required expression:

g(a) * lim_{m -> a+} g(m) + g(b) * (ln b / b)^2

Here, a = 0, b = 1.

g(0) = 1 (only one intersection at x = 0).

lim_{m -> 0+} g(m) = 2 (for m > 0, two intersections).

g(1) = 1 (tangency at x = 2).

(ln 1 / 1)^2 = 0.

So, the expression is:

1 * 2 + 1 * 0 = 2.

[/PYTHON]

[THOUGHT] The previous approach identified the discontinuities at ( m = 0 ) and ( m = 1 ).

• For ( m = 0 ), ( g(0) = 1 ) (only one intersection at ( x = 0 )).
• For ( m ) slightly greater than 0, ( g(m) = 2 ) (two intersections).
• For ( m = 1 ), ( g(1) = 1 ) (tangency at ( x = 2 )).
• The term ( \left( \frac{\ln 1}{1} \right)^2 = 0 ).

Thus, the expression evaluates to: [ g(0) \times \left(\lim_{m \to 0^+} g(m)\right) + g(1) \times \left( \frac{\ln 1}{1} \right)^2 = 1 \times 2 + 1 \times 0 = 2. ]

[/THOUGHT] \boxed{2}