[THOUGHT] The problem asks for the evaluation of an expression involving $g(m)$, where $g(m)$ is the number of intersection points between the line $y = m(x-1)$ and the curve $y = f(x)$. The function $f(x)$ is the inverse of $f^{-1}(x)$, which is given piecewise.

First, let's determine the explicit form of $y = f^{-1}(x)$. Since $f$ is increasing, $f^{-1}$ must also be increasing. Condition (A): For $|x| \le 1$, $4(f^{-1}(x))^2 = x^2(x^2-5)^2$. Taking the square root, $2|f^{-1}(x)| = |x(x^2-5)| = |x||x^2-5|$. Since $|x| \le 1$, $x^2 \le 1$, so $x^2-5$ is negative. Thus $|x^2-5| = 5-x^2$. So $2|f^{-1}(x)| = |x|(5-x^2)$. Since $f^{-1}$ is increasing, it must pass through the origin (or near it) and preserve sign. Let's check $f^{-1}(0)$. $4(f^{-1}(0))^2 = 0 \implies f^{-1}(0) = 0$. For $x > 0$, $f^{-1}(x)$ must be positive because $f^{-1}$ is increasing and $f^{-1}(0)=0$. So for $0 \le x \le 1$, $2 f^{-1}(x) = x(5-x^2) \implies f^{-1}(x) = \frac{1}{2}x(5-x^2)$. For $-1 \le x < 0$, let $x = -t$ where $t > 0$. $2|f^{-1}(-t)| = |-t|(5-t^2) = t(5-t^2)$. Since $f^{-1}$ is increasing, $f^{-1}(x)$ must be negative for $x < 0$. So $-2 f^{-1}(x) = -x(5-x^2) \implies f^{-1}(x) = \frac{1}{2}x(5-x^2)$. Thus, for $|x| \le 1$, $f^{-1}(x) = \frac{1}{2}x(5-x^2) = \frac{5}{2}x - \frac{1}{2}x^3$. Let's check the derivative to ensure it's increasing in this interval. $(f^{-1})'(x) = \frac{5}{2} - \frac{3}{2}x^2$. For $|x| \le 1$, $x^2 \le 1$, so $\frac{3}{2}x^2 \le 1.5$. Then $(f^{-1})'(x) \ge 2.5 - 1.5 = 1 > 0$. So it is increasing. At $x=1$, $f^{-1}(1) = \frac{1}{2}(1)(4) = 2$. At $x=-1$, $f^{-1}(-1) = -2$.

Condition (B): For $|x| > 1$, $|f^{-1}(x)| = e^{|x|-1} + 1$. Since $f^{-1}$ is increasing: For $x > 1$, $f^{-1}(x)$ must be positive and greater than $f^{-1}(1)=2$. $f^{-1}(x) = e^{x-1} + 1$. Check continuity at $x=1$: Limit from right is $e^0 + 1 = 2$. Value from left is 2. Continuous. Check derivative for $x > 1$: $(f^{-1})'(x) = e^{x-1} > 0$. Increasing. For $x < -1$, $f^{-1}(x)$ must be negative and less than $f^{-1}(-1)=-2$. $|f^{-1}(x)| = -f^{-1}(x) = e^{-x-1} + 1$. So $f^{-1}(x) = -(e^{-(x+1)} + 1) = -e^{-(x+1)} - 1$. Check continuity at $x=-1$: Limit from left is $-(e^0 + 1) = -2$. Value from right is -2. Continuous. Check derivative for $x < -1$: $(f^{-1})'(x) = -e^{-(x+1)} \cdot (-1) = e^{-(x+1)} > 0$. Increasing.

So $f^{-1}(x)$ is defined as: $f^{-1}(x) = \begin{cases} -e^{-(x+1)} - 1 & x < -1 \\ \frac{5}{2}x - \frac{1}{2}x^3 & -1 \le x \le 1 \\ e^{x-1} + 1 & x > 1 \end{cases}$

The problem involves the intersection of the line $y = m(x-1)$ and the curve $y = f(x)$. The line passes through $(1,0)$. Let's analyze the intersection in the $xy$-plane. $y = f(x) \iff x = f^{-1}(y)$. The line equation is $y = m(x-1) \iff x = \frac{y}{m} + 1$ (for $m \ne 0$). If $m=0$, $y=0$. Since $f(1)=0$ (because $f^{-1}(0)$ is not 1? Wait. $f^{-1}(0) = 0$. So $f(0)=0$. The line passes through $(1,0)$. $f(1)$ is not 0. Let's check $f(1)$. $y=1 \implies x = f^{-1}(1) = 2$. So $(2,1)$ is on $y=f(x)$. $y=0 \implies x = f^{-1}(0) = 0$. So $(0,0)$ is on $y=f(x)$. The line passes through fixed point $P(1,0)$. This point is NOT on the curve $y=f(x)$. We are looking for the number of solutions to $f(x) = m(x-1)$. Let $x$ be the variable on the curve $y=f(x)$. Alternatively, let's switch to the $Y$-domain (which is the domain of $f^{-1}$). Let $y$ be the variable. $x = f^{-1}(y)$. The intersection condition is $y = m(f^{-1}(y) - 1)$. So we need to find the number of roots $y$ for the equation $y = m(f^{-1}(y) - 1)$. Let $h(y) = f^{-1}(y) - 1$. Then we are solving $y = m \cdot h(y)$, or $\frac{y}{h(y)} = m$. Wait, we need to be careful with $h(y)=0$. $h(y) = 0 \implies f^{-1}(y) = 1$. From the definition of $f^{-1}(y)$: For $|y| \le 1$, $f^{-1}(y) = \frac{5}{2}y - \frac{1}{2}y^3$. If $f^{-1}(y) = 1$, then $\frac{5}{2}y - \frac{1}{2}y^3 = 1 \implies 5y - y^3 = 2 \implies y^3 - 5y + 2 = 0$. Let's check roots. $y=2$ is not in interval. $y \approx 0.4$. Actually, let's use Python to analyze the function $k(y) = \frac{y}{f^{-1}(y) - 1}$ and find the range of values of $k(y)$. The number of intersections $g(m)$ corresponds to the number of times the horizontal line $Y = m$ intersects the graph of $k(y)$. However, we must handle the vertical asymptote where $f^{-1}(y) = 1$. Also, consider the case $m=0$ separately. $y=0$. $f^{-1}(0)=0 \ne 1$. So $y=0$ is a solution if $0 = m(0-1) \implies 0=0$. Yes. So $y=0$ is always a solution for any $m$? $y=0 \implies f^{-1}(0)=0$. Equation: $0 = m(0-1) \implies 0 = -m$. So $y=0$ is a solution ONLY if $m=0$. Wait. The equation is $y = m(f^{-1}(y) - 1)$. If $y=0$, LHS=0. RHS = $m(0-1) = -m$. So $m=0$. So for $m \ne 0$, $y=0$ is not a solution. So we can analyze $m = \frac{y}{f^{-1}(y) - 1}$ for $y \ne 0$ and $f^{-1}(y) \ne 1$.

Let's define $K(y) = \frac{y}{f^{-1}(y) - 1}$. We need to plot or analyze $K(y)$ to find the number of solutions for a given $m$. The values of $m$ where $g(m)$ is discontinuous are the local extrema of $K(y)$ or asymptotic values. The problem states $g(m)$ is discontinuous at $m=a$ and $m=b$ with $a < b$. This suggests there are two critical values of $m$. I will use Python to:

1. Define $f^{-1}(y)$.
2. Define $K(y)$.
3. Find critical points of $K(y)$ (where derivative is zero).
4. Find limits at infinity and at vertical asymptotes.
5. Determine $a$ and $b$.
6. Calculate the required expression.

The expression is $g(a) \times (\lim_{m \to a^+} g(m)) + g(b) \times (\frac{\ln b}{b})^2$. Note: $g(m)$ is the number of intersections. $a$ and $b$ are likely the local maximum/minimum values of $K(y)$.

Let's start coding. [/THOUGHT] [PYTHON] import numpy as np import sympy as sp

def f_inv(y): if y < -1: return -np.exp(-(y + 1)) - 1 elif -1 <= y <= 1: return 2.5 * y - 0.5 * y**3 else: return np.exp(y - 1) + 1

def K(y): val = f_inv(y) if val == 1: return np.inf # Handle division by zero return y / (val - 1)

Let's use sympy to find exact derivatives and critical points

y = sp.symbols('y', real=True)

Piece 1: y < -1

f1 = -sp.exp(-(y + 1)) - 1 K1 = y / (f1 - 1)

Piece 2: -1 <= y <= 1

f2 = sp.Rational(5, 2) * y - sp.Rational(1, 2) * y**3 K2 = y / (f2 - 1)

Piece 3: y > 1

f3 = sp.exp(y - 1) + 1 K3 = y / (f3 - 1)

Derivatives

dK1 = sp.diff(K1, y) dK2 = sp.diff(K2, y) dK3 = sp.diff(K3, y)

print("Derivative Piece 1:", sp.simplify(dK1)) print("Derivative Piece 2:", sp.simplify(dK2)) print("Derivative Piece 3:", sp.simplify(dK3))

Solve for critical points where derivative is 0

crit1 = sp.solve(dK1, y) crit2 = sp.solve(dK2, y) crit3 = sp.solve(dK3, y)

print("Critical points 1:", crit1) print("Critical points 2:", crit2) print("Critical points 3:", crit3)

Also need to check the values at boundaries and asymptotes

Asymptote is where f_inv(y) = 1.

In piece 2: 2.5y - 0.5y^3 = 1 => y^3 - 5y + 2 = 0

roots_asymp = sp.solve(y**3 - 5*y + 2, y) print("Asymptote roots (Piece 2):", [r.evalf() for r in roots_asymp])

Check limits at infinity

lim_inf_neg = sp.limit(K1, y, -sp.oo) lim_inf_pos = sp.limit(K3, y, sp.oo) print("Limit y -> -inf:", lim_inf_neg) print("Limit y -> +inf:", lim_inf_pos)

Check values at boundaries y = -1 and y = 1

val_neg1 = K(-1) val_pos1 = K(1) print("K(-1):", val_neg1) print("K(1):", val_pos1)

[/PYTHON]

[THOUGHT] Let's analyze the function $K(y) = \frac{y}{f^{-1}(y) - 1}$ to determine the number of intersections $g(m)$. The function $K(y)$ determines the value of $m$ for a given $y$. We are looking for the number of $y$ values for a given $m$. The domain of $f^{-1}(y)$ is all real numbers. Let's break down the behavior of $K(y)$ in different regions.

Piece 1 ($y < -1$): $f^{-1}(y) = -e^{-(y+1)} - 1$. $K(y) = \frac{y}{-e^{-(y+1)} - 2}$. Derivative is zero at $y = -1 - W(2)$. $W(2) \approx 0.85$. So $y_c \approx -1.85$. Let's evaluate $K(y_c)$. $y_c = -1 - W(2)$. Note that $W(2)e^{W(2)} = 2 \implies e^{W(2)} = 2/W(2)$. $e^{-(y_c+1)} = e^{W(2)} = 2/W(2)$. $K(y_c) = \frac{-1-W(2)}{-2/W(2) - 2} = \frac{-(1+W(2))}{-2(1+W(2))/W(2)} = \frac{1}{2/W(2)} = \frac{W(2)}{2}$. Since $W(2) \approx 0.85$, $K(y_c) \approx 0.425$. Limit as $y \to -\infty$: $K(y) \to 0$ (from positive side, since numerator is neg and denom is neg). At $y = -1$: $K(-1) = \frac{-1}{-1-1-1} = \frac{-1}{-3} = 1/3 \approx 0.333$. So in this region, $K(y)$ goes from 0 up to $W(2)/2$ and then down to $1/3$. Wait, let's check the sign of derivative. Numerator of derivative is proportional to $-y - 2e^{y+1} - 1$. At $y \to -\infty$, large positive. At $y=-1$, $1 - 2 - 1 = -2$ negative. So it increases then decreases. Max at $y_c$. Local max value $m_1 = W(2)/2$.

Piece 2 ($-1 \le y \le 1$): $f^{-1}(y) = \frac{5}{2}y - \frac{1}{2}y^3$. $K(y) = \frac{y}{\frac{5}{2}y - \frac{1}{2}y^3 - 1} = \frac{2y}{5y - y^3 - 2} = \frac{-2y}{y^3 - 5y + 2}$. Roots of denominator (asymptotes): $y^3 - 5y + 2 = 0$. From Python output: $2, \approx 0.414, \approx -2.414$. Inside $[-1, 1]$, the only root is $y_0 \approx 0.414$. Let's call this root $\alpha$. So there is a vertical asymptote at $y = \alpha$. Let's check behavior around $\alpha$. $y^3 - 5y + 2$. Derivative $3y^2 - 5$. At $\alpha \approx 0.4$, deriv is negative. So for $y < \alpha$, denom is positive. For $y > \alpha$, denom is negative. Numerator is $-2y$. For $y \in [-1, 0)$, num > 0, denom > 0. $K(y) > 0$. For $y \in (0, \alpha)$, num < 0, denom > 0. $K(y) < 0$. For $y \in (\alpha, 1]$, num < 0, denom < 0. $K(y) > 0$. Values at boundaries: $K(-1) = 1/3$. $K(0) = 0$. $K(1) = \frac{-2}{1-5+2} = \frac{-2}{-2} = 1$. Derivative of $K(y)$ in this region: From Python: $4(y^3-1)/(y^3-5y+2)^2$. Critical points where $y^3=1 \implies y=1$. So $y=1$ is a critical point? Let's check derivative at $y=1$: $4(0)/... = 0$. Yes. Is it a local extremum? For $y < 1$ (close to 1), $y^3 < 1$, deriv < 0. So $K(y)$ is decreasing. So at $y=1$, we have a local minimum? Wait, for $y \in (\alpha, 1)$, $K(y) > 0$. As $y \to \alpha^+$, $K(y) \to +\infty$ (num negative, denom negative small). So $K(y)$ decreases from $+\infty$ to $K(1)=1$. So $y=1$ is a local minimum with value 1. For $y \in [-1, \alpha)$: $K(-1) = 1/3$. $K(0) = 0$. Derivative zero only at $y=1$ (not in this sub-interval). So $K(y)$ decreases from $1/3$ to $0$ at $y=0$, then decreases to $-\infty$ as $y \to \alpha^-$. Wait, check derivative sign. $y^3-1$ is negative for $y < 1$. So derivative is always negative in $[-1, \alpha)$. So $K(y)$ decreases from $1/3$ to $-\infty$.

Piece 3 ($y > 1$): $f^{-1}(y) = e^{y-1} + 1$. $K(y) = \frac{y}{e^{y-1}}$. Derivative: $(1-y)e^{1-y}$. Zero at $y=1$. For $y > 1$, $1-y < 0$, so derivative is negative. $K(1) = 1$. Limit as $y \to \infty$: $0$. So $K(y)$ decreases from $1$ to $0$.

Summary of $K(y)$ graph:

1. $y \in (-\infty, -1)$: Increases from 0 to $W(2)/2$ (approx 0.425), then decreases to $1/3$ (approx 0.333).
2. $y \in [-1, \alpha)$: Decreases from $1/3$ to $-\infty$. (Passes through 0 at $y=0$).
3. $y \in (\alpha, 1]$: Decreases from $+\infty$ to $1$.
4. $y \in (1, \infty)$: Decreases from $1$ to $0$.

Let's collect the ranges covered by $K(y)$ (which corresponds to $m$):

• Region 1 ($y < -1$): Range $(0, W(2)/2]$. Max is $W(2)/2$. End value $1/3$.
• Region 2a ($y \in [-1, \alpha)$): Range $(-\infty, 1/3]$.
• Region 2b ($y \in (\alpha, 1]$): Range $[1, \infty)$.
• Region 3 ($y > 1$): Range $(0, 1)$.

Let's visualize the number of solutions $g(m)$ for a given $m$. We draw the horizontal line $Y=m$ and count intersections with graph of $K(y)$.

• $m < 0$: Only intersects Region 2a. (1 solution).
• $m = 0$: $y=0$ is a solution. Also limits at $\pm \infty$ are 0. But $K(y)=0$ only at $y=0$. (1 solution).
• $0 < m < 1/3$:
  • Region 1: Intersects twice (once going up, once going down).
  • Region 2a: Intersects once (between -1 and 0).
  • Region 3: Intersects once.
  • Total: 4 solutions.
• $m = 1/3$:
  • Region 1: Intersects once (at $y=-1$ boundary? No, $y=-1$ is boundary. $K(-1)=1/3$. In Region 1, value approaches $1/3$ at boundary. In Region 2a, starts at $1/3$. So $y=-1$ is one point).
  • Region 1 peak is $W(2)/2 \approx 0.425 > 1/3$. So there is another intersection in Region 1.
  • Region 3: Intersects once.
  • Total: 3 solutions.
• $1/3 < m < W(2)/2$:
  • Region 1: Intersects twice.
  • Region 2a: No intersection (max is 1/3).
  • Region 3: Intersects once.
  • Total: 3 solutions.
• $m = W(2)/2$:
  • Region 1: Tangent at peak (1 solution).
  • Region 3: Intersects once.
  • Total: 2 solutions.
• $W(2)/2 < m < 1$:
  • Region 1: No intersection.
  • Region 3: Intersects once.
  • Total: 1 solution.
• $m = 1$:
  • Region 2b: Minimum at $y=1$. Value 1. (1 solution).
  • Region 3: Starts at 1. (Continuous at 1). So $y=1$ is the point.
  • Total: 1 solution.
• $m > 1$:
  • Region 2b: Intersects twice? No, monotonic decreasing from $\infty$ to 1. So 1 intersection.
  • Region 3: Max is 1 (at boundary), decreases to 0. No intersection for $m>1$.
  • Total: 1 solution.

Wait, let's re-verify the monotonicity in Region 2b. $K(y)$ decreases from $+\infty$ to 1. So for $m > 1$, there is exactly 1 solution in $(\alpha, 1)$. So for $m > 1$, $g(m) = 1$. At $m=1$, $g(1) = 1$ (at $y=1$). For $m \in (W(2)/2, 1)$, $g(m) = 1$ (in Region 3). So $g(m)$ is constant 1 for $m > W(2)/2$. Wait, is $W(2)/2 < 1$? $W(2) \approx 0.85$. $W(2)/2 \approx 0.425$. Yes.

Let's check discontinuities. $g(m)$ values:

• $m < 0$: 1
• $m = 0$: 1
• $0 < m < 1/3$: 4
• $m = 1/3$: 3
• $1/3 < m < W(2)/2$: 3
• $m = W(2)/2$: 2
• $W(2)/2 < m < 1$: 1
• $m = 1$: 1
• $m > 1$: 1

Discontinuities are at:

1. $m = 0$: jumps from 1 to 4.
2. $m = 1/3$: jumps from 4 to 3.
3. $m = W(2)/2$: jumps from 3 to 2, then to 1.

Wait, let's trace carefully. Range of Region 1: $(0, W(2)/2]$. Range of Region 2a: $(-\infty, 1/3]$. Range of Region 2b: $[1, \infty)$. Range of Region 3: $(0, 1)$.

Let's sum the number of solutions for each interval of $m$:

1. $m < 0$:
   • R1: 0
   • R2a: 1
   • R2b: 0
   • R3: 0
   • Total: 1
2. $m = 0$:
   • Total: 1 ($y=0$)
3. $0 < m < 1/3$:
   • R1: 2 (values go 0 -> max -> 1/3. Since $1/3 < max$, any value in $(0, 1/3)$ is hit twice)
   • R2a: 1 (values go 1/3 -> 0 -> -inf. Hits $(0, 1/3)$ once)
   • R2b: 0
   • R3: 1 (values 1 -> 0. Hits once)
   • Total: 2 + 1 + 1 = 4.
4. $m = 1/3$:
   • R1: 1 (at $y=-1$, value is 1/3. But wait, R1 is $y<-1$. Limit is 1/3. R2a is $y \ge -1$. Value is 1/3. So $y=-1$ is the solution. It counts once.)
   • But does R1 hit 1/3 again? R1 goes $0 \to max \to 1/3$. If $max > 1/3$, it hits 1/3 somewhere else?
   • $max = W(2)/2 \approx 0.425$. $1/3 \approx 0.333$. Yes. So there is a solution in $y < -1$ where $K(y) = 1/3$.
   • So R1 contributes 1 solution strictly inside, plus the boundary $y=-1$.
   • R2a starts at $1/3$ (at $y=-1$). So $y=-1$ is the solution.
   • R3 contributes 1 solution ($1/3 \in (0, 1)$).
   • Total: 1 (inside R1) + 1 (boundary) + 1 (R3) = 3.
5. $1/3 < m < W(2)/2$:
   • R1: 2 solutions.
   • R2a: 0 solutions (max is 1/3).
   • R2b: 0.
   • R3: 1 solution.
   • Total: 3.
6. $m = W(2)/2$:
   • R1: 1 solution (peak).
   • R3: 1 solution.
   • Total: 2.
7. $W(2)/2 < m < 1$:
   • R1: 0.
   • R3: 1.
   • Total: 1.
8. $m = 1$:
   • R2b: 1 (at $y=1$).
   • R3: Limit is 1. $y=1$ is the solution.
   • Total: 1.
9. $m > 1$:
   • R2b: 1.
   • R3: 0.
   • Total: 1.

So discontinuities are at:

• $m=0$ (1 to 4)
• $m=1/3$ (4 to 3)
• $m=W(2)/2$ (3 to 1)
• $m=1$ (constant 1? No, 1 to 1. Continuous?)

Let's check $m=1$. Left limit ($m \to 1^-$): 1. Value ($m=1$): 1. Right limit ($m \to 1^+$): 1. So continuous at $m=1$.

Discontinuities are at $m=0$, $m=1/3$, $m=W(2)/2$. The problem says "discontinuous at $m=a$ and $m=b$ with $a < b$". This implies exactly two discontinuities. Let's re-evaluate $m=0$. Limit $m \to 0^-$ is 1. Value $g(0) = 1$. Limit $m \to 0^+$ is 4. So discontinuous at $a=0$.

Now check $m=1/3$ and $m=W(2)/2$. At $m=1/3$: Left limit 4. Value 3. Right limit 3. Discontinuous. At $m=W(2)/2$: Left limit 3. Value 2. Right limit 1. Discontinuous.

So there are 3 discontinuities? $0, 1/3, W(2)/2$. The problem says "discontinuous at $m=a$ and $m=b$". Maybe one of them is not a discontinuity? Or maybe I miscounted solutions. Let's check $m=W(2)/2$. $W(2)/2 \approx 0.425$. Is it possible that $W(2)/2 = 1/3$? No. Is it possible that $W(2)/2 = 1$? No.

Let's re-read carefully. "Suppose that $g(m)$ is discontinuous at $m=a$ and $m=b$". This implies ONLY two points. Let's check the $m=0$ case again. $y=0$ is always a root for $m=0$. For $m \to 0^+$, we found 4 roots. For $m \to 0^-$, we found 1 root. Definitely discontinuous at 0. So $a=0$.

Now between $1/3$ and $W(2)/2$. Maybe $g(m)$ is not discontinuous at one of them? At $m=1/3$: $m \to (1/3)^-$: 4 solutions. $m = 1/3$: 3 solutions. $m \to (1/3)^+$: 3 solutions. Jump from 4 to 3. Discontinuous.

At $m=W(2)/2$: $m \to (W(2)/2)^-$: 3 solutions. $m = W(2)/2$: 2 solutions. $m \to (W(2)/2)^+$: 1 solution. Jump from 3 to 1. Discontinuous.

Why would the problem imply only two? Maybe my analysis of Region 1 max is wrong? $K(y) = \frac{y}{-e^{-(y+1)} - 2}$. Derivative zero at $-y - 2e^{y+1} - 1 = 0$. Let $t = y+1$. $- (t-1) - 2e^t - 1 = 0 \implies -t + 1 - 2e^t - 1 = 0 \implies -t - 2e^t = 0 \implies t = -2e^t$. $-t e^{-t} = 2$. $(-t) e^{(-t)} = 2$. So $-t = W(2)$. $t = -W(2)$. $y = -1 - W(2)$. Max value: $K(y_c) = W(2)/2$. Is it possible $W(2)/2 = 1/3$? $W(2) \approx 0.8526$. $W(2)/2 \approx 0.4263$. $1/3 \approx 0.3333$. They are distinct.

Is it possible I missed a region? $y \in (-\infty, -1)$: Range $(0, 0.426]$. $y \in [-1, \alpha)$: Range $(-\infty, 1/3]$. $y \in (\alpha, 1]$: Range $[1, \infty)$. $y \in (1, \infty)$: Range $(0, 1)$.

Let's re-evaluate the overlaps. Intervals of $m$ where solutions exist: R1: $(0, 0.426]$ R2a: $(-\infty, 0.333]$ R2b: $[1, \infty)$ R3: $(0, 1)$

Summing up counts: For $m < 0$: R2a (1). Total 1. For $m = 0$: 1. For $0 < m < 0.333$: R1 (2) + R2a (1) + R3 (1) = 4. For $m = 0.333$: R1 (2) + R2a (1 boundary) + R3 (1) = 4? Wait. At $m=1/3$: R1 intersects twice? $1/3 < 0.426$. Yes. One intersection is at $y=-1$. The other is inside $y < -1$. R2a intersects at $y=-1$. So $y=-1$ is counted in both R1 and R2a? No, $y=-1$ is a single point. So we have:

1. One root in R1 (strict inequality $y < -1$).
2. One root at $y = -1$.
3. One root in R3. Total 3 roots. Wait. Limit $m \to 1/3^-$: 4 roots. Limit $m \to 1/3^+$: For $m > 1/3$: R1 has 2 roots (since $m < 0.426$). R2a has 0 roots (max is 1/3). R3 has 1 root. Total 3 roots. So at $m=1/3$, $g(m)$ jumps from 4 to 3. So it is discontinuous.

For $m = 0.426$ ($W(2)/2$): R1 has 1 root. R3 has 1 root. Total 2 roots. Limit $m \to 0.426^-$: 3 roots. Limit $m \to 0.426^+$: R1 has 0 roots. R3 has 1 root. Total 1 root. Discontinuous.

So we have 3 points of discontinuity: $0, 1/3, W(2)/2$. Why does the problem say $a$ and $b$? Maybe $a=0$ is not considered? "For a real number $m$..." Usually $m$ is slope. Maybe the domain of $g$ is restricted? No. Maybe I should check the $m=1$ point again. At $m=1$: R2b has min 1 at $y=1$. R3 has sup 1 at $y=1$. So $y=1$ is the solution. For $m < 1$ (close to 1): R2b has no solution. R3 has 1 solution. Total 1. For $m > 1$: R2b has 1 solution. R3 has 0 solutions. Total 1. So $g(m)$ is continuous at $m=1$. Correct.

Is it possible that $W(2)/2$ is not a discontinuity? No, jump from 3 to 1. Is it possible $1/3$ is not? Jump from 4 to 3.

Could $a$ and $b$ be $1/3$ and $W(2)/2$? If so, what about 0? Maybe the question implies $a$ and $b$ are the only discontinuities? Or maybe $a$ and $b$ are positive? "Suppose that $g(m)$ is discontinuous at $m=a$ and $m=b$ with $a < b$." This phrasing usually implies these are the only ones. Let's re-read condition A. $4(f^{-1}(x))^2 = x^2(x^2-5)^2$. $2|f^{-1}(x)| = |x(x^2-5)|$. For $|x| \le 1$, $x^2-5 < 0$. So $|x^2-5| = 5-x^2$. $2|f^{-1}(x)| = |x|(5-x^2)$. $f^{-1}(x)$ is increasing. $f^{-1}(0)=0$. For $x>0$, $f^{-1}(x) > 0$. So $2f^{-1}(x) = x(5-x^2)$. For $x<0$, $f^{-1}(x) < 0$. So $-2f^{-1}(x) = -x(5-x^2) \implies f^{-1}(x) = x(5-x^2)/2$. So $f^{-1}(x) = \frac{5}{2}x - \frac{1}{2}x^3$ for all $|x| \le 1$. This matches my derivation.

Let's check the derivative at $y=-1$. $f^{-1}(y) = 2.5y - 0.5y^3$. Derivative: $2.5 - 1.5y^2$. At $y=-1$: $2.5 - 1.5 = 1$. Left side: $f^{-1}(y) = -e^{-(y+1)} - 1$. Derivative: $e^{-(y+1)}$. At $y=-1$: $e^0 = 1$. So $f^{-1}$ is differentiable at $y=-1$. This means the curve $x = f^{-1}(y)$ is smooth at $y=-1$. The function $K(y) = y / (f^{-1}(y) - 1)$. At $y=-1$, $f^{-1}(-1) = -2$. $K(-1) = -1 / (-3) = 1/3$. $K'(y) = \frac{1 \cdot (f^{-1}-1) - y (f^{-1})'}{(f^{-1}-1)^2}$. At $y=-1$: Num = $(-2-1) - (-1)(1) = -3 + 1 = -2$. Denom = $(-3)^2 = 9$. $K'(-1) = -2/9$. So $K(y)$ is smooth at $y=-1$ and decreasing. So $K(y)$ passes through $1/3$ with non-zero slope. This means $K(y)$ takes values slightly above $1/3$ for $y < -1$ and slightly below $1/3$ for $y > -1$. Wait. In Region 1 ($y < -1$), $K(y)$ goes up to $W(2)/2$ then down to $1/3$. In Region 2a ($y > -1$), $K(y)$ goes down from $1/3$. So $1/3$ is NOT a local maximum. It is just a value passed through. Wait. Region 1: $y \in (-\infty, -1)$. $K(y)$ ends at $1/3$. Region 2a: $y \in [-1, \alpha)$. $K(y)$ starts at $1/3$ and decreases. So around $y=-1$, $K(y)$ is decreasing. Values of $K(y)$ for $y$ near $-1$: $y < -1$: $K(y) > 1/3$. $y = -1$: $K(y) = 1/3$. $y > -1$: $K(y) < 1/3$. So as we vary $m$ across $1/3$: For $m > 1/3$ (slightly):