Problem Analysis #33

[THOUGHT] I will use Python to verify the solution found manually and check for other possibilities. The manual derivation suggests $t=2$, $\alpha=11/3$, $k=3/14$. I need to check if there are other tuples $(t, \alpha, k)$ that satisfy the conditions. Especially the case where $t=\alpha$.

The conditions are:

1. $f(x) = k x (x-2) (x-\alpha)$ with $k>0$.
2. $t \in \{0, 2, \alpha\}$.
3. $g(x) = -f(x)$ for $x < t$, $f(x)$ for $x \ge t$.
4. $A = g(-1) \in \mathbb{N}$.
5. $B = -3.5 g(1) \in \mathbb{N}$.
6. $S = \{m \in \mathbb{N} \mid L(m) < 0\}$.
7. $S = \{A, B\}$.
8. $A \neq B$.

I will write a script to sweep through possible integer sets $S$ (since $A, B$ are integers) or solve the equations for different cases of $t$. Since $A, B$ are determined by parameters, I can just solve for parameters given the structure of $S$. Possible structures for $S$: If $t=2$, $S=\{2, 3, \dots\}$. Since $|S|=2$, $S=\{2, 3\}$. This led to a solution. If $t=0$, $S=\{2, \dots\}$? If $t=\alpha$, the sign of $L(m)$ changes behavior.

Let's define a function check_solution(k, alpha, t) that returns $S, A, B$ and checks consistency. Then I will solve for $k, \alpha$ for different $t$ and set $S$. [/THOUGHT] [PYTHON] import sympy from sympy import symbols, solve, S

def solve_problem(): # Variables k, alpha = symbols('k alpha', real=True)

# We identified t must be a root. Roots are 0, 2, alpha.
# Case 1: t = 2
# We derived alpha = 11/3, k = 3/14. Let's verify.

solutions = []

# Function to evaluate g(x)
def get_g(x_val, k_val, alpha_val, t_val):
    f_val = k_val * x_val * (x_val - 2) * (x_val - alpha_val)
    if x_val < t_val:
        return -f_val
    else:
        return f_val

# Function to check limit sign
def get_limit_sign(m, k_val, alpha_val, t_val):
    # L(m) = lim_{x->m+} g(x)/(x(x-2))
    # If m != 0, 2: sign is sign of g(m)
    if m != 0 and m != 2:
        g_val = get_g(m, k_val, alpha_val, t_val)
        return 1 if g_val > 0 else (-1 if g_val < 0 else 0)
    
    # If m = 2
    if m == 2:
        # x -> 2+. x > 2.
        # If t <= 2, g(x) = f(x). L(2) = f'(2)/2 = k(2)(2-alpha)/2 = k(2-alpha) ? No.
        # f(x)/(x(x-2)) = k(x-alpha). Limit is k(2-alpha).
        # If t > 2, for x in (2, t), g(x) = -f(x). Limit is -k(2-alpha) = k(alpha-2).
        if t_val <= 2:
            val = k_val * (2 - alpha_val)
        else:
            val = k_val * (alpha_val - 2)
        return 1 if val > 0 else (-1 if val < 0 else 0)
    
    # If m = 1 (handled above)
    return 0

# Let's generalize the solver
# Possible values for t: 0, 2, alpha
# We need A = g(-1) and B = -3.5 g(1) to be integers.
# S = {A, B}.

# Let's try to solve algebraically for each case of t and set S.
# Since S has 2 elements, let's denote them {n1, n2}.
# We derived for t=2, S={2, 3}.

# Case t=2
# S = {2, 3}
# A, B are 2, 3 in some order.
# Equations:
# g(-1) = 3k(1+alpha)
# -3.5 g(1) = 3.5 k (alpha-1)
# We found alpha=11/3, k=3/14.

# Let's check this specific solution first.
k1 = 3/14
a1 = 11/3
t1 = 2

print(f"Checking t=2, alpha={a1}, k={k1}")
A = get_g(-1, k1, a1, t1)
B = -3.5 * get_g(1, k1, a1, t1)
print(f"A = {A}, B = {B}")

# Check S
S_set = []
for m in range(1, 10):
    sign = get_limit_sign(m, k1, a1, t1)
    if sign < 0:
        S_set.append(m)
print(f"S = {S_set}")

if {A, B} == set(S_set):
    print("Valid solution found!")
    # Calculate g(-5)
    g_neg5 = get_g(-5, k1, a1, t1)
    print(f"g(-5) = {g_neg5}")
    solutions.append(g_neg5)

# Are there other cases?
# What if t = alpha?
# If t = alpha, we need to check the structure of S.
# L(m) signs:
# m=1: -g(1).
# m=2: limit.
# m >= 3: g(m).

# Let's try to find other solutions by brute forcing integer pairs for A, B?
# Or solving symbolically.

# Let's check t=alpha case symbolically.
# g(-1) depends on -1 < alpha.
# g(1) depends on 1 < alpha.

# Subcase: alpha > 2. Then t > 2.
# g(-1) = -f(-1) = 3k(1+alpha) (since -1 < t)
# g(1) = -f(1) = -k(alpha-1) = k(1-alpha) (since 1 < t)
# B = -3.5 g(1) = -3.5 k (1-alpha) = 3.5 k (alpha-1).
# Note: For t=2, B was 3.5 k (alpha-1). Same formula.
# A = 3k(1+alpha). Same formula.
# So A and B formulas are same as t=2 case if alpha > 2.

# Now check S for t=alpha (>2).
# L(1) = -g(1) = -k(1-alpha) = k(alpha-1). Since alpha > 2, L(1) > 0. 1 not in S.
# L(2): t > 2. Limit is k(alpha-2). Since alpha > 2, L(2) > 0. 2 not in S.
# m >= 3.
# If m < t: g(m) = -f(m) = -k m (m-2) (m-alpha).
#   Since m >= 3, m-alpha < 0 (if m < alpha).
#   So f(m) < 0. -f(m) > 0. L(m) > 0.
# If m >= t: g(m) = f(m).
#   Since m >= alpha, f(m) >= 0. L(m) >= 0.
# So for t=alpha > 2, S is empty!
# Thus t=alpha > 2 yields no solution.

# Subcase: 0 < alpha < 2. Then t = alpha is in (0, 2).
# g(-1) = -f(-1) = 3k(1+alpha). (since -1 < t)
# g(1): 1 vs t.
# If 1 < t (1 < alpha < 2):
#   g(1) = -f(1) = k(1-alpha).
#   B = -3.5 k (1-alpha).
#   L(1) = -g(1) = -k(1-alpha) < 0. So 1 in S.
#   L(2): t < 2. Limit is k(2-alpha). Since alpha < 2, L(2) > 0. 2 not in S.
#   m >= 3: g(m) = f(m) (since m > t).
#   f(m) = k m (m-2) (m-alpha). All factors positive. L(m) > 0.
#   So S = {1}. But we need |S|=2. No solution.

# If 1 >= t (0 < alpha <= 1):
#   g(1) = f(1) = k(alpha-1).
#   B = -3.5 k (alpha-1) = 3.5 k (1-alpha).
#   L(1) = -g(1) = k(1-alpha) > 0. 1 not in S.
#   L(2): t < 2. Limit k(2-alpha) > 0. 2 not in S.
#   m >= 3: L(m) > 0.
#   S empty.

# Subcase: alpha < 0. Then t = alpha < 0.
# g(-1): -1 vs t.
# If -1 < t (alpha > -1): Contradiction with alpha < 0.
# If -1 >= t (alpha <= -1):
#   g(-1) = f(-1) = -3k(1+alpha) = 3k(-1-alpha).
#   g(1) = f(1) = k(alpha-1).
#   B = -3.5 k (alpha-1) = 3.5 k (1-alpha).
#   L(1) = -g(1) = k(1-alpha) > 0.
#   L(2): t < 2. Limit k(2-alpha) > 0.
#   m >= 3: L(m) > 0.
#   S empty.

# Case t=0.
# g(-1) = -f(-1) = 3k(1+alpha).
# g(1) = f(1) = k(alpha-1).
# B = 3.5 k (1-alpha).
# L(1) = -g(1) = k(1-alpha).
# If alpha < 1, L(1) > 0.
# L(2) = k(2-alpha).
# If alpha < 1, L(2) > 0.
# m >= 3: L(m) > 0.
# S empty.

# It seems t=2 is the only viable case.

solve_problem() [/PYTHON]

The function $f(x)$ is a cubic with positive leading coefficient $k$. Let $f(x) = kx(x-2)(x-\alpha)$. The function $g(x)$ is defined as $-f(x)$ for $x < t$ and $f(x)$ for $x \ge t$. Condition (A) requires the right-hand limit of $g(x)/(x(x-2))$ to exist everywhere. This is satisfied if $g(0)=0$ and $g(2)=0$. Since $f(0)=f(2)=0$, $g(0)$ and $g(2)$ are 0 regardless of $t$ (assuming $t$ doesn't break the root structure, but $g$ is continuous, so $g(t)=0$ implies $t \in \{0, 2, \alpha\}$). Condition (B) involves the set $S = \{m \in \mathbb{N} \mid \lim_{x \to m^+} \frac{g(x)}{x(x-2)} < 0\}$. We are given $S = \{g(-1), -3.5 g(1)\}$. Let $A = g(-1)$ and $B = -3.5 g(1)$. For the case $t=2$, we found $S = \{2, 3\}$. Matching $\{A, B\} = \{2, 3\}$ yields the solution $\alpha = 11/3$ and $k = 3/14$. With these parameters: $g(-1) = -f(-1) = 3$. $-3.5 g(1) = -3.5 (-f(1)) = 2$. So $A=3, B=2$, which matches $S=\{2, 3\}$. We need to compute $g(-5)$. Since $-5 < 2$, $g(-5) = -f(-5)$. Using $f(x) = \frac{3}{14}x(x-2)(x-\frac{11}{3})$: [ f(-5) = \frac{3}{14}(-5)(-7)\left(-5-\frac{11}{3}\right) = \frac{3}{14}(35)\left(-\frac{26}{3}\right) = \frac{3 \cdot 35 \cdot (-26)}{14 \cdot 3} = \frac{35 \cdot (-26)}{14} = \frac{5}{2} \cdot (-26) = -65 ] Thus, $g(-5) = -(-65) = 65$.

\boxed{65}