Let a, b, c, d be numbers in the set {1, 2, 3 | vedclass

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(B) Given the equations of the curves:$y = 2x^3 + ax + b$  $(i)$$y = 2x^3 + cx + d$  $(ii)$For the curves to have no point in common,the equation $2x^

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$5$

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(B) Given the equations of the curves:$y = 2x^3 + ax + b$  $(i)$$y = 2x^3 + cx + d$  $(ii)$For the curves to have no point in common,the equation $2x^3 + ax + b = 2x^3 + cx + d$ must have no real solution for $x$.This simplifies to $(a - c)x = d - b$.If $a - c 
eq 0$,then $x = \frac{d - b}{a - c}$ is always a real solution,which contradicts the condition.Therefore,we must have $a - c = 0$,which implies $a = c$.Substituting $a = c$ into the equation,we get $0 = d - b$,or $b = d$.However,the problem implies the curves have no common point,which means the equation $(a - c)x = d - b$ must be inconsistent.If $a = c$,then $0 = d - b$. For this to have no solution,we must have $d - b 
eq 0$.We want to maximize $(a - c)^2 + b - d$. Since $a = c$,this expression becomes $0 + b - d = b - d$.To maximize $b - d$ where $b, d \in \{1, 2, 3, 4, 5, 6\}$ and $b 
eq d$,we choose $b = 6$ and $d = 1$.Thus,the maximum value is $6 - 1 = 5$.

Let $a, b, c, d$ be numbers in the set $\{1, 2, 3, 4, 5, 6\}$ such that the curves $y = 2x^3 + ax + b$ and $y = 2x^3 + cx + d$ have no point in common. The maximum possible value of $(a - c)^2 + b - d$ is

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