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Let $a, b, c, d$ be numbers in the set $\{1,2,3,4,5,6\}$ such that the curves $y=2 x^{3}+a x+b$ and $y=2 x^{3}+c x+d$ have no point in common. The maximum possible value of $(a-c)^{2}+b-d$ is $-$
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Verified Answer
The correct answer is:
5
$y=2 x^{3}+a x+b$
$y=2 x^{3}+c x+d$
No Solution
$\begin{array}{l}
2 x^{3}+a x+b \neq 2 x^{3}+c x+d \\
a x+b \neq c x+d \quad \text { for no real } x \\
(a-c) x \neq d-b \\
x \neq \frac{d-b}{a-c} \quad a=c \\
(a-c)^{2}+(b-d)=0+6-1=5
\end{array}$
$y=2 x^{3}+c x+d$
No Solution
$\begin{array}{l}
2 x^{3}+a x+b \neq 2 x^{3}+c x+d \\
a x+b \neq c x+d \quad \text { for no real } x \\
(a-c) x \neq d-b \\
x \neq \frac{d-b}{a-c} \quad a=c \\
(a-c)^{2}+(b-d)=0+6-1=5
\end{array}$
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