Q57 / 90 · KVPY 2012 (SB/SX) · Mathematics

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 $-$

  • A
    0
  • Correct answer
    B
    5
  • C
    30
  • D
    36

Solution

$y=2 x^{3}+a x+b$ $y=2 x^{3}+c x+d$ No Solution $\begin{array}{l} 2 x^{3}+a x+b \neq 2 x^{3}+c x+d \\ a x+b \neq c x+d \quad \text { for no real } x \\ (a-c) x \neq d-b \\ x \neq \frac{d-b}{a-c} \quad a=c \\ (a-c)^{2}+(b-d)=0+6-1=5 \end{array}$

Hence, the correct answer is option: 'B'.

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