Let a, b, c, d be numbers in set {1,2,3,4,5,6} such that curves y=2 x^3+a x+b and y=2 x^3+c x+d have

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4-2.Quadratic Equations and Inequations

normal

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

$0$

$5$

$30$

$36$

(KVPY-2012)

Solution

(b)

We have,

$y=2 x^3+ a x +b$

$y=2 x^3+c x+ d$

$\text { From Eqs. (i) and (ii), we get }$

$\quad 2 x^3+a x+b=2 x^3+c x+ d$

$\Rightarrow \quad(a-c) x=(d-b)$

Since, Eqs. (i) and (ii) have no common

point.

$\therefore \quad(a-c) x \neq(d-b)$

If $(a-c)=0$ and $(d-b) \neq 0$

$\therefore$ Maximum value of $(a-c)^2+(b-d)$

$=0+(b-d)$

$=6-1=5 \quad[\because b, d \in\{1,2,3,4,5,6\}$

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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

Solution

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