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The number of integers in the range of 'c' such that there exists a line which intersects the curve $ y = x^4 – 6x^3 + 12x^2 + cx + 1$ at four distinct points.

My approach we need to intersect with line $y=mx+C$

Substituting we get $x^4 – 6x^3 + 12x^2 + cx + 1-mx-C=0$

Now this is an equation of polynomial of degree 4

$x^4 – 6x^3 + 12x^2 + (c-m)x + 1-C=0$

All four roots needs to be real.

I don't have any idea how to check whether all the roots are real or not.

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  • $\begingroup$ Just suggestion: after applying substitution $x=2t+3$, one can consider intersection of symmetric curve $z=2t^4-3t^2$ with line $z=At+B$, where $A,B$ depend on $c$ (or $c$ depends on $A,B$). $\endgroup$
    – Oleg567
    Oct 18, 2019 at 6:21
  • $\begingroup$ Wikipedia has an entry but I suspect there is a simpler solution. $\endgroup$ Oct 18, 2019 at 6:54

3 Answers 3

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Note that in order for a line to be able to meet this curve at four points, the curve must "change direction" thrice, which is to say that the derivative must have three roots.

The derivative in question is visibly $$g'(x) = 4x^3 -18x^2+24x+c$$ For it to have three roots, it must "change direction" twice, and on the same side of the $x\text{-axis}$. To check that, see its derivative: $$g''(x)= 12x^2 - 36x + 24 = 12(x^2-3x+2)$$ which visibly has roots at $x = 2, 1$. Now all that is left is to ensure that these occur on both sides of the $x$ axis in $g'(x)$.

So, we have: $$g'(2) = 4\cdot 8 - 18 \cdot 4 + 24 \cdot 2 + c = c+8$$ $$g'(1) = 4-18+24+c = c+10$$

So, $c+10$ and $c+8$ have to be on different sides of the $x$ axis. Obviously $c+10>c+8$ so the former would lie above the axis. So, we have $c+10>0$ and $c+8<0$, so our range is $$\boxed{c\in(-10,-8)}$$

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    $\begingroup$ It doesn't need to change “direction thrice”. $y=2x+\cos x$ never changes direction. However, the line $y=2x$ intersects it infinetely many times. $\endgroup$ Oct 18, 2019 at 12:37
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    $\begingroup$ Well, the question is about intersecting four times. But that is a technicality, I should clarify. $\endgroup$ Oct 18, 2019 at 12:39
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The second derivative $y''(x)=12x^2-36x+24=12(x-1)(x-2)$ has roots $\{1,2\}$ independently of $c$. Thus, $y(x)$ has two points of inflection at $(1, 8+c)$ and $(2, 17+2c)$. A line that goes through these two points will always intersect $y(x)$ at 4 points (two points of indflection and $\{(3\pm\sqrt5)/2\}$).

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Consider equation $$ x^4 – 6x^3 + 12x^2 - 9x + 2 = 0.\tag{1} $$ It is equivalent to $$ (x - 2) (x - 1) (x^2 - 3 x + 1) = 0,\tag{2} $$ and has $4$ real solutions: $$ x_{1} = 1,\\ x_{2} = 2, \\ x_{3} = \dfrac{3-\sqrt{5}}{2}, \\ x_{4} = \dfrac{3-\sqrt{5}}{2}.\tag{3} $$

Then for any $c\in\mathbb{R}$ we can rewrite $(1)$ in the form $$ 0 = x^4 – 6x^3 + 12x^2 -9x + 2 \\ = (x^4 – 6x^3 + 12x^2 +cx + 1) - ((9+c)x-1), $$ which means that equation $$ x^4 – 6x^3 + 12x^2 +cx + 1 = (9+c)x-1 $$ has those four real solutions: see $(3)$. Other words, given curve and line $$y=(9+c)x-1$$ intersect in four distinct points with abscissas from list $(3)$.

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