Here is the question:

If $f(x)=(x-\alpha)^n g(x)$, then $f(\alpha)=f'(\alpha)=f''(\alpha)=...=f^{n-1}(\alpha)=0$, where$f(x)$ and $g(x)$ are polynomials.
For a polynomial $f(x)$ with rational coefficients, answer the following questions:

  1. If $f(x)$ touches x-axis at only one point, then the point of touching is
    (a) always a rational number
    (b) may or may not be a rational number
    (c) never a rational number
    (d) none of these

  2. If $f(x)$ is of degree 3 and touches x-axis, then
    (a)all the roots of $f(x)$ are rational
    (b) only one root is rational
    (c) both (a) and (b) may be possible
    (d) none of these

  3. $f(\alpha)=f'(\alpha)=f''(\alpha)=0$,$f(\beta)=f'(\beta)=f''(\beta)=0$ and $f(x)$ is a polynomial of degree 6, then
    (a) all the roots of $f''(x)=0$ are real
    (b) at least two roots of $f''(x)=0$
    (c) exactly two roots of $f''(x)=0$ are real
    (d) none of these


  1. I got this one by contradiction. I assumed that the only point of touching is irrational so therefore $\alpha$ which will be the point should be irrational. This implies that $f(x)$ will have irrational coefficients which is not the case. Therefore, $\alpha$ has to be rational.

  2. I got stuck here. I only know that the point of touching has to be rational. That's it. I don't know about the other two roots.

  3. I think I got this one but I am not so sure.
    According to given condition, $f(x)= (x-\alpha)^3(x-\beta)^3$
    $\implies f'(x)=3(x-\alpha)^2(x-\beta)^2(2x-(\alpha+\beta))$
    Therefore the roots are $\alpha$ , $\beta$ and a root each in $(\alpha,\frac{\alpha+\beta}{2})$ and $(\frac{\alpha+\beta}{2},\beta)$

P.S. Please do correct me if I have done anything wrong in the ones that I have solved.

  • $\begingroup$ 1b is a sure bet. Does touching exclude crossing? $\endgroup$ – William Elliot Nov 12 '18 at 9:20
  • $\begingroup$ 1.) $(x^3-2)^2$. 2) reduce to $(x-a)^2(x+2a)=x^3-3a^2x+2a^3$, obvious from last two coefficients. 3) correct. $\endgroup$ – Lutz Lehmann Nov 12 '18 at 15:08

For 1.

In your approach, it is not true in general that

This implies that $f(x)$ will have irrational coefficients

As LutzL comments, $f(x)=(x^3-2)^2$ satisfies our conditions since this can be written as $f(x)=(x-\sqrt[3]{2})^2(x^2+\sqrt[3]2\ x+\sqrt[3]4)^2$ where $\sqrt[3]2$ is irrational. Also, $f(x)=(x-1)^2$ satisfies our conditions. So, (b) is correct.

For 2.

$f(x)$ has to be of the form $$f(x)=a(x-\alpha)^2(x-\beta)=ax^3-a(\beta+2\alpha)x^2+a\alpha(\alpha+2\beta)x-a\alpha^2\beta$$ where $a\ (\not=0),\beta+2\alpha,\alpha(\alpha+2\beta)$ and $\alpha^2\beta$ are rational with $\alpha\not=\beta$.

$\beta+2\alpha$ is rational, so if $\beta$ is rational, then $\alpha$ is rational. So, (b)(c) are incorrect.

Here, let $$s=\beta+2\alpha,\qquad t=\alpha(\alpha+2\beta),\qquad u=\alpha^2\beta$$ where $s,t,u$ are rational.

Since $\beta=s-2\alpha$, we get $$t=\alpha(\alpha+2s-4\alpha),\qquad u=\alpha^2(s-2\alpha),$$ i.e. $$\alpha^2=\frac{2s\alpha-t}{3},\qquad \alpha^2(2\alpha-s)+u=0$$ Substituting the former into the latter gives $$\frac{2s\alpha-t}{3}(2\alpha-s)+u=0,$$ i.e. $$4s\alpha^2-2s^2\alpha-2t\alpha+st+3u=0,$$ i.e. $$4s\frac{2s\alpha-t}{3}-2s^2\alpha-2t\alpha+st+3u=0,$$ i.e. $$(s^2-3t)\alpha=\frac{st-9u}{2}$$

If $s^2-3t=0$, then we have $st-9u=0$. So, $3t=s^2,27u=3st=s^3$ to have $$s=\beta+2\alpha,\qquad s^2=3\alpha(\alpha+2\beta),\qquad s^3=27\alpha^2\beta$$ So, we have $$(\beta+2\alpha)\times 3\alpha(\alpha+2\beta)=27\alpha^2\beta$$$$\implies 6\alpha(\alpha-\beta)^2=0\implies \alpha=0$$since $\alpha\not=\beta$. So, both $\alpha$ and $\beta$ are rational.

If $s^2-3t\not =0$, then both $\alpha$ and $\beta$ are rational.

Therefore, we see that both $\alpha$ and $\beta$ are rational.

It follows that (a) is correct.

For 3.

What you've done looks correct to me.

(It should be better to start with $f(x)=a(x-\alpha)^3(x-\beta)^3$ where $a\ (\not=0)$ is rational and write "we may assume that $\alpha\lt\beta$".)


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