Cfrgtkky

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

MY APPROACH:

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.

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 $$f(x)=a(x-alpha)^2(x-beta)=ax^3-a(beta+2alpha)x^2+aalpha(alpha+2beta)x-aalpha^2beta$$ where $a (not=0),beta+2alpha,alpha(alpha+2beta)$ and $alpha^2beta$ are rational with $alphanot=beta$.

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

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

Since $beta=s-2alpha$, we get
$$t=alpha(alpha+2s-4alpha),qquad u=alpha^2(s-2alpha),$$
i.e.
$$alpha^2=frac{2salpha-t}{3},qquad alpha^2(2alpha-s)+u=0$$
Substituting the former into the latter gives
$$frac{2salpha-t}{3}(2alpha-s)+u=0,$$
i.e.
$$4salpha^2-2s^2alpha-2talpha+st+3u=0,$$
i.e.
$$4sfrac{2salpha-t}{3}-2s^2alpha-2talpha+st+3u=0,$$
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
$$s=beta+2alpha,qquad s^2=3alpha(alpha+2beta),qquad s^3=27alpha^2beta$$
So, we have
$$(beta+2alpha)times 3alpha(alpha+2beta)=27alpha^2beta$$$$implies 6alpha(alpha-beta)^2=0implies alpha=0$$since $alphanot=beta$. So, both $alpha$ and $beta$ are rational.

If $s^2-3tnot =0$, 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 $alphaltbeta$".)