Mixing
For a polynomial $f(x)$ with rational coefficients:
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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:
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 theseIf $f(x)$ 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$f(alpha)=f'(alpha)=f''(alpha)=0$,$f(beta)=f'(beta)=f''(beta)=0$ 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:
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.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.
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.
calculus derivatives
asked Nov 12 at 8:43
Vaibhav
558
This question has an open bounty worth +50
reputation from Vaibhav ending in 6 days.
Looking for an answer drawing from credible and/or official sources.
I really need a precise answer. I have just received comments that are unsatisfactory too.
add a comment |
up vote
0
down vote
favorite
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:
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 theseIf $f(x)$ 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$f(alpha)=f'(alpha)=f''(alpha)=0$,$f(beta)=f'(beta)=f''(beta)=0$ 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:
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.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.
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.
calculus derivatives
asked Nov 12 at 8:43
Vaibhav
558
This question has an open bounty worth +50
reputation from Vaibhav ending in 6 days.
Looking for an answer drawing from credible and/or official sources.
I really need a precise answer. I have just received comments that are unsatisfactory too.
1b is a sure bet. Does touching exclude crossing?
– William Elliot
Nov 12 at 9:20
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.
– LutzL
Nov 12 at 15:08
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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:
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 theseIf $f(x)$ 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$f(alpha)=f'(alpha)=f''(alpha)=0$,$f(beta)=f'(beta)=f''(beta)=0$ 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:
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.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.
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.
calculus derivatives
asked Nov 12 at 8:43
Vaibhav
558
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:
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 theseIf $f(x)$ 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$f(alpha)=f'(alpha)=f''(alpha)=0$,$f(beta)=f'(beta)=f''(beta)=0$ 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:
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.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.
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.
calculus derivatives
calculus derivatives
asked Nov 12 at 8:43
Vaibhav
558
asked Nov 12 at 8:43
Vaibhav
558
asked Nov 12 at 8:43
Vaibhav
558
asked Nov 12 at 8:43
Vaibhav
558
asked Nov 12 at 8:43
Vaibhav
558
558
This question has an open bounty worth +50
reputation from Vaibhav ending in 6 days.
Looking for an answer drawing from credible and/or official sources.
I really need a precise answer. I have just received comments that are unsatisfactory too.
This question has an open bounty worth +50
reputation from Vaibhav ending in 6 days.
Looking for an answer drawing from credible and/or official sources.
I really need a precise answer. I have just received comments that are unsatisfactory too.
1b is a sure bet. Does touching exclude crossing?
– William Elliot
Nov 12 at 9:20
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.
– LutzL
Nov 12 at 15:08
add a comment |
1b is a sure bet. Does touching exclude crossing?
– William Elliot
Nov 12 at 9:20
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.
– LutzL
Nov 12 at 15:08
1b is a sure bet. Does touching exclude crossing?
– William Elliot
Nov 12 at 9:20
1b is a sure bet. Does touching exclude crossing?
– William Elliot
Nov 12 at 9:20
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.
– LutzL
Nov 12 at 15:08
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.
– LutzL
Nov 12 at 15:08
add a comment |
1 Answer
1
active
oldest
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up vote
0
down vote
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.
$beta+2alpha$ 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
$$(beta+2alpha)times 3alpha(alpha+2beta)=27alpha^2beta,$$
i.e.
$$6alpha(alpha-beta)^2=0,$$
i.e.
$$alpha=0qquad text{or}qquad alpha=beta$$
In either case, both $alpha$ and $beta$ are rational.
If $s^2-3tnot =0$, then both $alpha$ and $beta$ are rational.
Therefore, $alpha,beta$ 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 $alphaltbeta$".)
answered 17 hours ago
mathlove
91.1k880214
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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.
$beta+2alpha$ 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
$$(beta+2alpha)times 3alpha(alpha+2beta)=27alpha^2beta,$$
i.e.
$$6alpha(alpha-beta)^2=0,$$
i.e.
$$alpha=0qquad text{or}qquad alpha=beta$$
In either case, both $alpha$ and $beta$ are rational.
If $s^2-3tnot =0$, then both $alpha$ and $beta$ are rational.
Therefore, $alpha,beta$ 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 $alphaltbeta$".)
answered 17 hours ago
mathlove
91.1k880214
add a comment |
up vote
0
down vote
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.
$beta+2alpha$ 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
$$(beta+2alpha)times 3alpha(alpha+2beta)=27alpha^2beta,$$
i.e.
$$6alpha(alpha-beta)^2=0,$$
i.e.
$$alpha=0qquad text{or}qquad alpha=beta$$
In either case, both $alpha$ and $beta$ are rational.
If $s^2-3tnot =0$, then both $alpha$ and $beta$ are rational.
Therefore, $alpha,beta$ 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 $alphaltbeta$".)
answered 17 hours ago
mathlove
91.1k880214
add a comment |
up vote
0
down vote
up vote
0
down vote
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.
$beta+2alpha$ 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
$$(beta+2alpha)times 3alpha(alpha+2beta)=27alpha^2beta,$$
i.e.
$$6alpha(alpha-beta)^2=0,$$
i.e.
$$alpha=0qquad text{or}qquad alpha=beta$$
In either case, both $alpha$ and $beta$ are rational.
If $s^2-3tnot =0$, then both $alpha$ and $beta$ are rational.
Therefore, $alpha,beta$ 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 $alphaltbeta$".)
answered 17 hours ago
mathlove
91.1k880214
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.
$beta+2alpha$ 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
$$(beta+2alpha)times 3alpha(alpha+2beta)=27alpha^2beta,$$
i.e.
$$6alpha(alpha-beta)^2=0,$$
i.e.
$$alpha=0qquad text{or}qquad alpha=beta$$
In either case, both $alpha$ and $beta$ are rational.
If $s^2-3tnot =0$, then both $alpha$ and $beta$ are rational.
Therefore, $alpha,beta$ 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 $alphaltbeta$".)
answered 17 hours ago
mathlove
91.1k880214
edited 16 hours ago
answered 17 hours ago
mathlove
91.1k880214
answered 17 hours ago
mathlove
91.1k880214
answered 17 hours ago
mathlove
91.1k880214
91.1k880214
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1b is a sure bet. Does touching exclude crossing?
– William Elliot
Nov 12 at 9:20
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.
– LutzL
Nov 12 at 15:08