What is the necessary and sufficient condition?

[Helly123]

Homework Statement

|x-2| < a is a necessary condition for x^2 -3x -10 < 10 . What is the range value of a?
|x-2| < a is a sufficient condition for x^2 -3x -10 < 10. What is the range value of a?[/B]
The options are
a>= 4
a>=3
0<a<=2
0<a<=3
0<a<=5

The Attempt at a Solution

Range of x for x^2 -3x -10 < 0
2<x<5

And
|x-2| < a
Range of a is 0 < a < 3

Necessary = 0<a<=3
What is the necessary and sufficient conditions?

 

[Ray Vickson]

Homework Statement

|x-2| < a is a necessary condition for x^2 -3x -10 < 10 . What is the range value of a?
|x-2| < a is a sufficient condition for x^2 -3x -10 < 10. What is the range value of a?[/B]
The options are
a>= 4
a>=3
0<a<=2
0<a<=3

The Attempt at a Solution

Range of x for x^2 -3x -10 < 10
2<x<5

And
|x-2| < a
Range of a is 0 < a < 3

Necessary = 0<a<=3
What is the necessary and sufficient conditions?

Your x-range is incorrect: the range of x for $x^2 - 3x - 10 < 10$ is not $2 < x < 5$.

Are you sure you copied the problem correctly?

 

[Helly123]

Your x-range is incorrect: the range of x for $x^2 - 3x - 10 < 10$ is not $2 < x < 5$.

Are you sure you copied the problem correctly?

Sorry. I have corrected it

 

[Ray Vickson]

Sorry. I have corrected it

So, where is the correction? All I see is the original problem, not corrected.

 

[Helly123]

So, where is the correction? All I see is the original problem, not corrected.

Sorry. I forgot to correct the up side of question. The equation is x^2 - 3x -10 < 0

 

[Ray Vickson]

Sorry. I forgot to correct the up side of question. The equation is x^2 - 3x -10 < 0

It is certainly possible to give $|x-2| < a$ as a sufficient condition, and there are many values of $a$ that will work; however, there is no value of $a$ that will have $|x-2| < a$ as a necessary condition.

 

[SammyS]

Sorry. I forgot to correct the up side of question. The equation is x^2 - 3x -10 < 0

You have not solved this correctly.

In the OP, when you make statement about "Range" I assume that you mean "solution".

The Attempt at a Solution

Range of x for x² -3x -10 < 0
2<x<5

The solution to $\ x^2 -3x -10 < 0\$ is not $\ 2<x<5\$.

For example x = 1 is also a solution to $\ x^2 -3x -10 < 0\$.

By the way, one of the given choices does give a necessary condition.

 

[nuuskur]

If $P\implies Q$, then we say $P$ is a sufficient condition for $Q$ and $Q$ is a necessary condition for $P$
The first problem is to find $a$ such that
<br /> x^2 -3x-10&lt;0\implies \lvert x-2\rvert &lt;a<br />
One may note that $x^2 - 3x -10 = (x-5)(x+2)<0$, which is true for every $x$ satisfying $-2<x<5$.
Now examine a condition $\lvert x-2\rvert <a$ which is the same as
<br /> -a &lt; x-2 &lt; a \iff 2-a &lt;x&lt;2+a<br />
Now, for which largest possible $a$ do you not exceed the bounds? (NOT a necessary condition, however!)

Conversely, consider the reversed implication
<br /> \lvert x-2\rvert &lt;a \implies x^2 -3x-10&lt;0<br />
Only now, your premise is the given condition $2-a<x<2+a$. You must guarantee $(x+2)(x-5)<0$. Which values of $a$ guarantee that?There is a problem with terminology, though. You can only find sufficient conditions of the form $2-a<x<2+a$. There are no necessary conditions.
To quickly illustrate what I mean. If $Q$ is a necessary condition for $P$, then $\neg Q$ is a sufficient condition for $\neg P$. In other words, if we did, indeed have a necessary condition $2-a<x<2+a$ for some $a$, then $x\leq 2-a\lor x\geq 2+a$ must be a sufficient condition for $(x+2)(x-5)\geq 0$, but that is not true of any $a$, therefore a necessary condition of the given form cannot exist.

 

[SammyS]

Not only is there a sufficient condition given, but there is also a necessary condition given.

It's rather futile to discuss either without the OP giving the correct solution to the given quadratic inequality.

$\ x^2 -3x -10 < 0\$​

 

[Ray Vickson]

Not only is there a sufficient condition given, but there is also a necessary condition given.

It's rather futile to discuss either without the OP giving the correct solution to the given quadratic inequality.

$\ x^2 -3x -10 < 0\$​

NO. If you look at the correct interval $\alpha < x < \beta$ for $x^2 - 3x - 10 < 0$ it just cannot be written as $|x - 2 | < a$ for some $a$. Basically, $2$ is not the mid-point of the interval $(\alpha, \beta)$.

 

[SammyS]

NO. If you look at the correct interval $\alpha < x < \beta$ for $x^2 - 3x - 10 < 0$ it just cannot be written as $|x - 2 | < a$ for some $a$. Basically, $2$ is not the mid-point of the interval $(\alpha, \beta)$.

It doesn't need to be centered on any particular point..

I think it best to wait for OP to correctly solve the given quadratic inequality before further discussing this point.

 

[Ray Vickson]

It doesn't need to be centered on any particular point..

I think it best to wait for OP to correctly solve the given quadratic inequality before further discussing this point.

I agree. Certainly a sufficient condition can be written as $|x-2| < a$---and that for lots of different values of $a$. And, of course, if we let $a = +\infty$ we can write a necessary condition $x \in \mathbb{R}$ as $|x-2| < a.$ I don't think that is what the questioner meant, although I could be wrong.

 

[Helly123]

Sorry again.. the value of x for x^2 -3x -10<0 is -2<x<5

When -2<x<2
a value is 0<a<4

When 2<=x<5
0<=a<3

But i cannot decide what is the sufficient and necessary conditions

My answer now, sufficient 0<=a<3

 

[SammyS]

Sorry again.. the value for x is −2 < x < 5

The above is the correct solution to the given quadratic inequality.

The following makes little to no sense.

For −2<x<2
0<a<4

If a = 2, then |x − 2| < a is equivalent to 0 < x < 4.

For 2<=x<5
0<=a<3

But i cannot decide what is the sufficient and necessary conditions

 

[Helly123]

The above is the correct solution to the given quadratic inequality.

The following makes little to no sense.

If a = 2, then |x − 2| < a is equivalent to 0 < x < 4.

I have corrected it

 

[nuuskur]

And, of course, if we let $a = +\infty$ we can write a necessary condition $x \in \mathbb{R}$ as $|x-2| < a.$

That's basically saying if $f(x)< 0$, then $x\in\mathbb R$, a vacuously true statement. That said, I doubt they operate in $\bar{\mathbb R}$

 

[Ray Vickson]

That's basically saying if $f(x)< 0$, then $x\in\mathbb R$, a vacuously true statement. That said, I doubt they operate in $\bar{\mathbb R}$

Well, of course it is vacuous---that is exactly the point! Anyway, $\bar{\mathbb{R}}$ is not involved, just $\mathbb{R}$.

 

[nuuskur]

$a=+\infty$ is meaningless in that instance, if one insists on being overly pedantic

 

[Ray Vickson]

$a=+\infty$ is meaningless in that instance, if one insists on being overly pedantic

No: it is just notation. Saying that $|x-2| < \infty$ is just saying that $|x-2|$ can be any real number. Personally, I would never voluntarily say it that way, but it would certainly be a way to put the statement in the format requested by the problem.

 

[SammyS]

Sorry again.. the value of x for x^2 -3x -10<0 is -2<x<5

When -2<x<2
a value is 0<a<4

When 2<=x<5
0<=a<3

But i cannot decide what is the sufficient and necessary conditions

My answer now, sufficient 0<=a<3

Yes.
For | x − 2 | < a to be a sufficient condition so that x² − 3x − 10 < 0, the range for a is 0 < a ≤ 3.
You have not been very clear in showing that this is the case.

This does not give all of the values of x which satisfy the given quadratic inequality. Rather any x fulfilling | x − 2 | < a will satisfy the quadratic inequality.for the specified range of a values. So being in that interval is a sufficient condition for x to satisfy the quadratic inequality..

 

[Helly123]

Yes.
For | x − 2 | < a to be a sufficient condition so that x² − 3x − 10 < 0, the range for a is 0 < a ≤ 3.
You have not been very clear in showing that this is the case.

This does not give all of the values of x which satisfy the given quadratic inequality. Rather any x fulfilling | x − 2 | < a will satisfy the quadratic inequality.for the specified range of a values. So being in that interval is a sufficient condition for x to satisfy the quadratic inequality..

I don't get it why is 0 < a ≤ 3? Why including 3?

 

[SammyS]

I don't get it why is 0 < a ≤ 3? Why including 3?

You beat me to that question.

For one thing, if you look at the problem statement you gave, that choice includes the 3.

The options are
a>= 4
a>=3
0<a<=2
0<a<=3
0<a<=5

As a teacher I would be inclined to ask you , "Why does that include the 3?"

Plug a = 3 into the inequality |x−2| < a and solve for x.
What do you get?

Then we can tackle the "necessary" condition.

 

[Helly123]

You beat me to that question.

For one thing, if you look at the problem statement you gave, that choice includes the 3.

As a teacher I would be inclined to ask you , "Why does that include the 3?"

Plug a = 3 into the inequality |x−2| < a and solve for x.
What do you get?

Then we can tackle the "necessary" condition.

If i plug a = 3. I get x = 5. But, the range for x^2-3x-10<0 is x less than 5. Not including 5

 

[SammyS]

If i plug a = 3. I get x = 5.

Really?

You shouldn't.

| x − 2 | < 3

gives

−3 < x − 2 < 3 .

Right?

 

[Helly123]

Really?

You shouldn't.

| x − 2 | < 3

gives

−3 < x − 2 < 3 .

Right?

Hm yes. Then, i am confused how to get 0 < a<= 3 itself..
How i get it is when x=-2, a must be 4
When x = 5, a must be 3
When x = 2, a must be 0
The range for a must be between 0 and 4

 

[SammyS]

Hm yes. Then, i am confused how to get 0 < a<= 3 itself..
How i get it is when x=-2, a must be 4
When x = 5, a must be 3
When x = 2, a must be 0
The range for a must be between 0 and 4

You seem to be working this backwards.

If a = 4, then | x − 2 | < a gives −4 < x − 2 < 4, which includes values of x which do not satisfy the quadratic inequality.

Also, for a ≤ 0: There is no value of x for which | x − 2 | ≤ 0 . So we must have a > 0 .

Hm yes. Then, I am confused how to get 0 < a<= 3 itself.

@nuuskur outlined some of this in post #8.

If a ≤ 3 and | x − 2 | < a , then | x − 2 | < 3 .
...

You need to do some thoughtful work on these issues. We can't do it all for you.

 

[Helly123]

You seem to be working this backwards.

If a = 4, then | x − 2 | < a gives −4 < x − 2 < 4, which includes values of x which do not satisfy the quadratic inequality.

Also, for a ≤ 0: There is no value of x for which | x − 2 | ≤ 0 . So we must have a > 0 .@nuuskur outlined some of this in post #8.

If a ≤ 3 and | x − 2 | < a , then | x − 2 | < 3 .
...

You need to do some thoughtful work on these issues. We can't do it all for you.

Ok i will try. Thank you

 

[Helly123]

Is the necessary condition a>=4 or 0<a<=4 ?
Both is possible, i think

 

[nuuskur]

I must apologise for #8, it contains some false statements at the end.
A necessary condition is something that simply has to be true given some assumptions. In this case, we assume $(x+2)(x-5)<0$
It does not imply that $-1<x<5$, for instance, but it does imply $-2<x<6$.

A quick way to check this, as I mentioned, is making use of the logical equivalence of two statements
<br /> P\implies Q\iff \neg Q\implies \neg P<br />
In this example, as $P$ we consider $(x+2)(x-5)<0$ and as $Q$ we consider $\lvert x-2 \rvert <a$.
If we want $Q$ to be a necessary condition, we can also pick $a$ such that $\neg Q = \lvert x-2\rvert \geq a$ is a sufficient condition for $\neg P = (x+2)(x-5)\geq 0$.

It suffices to pick $x\leq -2\lor x\geq 6$, then it's guaranteed that $(x+2)(x-5)\geq 0$. Reverse the order now, if $(x+2)(x-5)<0$, then it is necessarely true that $-2<x<6$ (i.e you are correct about $a\geq 4$)

However, $0<a<4$ will not yield a necessary condition. If you pick, say, $a=1$, then the implication
<br /> (x+2)(x-5)&lt;0 \implies -1&lt;x&lt;3<br />
is false. We can pick $x = -1.5$, it still stands that $(x+2)(x-5)<0$, but at the same time $x\leq -1$.

I think I got confused by an earlier post, but I am certain now that this is correct.

 

[Helly123]

I must apologise for #8, it contains some false statements at the end.
A necessary condition is something that simply has to be true given some assumptions. In this case, we assume $(x+2)(x-5)<0$
It does not imply that $-1<x<5$, for instance, but it does imply $-2<x<6$.

A quick way to check this, as I mentioned, is making use of the logical equivalence of two statements
<br /> P\implies Q\iff \neg Q\implies \neg P<br />
In this example, as $P$ we consider $(x+2)(x-5)<0$ and as $Q$ we consider $\lvert x-2 \rvert <a$.
If we want $Q$ to be a necessary condition, we can also pick $a$ such that $\neg Q = \lvert x-2\rvert \geq a$ is a sufficient condition for $\neg P = (x+2)(x-5)\geq 0$.

It suffices to pick $x\leq -2\lor x\geq 6$, then it's guaranteed that $(x+2)(x-5)\geq 0$. Reverse the order now, if $(x+2)(x-5)<0$, then it is necessarely true that $-2<x<6$ (i.e you are correct about $a\geq 4$)

However, $0<a<4$ will not yield a necessary condition. If you pick, say, $a=1$, then the implication
<br /> (x+2)(x-5)&lt;0 \implies -1&lt;x&lt;3<br />
is false. We can pick $x = -1.5$, it still stands that $(x+2)(x-5)<0$, but at the same time $x\leq -1$.

I think I got confused by an earlier post, but I am certain now that this is correct.

I must apologise for #8, it contains some false statements at the end.
A necessary condition is something that simply has to be true given some assumptions. In this case, we assume $(x+2)(x-5)<0$
It does not imply that $-1<x<5$, for instance, but it does imply $-2<x<6$.

A quick way to check this, as I mentioned, is making use of the logical equivalence of two statements
<br /> P\implies Q\iff \neg Q\implies \neg P<br />
In this example, as $P$ we consider $(x+2)(x-5)<0$ and as $Q$ we consider $\lvert x-2 \rvert <a$.
If we want $Q$ to be a necessary condition, we can also pick $a$ such that $\neg Q = \lvert x-2\rvert \geq a$ is a sufficient condition for $\neg P = (x+2)(x-5)\geq 0$.

It suffices to pick $x\leq -2\lor x\geq 6$, then it's guaranteed that $(x+2)(x-5)\geq 0$. Reverse the order now, if $(x+2)(x-5)<0$, then it is necessarely true that $-2<x<6$ (i.e you are correct about $a\geq 4$)

However, $0<a<4$ will not yield a necessary condition. If you pick, say, $a=1$, then the implication
<br /> (x+2)(x-5)&lt;0 \implies -1&lt;x&lt;3<br />
is false. We can pick $x = -1.5$, it still stands that $(x+2)(x-5)<0$, but at the same time $x\leq -1$.

I think I got confused by an earlier post, but I am certain now that this is correct.

Where the -2<x<6 comes from?