Find the values of a real number ##a## for an inequality to hold

[brotherbobby]

Homework Statement

Find the set of all possible real values of $a$ such that the inequality $(x-(a-1))(x-(a^2+2))<0$ holds for all $x \in (-1,3)$

Relevant Equations

If $\boldsymbol{(x-a)(x-b)<0\Rightarrow x \in (a,b)}$, where $a<b$. We remember that the "interval" $(a,b)$ stands for the set of all real numbers $x$ such that $a < x < b$.

Problem statement : Let me copy and paste the problem as it appears in the text.
Attempt : From the "Relevant Equations" given above, we can compare to see that $a-1 = -1$ and $a^2+2=3$. These lead (after some algebra) to the three values of $\boxed{a=0, \pm 1}$.

Issue : The book has a different answer. It says $\boxed{a\le -1}$.

Book's solution : I copy and paste the solution given in the book below.

Doubt : Besides not following the solution in the text, let's investigate its answer to the problem, viz. $\boxed{a\le -1}$. So let me take a value of $a$, say $a=-3$. That would make the inequality read $(x+4)(x-11)<0$. This would lead to the following interval for $x \in (-4,11)$, clearly different from the given interval [$x \in (-1,3)$].

Is the book mistaken? A hint or a suggestion would be welcome.

 

[PeroK]

Homework Statement:: Find the set of all possible real values of $a$ such that the inequality $(x-(a-1))(x-(a^2+2))<0$ holds for all $x \in (-1,3)$

Attempt : From the "Relevant Equations" given above, we can compare to see that $a-1 = -1$ and $a^2+2=3$. These lead (after some algebra) to the three values of $\boxed{a=0, \pm 1}$.

If we take $a = 0$ (one of your solutions), then we need:
$$\forall x \in (-1, 3): (x+1)(x-2) < 0 $$This fails for $x = 2.5$, say. So $a=0$ is not a solution.

Likewise for $a = 1$ we need:
$$\forall x \in (-1, 3): (x)(x-3) < 0 $$This fails for $x = 0$, say. So $a=1$ is not a solution.

For $a = -1$ we need:
$$\forall x \in (-1, 3): (x+2)(x-3) < 0 $$The inequality holds on that interval. So $a=-1$ is a solution.

Doubt : Besides not following the solution in the text, let's investigate its answer to the problem, viz. $\boxed{a\le -1}$. So let me take a value of $a$, say $a=-3$. That would make the inequality read $(x+4)(x-11)<0$. This would lead to the following interval for $x \in (-4,11)$, clearly different from the given interval [$x \in (-1,3)$].

The interval $(-4, 11)$ contains the interval $(-1, 3)$, so the inequality holds on this interval. As required. The question does not require that the inequality holds only on the specified interval.

 

[pasmith]

You cannot require both a -1 = -1 and a^2 + 2 = 3: The first requires a = 0, the second requires a = \pm 1. You do not get three possible values for a from these conditions; you get zero possible values of a.

So in fact there is no choice of a for which (x - (a-1))(x - (a^2 + 2)) &lt; 0 holds only for x \in (-1,3) and not otherwise; the book's solution that a must satisfy both a - 1 \leq -1 and a^2 + 2 \geq 3 and the inequality hold for some larger interval (which cannot be made smaller than (-2,3) when a = -1) is correct.

 

[brotherbobby]

You cannot require both a -1 = -1 and a^2 + 2 = 3: The first requires a = 0, the second requires a = \pm 1. You do not get three possible values for a from these conditions; you get zero possible values of a.

So in fact there is no choice of a for which (x - (a-1))(x - (a^2 + 2)) &lt; 0 holds only for x \in (-1,3) and not otherwise; the book's solution that a must satisfy both a - 1 \leq -1 and a^2 + 2 \geq 3 and the inequality hold for some larger interval (which cannot be made smaller than (-2,3) when a = -1) is correct.

It's with the first paragraph of your response that I am struggling with. If a variable like $a$ can have values less than or equal to 1 ($a\le 1$) which is the answer, surely $a$ can take a range of values and yet satisfy the given problem situation. Why then can it not be 0 and $\pm 1$? Surely it could have taken those three values. It doesn't, for as @PeroK has shown in post #2, they would violate the problem conditions. My question is that there is nothing intrinsically wrong with a variable having two or more values, is there?

 

[brotherbobby]

The question does not require that the inequality holds only on the specified interval.

Thank you @PeroK. I suppose this is the main place where I was mistaken. The inequality would hold in an interval beyond the one that is given, viz. $(-1,3)$.

 

[PeroK]

Thank you @PeroK. I suppose this is the main place where I was mistaken. The inequality would hold in an interval beyond the one that is given, viz. $(-1,3)$.

Suppose you had a train ticket and wanted to travel on a Wednesday. So, you ask whether the ticket is valid on Wednesday? Suppose the ticket is valid Monday to Friday. What answer would you expect:

A) No. The ticket is not valid on Wednesday- it's valid Monday to Friday.

B) Yes. The ticket is valid on Wednesday- because Wednesday is a subset of Monday to Friday.

From my point of view, "no" would be a strange, illogical answer

 

[brotherbobby]

Suppose you had a train ticket and wanted to travel on a Wednesday. So, you ask whether the ticket is valid on Wednesday? Suppose the ticket is valid Monday to Friday. What answer would you expect:

A) No. The ticket is not valid on Wednesday- it's valid Monday to Friday.

B) Yes. The ticket is valid on Wednesday- because Wednesday is a subset of Monday to Friday.

From my point of view, "no" would be a strange, illogical answer

Yes, it would be B), in answer to your question.

What made my given problem above in post# 1 a bit tricky is that there was apparently "two" intervals, one in the variable $a$ [$(a-1, a^2+2)$] and another in integers [$(-1,3)$]. A student might equate them, as I did. Doing so however gave wrong answers for the value of $a$. The trick is to realize that the interval given by variable $a$ includes that given with the integers.

 

[pasmith]

It's with the first paragraph of your response that I am struggling with. If a variable like $a$ can have values less than or equal to 1 ($a\le 1$) which is the answer, surely $a$ can take a range of values and yet satisfy the given problem situation. Why then can it not be 0 and $\pm 1$?

a must lie in a set of permissible values. When you impose the constraint a -1= -1 you limit that set to just \{0\}. None of the members of this set satisfy the further constraint a^2 +2= 3. That \pm 1 do satisfy this constraint does not assist you: the first constraint already excluded them from consideration. The set of permissible values is empty.

It is in this sense that you cannot impose both constraints: they do not lead to a solution.