But they are both variables!

It is given that: $\space f(x) = |x|^{x^2-x-2} = |x|^{(x+1)(x-2)} < 1$

The trivial solutions $f(x)=1$ are when:

$\begin{cases} |x| = 1 & \Rightarrow \begin{cases} x = -1 \\ x = 1 \end{cases} \\ x^2-x-2 = 0 & \Rightarrow \begin{cases} x = -1 \\ x = 2 \end{cases} \end{cases}$

Let us consider when $x = -1, 1,2$ . First consider $x = -1^-$ slightly less than $-1$ and $x=-1^+$ slightly more than $-1$ .

$\text{Case } x=-1 \Rightarrow \begin{cases} f(-1^-) = (>1)^{(>0)} & > 1 \\ f(-1) = (1)^{(0)} & = 1 \\ f(-1^+) = (<1)^{(<0)} & > 1\end{cases}$ is a local minimum.

$\text{Case } x=1 \Rightarrow \begin{cases} f(1^-) = (<1)^{(<0)} & > 1 \\ f(1) = (1)^{(-2)} & = 1 \\ f(1^+) = (>1)^{(<0)} & < 1\end{cases}$

$\text{Case } x=2 \Rightarrow \begin{cases} f(2^-) = (>1)^{(<0)} & < 1 \\ f(2) = (2)^{(0)} & = 1 \\ f(1^+) = (>1)^{(>0)} & > 1\end{cases}$

Therefore, $|x|^{x^2-x-2} <1$ , when $x \in (1,2)\quad \Rightarrow A+B = 1+2 = \boxed{3}$

The solution can be easily shown with a graph as below:

For some value $a,b \in \mathbb{R}, a^b<1$ if:

(1) $|a| > 1$ and $b<0$

OR

(2) $|a|< 1$ and $b>0$

In this problem, $a = |x|$ and $b = x^2-x-2$ .

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For case (1) :

$x^2-x-2<0$ in the interval $(-1, 2)$ .

$||x|| = |x| > 1$ in the interval $(-\infty, -1)\cup(1, +\infty)$ .

$(-1, 2) \cap ((-\infty, -1)\cup(1, +\infty))=(1,2)$ .

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For case (2) :

$x^2-x-2>0$ in the interval $(-\infty,-1)\cup(2,+\infty)$ .

$||x|| = |x| < 1$ in the interval $(-1, 1)$

$((-\infty,-1)\cup(2,+\infty))\cap(-1, 1)=\emptyset$

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$(1,2)\cup\emptyset=(1,2)$

$1+2=\boxed{3}$

A=1;B=2 => A+B=3

As, $\large |x|^{(x^2-x-2)} < 1$ $\large |x|^{(x^2-x-2)} < |x|^{0}$ so, $x^{2}-x-2 < 0$ => $x^{2}-2x+x-2 < 0$ => $x(x-2)+1(x-2) < 0$ => $(x-2)(x+1)< 0$ Therefore, $x < |2|$ and $x< | -1|$ ; so values so A and B will be 2, 1 . And A+B = 3 . Is my method is wrong, need some suggestion!

Case 1: $x>1:$ In this case: $x^2-x-2<0$ to yield a power of $abs(x)$ that is less than 1. By solving $x^2-x-2=0$ , we get $-1<x<2$ We originally defined x to be greater than 1 so we get the interval: $1<x<2$

Case 2: $x=1: abs(1)^(1^2-1-2)=1$ which is not less than 1 so x does not equal 1.

Case 3: $abs(x)<1$ To yield a power of $abs(x)$ less than one we need a positive power so: $x^2-x-2>0$ . By solving $x^2-x-2=0$ , we get $x<-1$ and $x>2$ We originally defined $abs(x)>1$ which it is not in this interval so we disregard this case and we stick with the case 1 interval of:

$1<x<2$

We are given that |x|^(x^2 - x - 2) < 1

This means that on the interval (A,B) which satisifies the above equation, the function should obtain a local maximum at either of the excluded endpoints.

Hence, we must satisfy this maximum at |x|^(x^2 - x - 2) = 1

This occurs when either x = -1 (with positive power) or 1 (with any power) or the power (x^2 -x - 2) = 0 with any base.

It is simplest to begin with solving the quadratic which yields:

(x -2)(x + 1) = 0 ==> x ={-1,0}

This takes care of the "-1" case.

Now, plugging in 1 to the quadratic we find that we end up with:

(1-2)(1+1)= -2. Which is valid. So, we can consider 4 cases through breaking up the interval (-1,2).

Case 1: -1 < x <= 0

Right away, x = 0 is forbidden since the quadratic becomes -2 and 0^(-2) is undefined.

Now, we know that any fractional value greater than -1 will require a positive power for the function to remain less than 1.

But if x<0 then x^2 -x >0

Which means that x^2 -x >2

Which will never occur for x > -1.

Hence, there are no values in the interval (-1,0) which will satisfy the inequality.

Case 2: 0 < x < 1

Arguing similarly as in the previous case, to ensure that the inequality never becomes greater than 1, we must have rational x with a positive power.

Looking at the quadratic this means that once again, x^2 -x > 2

But, with x less than one this will never be satisfied.

Hence, x cannot lie within (0,1)

Case 3: 1<= x <2

We already showed that x = 1 yields valid results.

Focusing on values falling in the open interval and focusing on rational values once again, we see that since said values are greater than 1, we must have negative powers (or rational and irrational powers less than 1) for the inequality to remain valid.

But, analysing the quadratic we see that x^2 -x can be analyzed for integers p and q such that x = p/q we have

(p/q)((p/q)-1) but, (p/q) < 2 ==> (p/q) -1 < 1

==> (p/q)((p/q)-1) < (p/q) < 2

And this automatically takes care of irrational values.

Hence, all values in this interval yield valid results.

This means the interval not containing the max of the inequality is (1,2) = (A,B). So, A+B = 3.

QED.

x can't be bigger than 2, or less than -2, by calculating 2^(2^2-2-2) = 1 we can see that the number must be less than 2. Numbers less than 1 (0.9, 0.8) give solutions that are bigger than 1. So we are left with the numbers bigger than 1 and less than 2. f(x) = (1.2)

But X can also be 0