<p>1, 2, 3</p>
<li>If m, n, and k are to be assigned different values from the list above, how many different values will be possible for the expression (m + n)^k?</li>
</ol>
<p>(A) Three
(B) Four
(C) Five
(D) Eight
(E) Nine</p>
<li>Each of the following inequalities is true for some values of x EXCEPT:
(A) x<x^2<x^3
(B) x<x^3<x^2
(C) x^2<x^3<x
etc…
The answer is C but why?</li>
</ol>
<ol>
<li>plug in random values for m n and k
so try m = 1 n =2 k =3 ( remember they MUST be different)
Keep doing this till you get an answer... i don't have time right now but this is how you do it</li>
<li>
ok to disprove B try putting in x = -.5
So B is gone
To disprove A put in x = .5.
There you go.</li>
</ol>
<p>10.) If you notice by counting principle, there are only 6 different possibility. So, try them all:
(1 + 2 )^3 = 27
(1 + 3 )^2 = 16
(2 + 1 )^3 = 27
(2 + 3 )^1 = 5
(3 + 2 )^1 = 5
(3 + 1 )^2 = 16</p>
<p>There are THREE different values (A). </p>
<p>Another way of thinking about it is remember that commutative property of addition still applies and the exponent number depends on the two numbers in the addition, so that you should halve the number of possibilities to get the answer.</p>
<p>15.) The best way to tackle this problem is make examples where each of these are true.</p>
<p>(A) x < x^2 < x^3
Well, an obvious case is a positive number > 1, like 2. 2 < 4 < 8. </p>
<p>(B) x < x^3 < x^2
This is a little tricky. The trick is to split it into two parts, and then combine. </p>
<p>First, consider when x^3 < x^2 --> 0 < x^2(1 - x) The domain of this inequality is when x < 1 and x != 0. </p>
<p>Let's consider x < x^3 --> 0 < x(x + 1)(x - 1). The domain for this inequality is when -1 < x < 0 and 1 < x.</p>
<p>If you intersect both domains, you can see that x < x^3 < x^2 is true when -1 < x < 0. Let's take the example -0.5: -0.5 < -0.125 < 0.25. It works.</p>
<p>(C)x^2 < x^3 < x
Let's use the same method as B.
First observe x^2 < x^3 --> 0 < x^2(x - 1).The inequality works only when x > 1.
Now let's observe x^3 < x --> 0 < x(1 - x)(1 + x). The inequality works only when x < -1 and 0 < x < 1.</p>
<p>Let's try to combine the two inequalities to form x^2 < x^3 < x. The domain of this is the intersection of the domains of its two separate pieces. Try to find the intersection of (x > 1) and (x < -1 and 0 < x < 1). It turns out to be null-set. There is NO VALUE of X that will work for this inequality. This is the right answer.</p>
<p>I assume there is an example available for part D and E.</p>
<p>10.</p>
<p>(m+n)^k.</p>
<p>Let k be dependent on the choices of m and n.</p>
<p>m and n are considered combinations not permutations since the communicative property allows m+n=n+m</p>
<p>3 x 2 (x 1)=6</p>
<p>15.</p>
<p>Take case by case with the domains (-1,0), (0,1), (-1,oo) and (1, oo)</p>
<p>A) Obviously works for (1,oo)
B) Works for (-1,0)
C) Never works.
Since, x^2>0, this choice implies that x^3>0 and x>0.</p>
<p>For (0,1) x^2>x^3.</p>
<p>(1, oo) x^3>x</p>
<p>No other case will x^2>0, x^3>0 and x>3.</p>
<p>10.) in case you didn't read my explanation, because of the commutative property of addition, it's exactly half that (remember they want UNIQUE values)</p>
<p>11.) for b, (-1,0) does NOT work, because x is strictly less than x^3, not equal to.</p>