Given −2 < a < 1 and 1 < b < 2, which are correct? A: −4 < a − b < 0. B: −2 < ab < 2. C: −2 < a/b < 1. D: ab − 2 < a − 2b.

Step-by-step solution

1. 1

   Fix the correct rule for subtraction. Inequalities may be added but not subtracted directly; instead negate the second one first. From $1<b<2$ we get $-2<-b<-1$, and adding to $-2<a<1$ gives

   $-4<a-b<0$

   So A is correct, and the bounds are sharp (approached as $a\to-2,b\to 2$ and as $a\to 1,b\to 1$).

2. 2

   B is wrong — check the extreme corners of the product. Because $a$ can be negative, the extremes of $ab$ occur at the corners of the rectangle: $ab$ ranges over $(-2)(2)=-4$ up to $(1)(2)=2$, so

   $-4<ab<2$

   Values below $-2$ really occur: $a=-1.8$, $b=1.9$ gives $ab=-3.42$. Incorrect.

3. 3

   C is correct. Since $b>0$, dividing preserves directions and the quotient is largest when $a$ is largest and $b$ smallest, smallest when $a$ is most negative and $b$ smallest:

   $\frac{a}{b}>\frac{-2}{1}=-2,\qquad \frac{a}{b}<\frac{1}{1}=1$

   so $-2<\dfrac ab<1$. Correct.

4. 4

   D is wrong — rearrange into a product. The claim $ab-2<a-2b$ is equivalent to

   $ab-a+2b-2<0\quad\Longleftrightarrow\quad a(b-1)+2(b-1)<0\quad\Longleftrightarrow\quad (a+2)(b-1)<0$

   But $a>-2$ gives $a+2>0$, and $b>1$ gives $b-1>0$, so the product is positive for every allowed pair. The stated inequality is therefore always false. Incorrect.

5. 5

   Conclude and verify with a sample point.

   $\boxed{\text{A and C}}$

   Take $a=0$, $b=1.5$: $a-b=-1.5\in(-4,0)$ ✓; $ab=0\in(-4,2)$ (and also in $(-2,2)$, which is why B needs a corner test to disprove); $a/b=0\in(-2,1)$ ✓; and $ab-2=-2$ versus $a-2b=-3$, so $-2<-3$ is false ✓, confirming D fails.