1. Given function

$f(x)=\frac{x^2-3x-6}{x^2+2x+4}$

We need to check monotonicity, onto-ness, and range.

2. Domain

Denominator: $x^2+2x+4=(x+1)^2+3>0 \quad \forall x\in\mathbb R$ So domain is all real numbers.

3. Find derivative to test increasing/decreasing

Let $N=x^2-3x-6,\qquad D=x^2+2x+4$ Then $f'(x)=\frac{N'D-ND'}{D^2}$

Now, $N'=2x-3,\qquad D'=2x+2$

So $f'(x)=\frac{(2x-3)(x^2+2x+4)-(x^2-3x-6)(2x+2)}{(x^2+2x+4)^2}$

Simplify numerator:

$(2x-3)(x^2+2x+4)=2x^3+x^2+2x-12$

$(x^2-3x-6)(2x+2)=2x^3-4x^2-18x-12$

Therefore, $N'D-ND'=(2x^3+x^2+2x-12)-(2x^3-4x^2-18x-12)$ $=5x^2+20x=5x(x+4)$

Hence $f'(x)=\frac{5x(x+4)}{(x^2+2x+4)^2}$

Since denominator is always positive, sign of $f'(x)$ depends on $x(x+4)$

Critical points: $x=-4,0$.

Sign chart:

• For $x<-4$, $f'(x)>0$  increasing
• For $-4<x<0$, $f'(x)<0$  decreasing
• For $x>0$, $f'(x)>0$  increasing

4. Check option A

Interval $(-2,-1)$ lies inside $(-4,0)$, where $f'(x)<0$. So $f$ is decreasing on $(-2,-1)$.

✅ A is true

5. Check option B

Interval $(1,2)$ lies inside $(0,\infty)$, where $f'(x)>0$. So $f$ is increasing on $(1,2)$.

✅ B is true

6. Find range of $f$

Since function is continuous on $\mathbb R$ and has local extrema at $x=-4$ and $x=0$, compute values there.

$f(-4)=\frac{16+12-6}{16-8+4}=\frac{22}{12}=\frac{11}{6}$

$f(0)=\frac{-6}{4}=-\frac32$

Also, $\lim_{x\to\pm\infty}f(x)=1$

Because:

• increasing on $(-\infty,-4)$,
• decreasing on $(-4,0)$,
• increasing on $(0,\infty)$,

we get:

• maximum value $=\dfrac{11}{6}$ at $x=-4$,
• minimum value $=-\dfrac32$ at $x=0$.

Thus range is $\left[-\frac32,\frac{11}{6}\right]$

So option D claims range is $\left[-\frac32,2\right]$, which is incorrect.

❌ D is false

7. Check option C (onto?)

For $f:\mathbb R\to\mathbb R$ to be onto, its range must be all real numbers. But actual range is $\left[-\frac32,\frac{11}{6}\right]$ which is not all of $\mathbb R$.

❌ C is false

8. Final conclusion

The true statements are: $\boxed{A,\ B}$

9. Comparison with stored correct answer

Stored correct answer: A, B

Our derived answer matches exactly.