Analysis of Function Monotonicity Range and Surjectivity

To analyze the properties of the given function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{x^2 - 3x - 6}{x^2 + 2x + 4}$

Step 1: Monotonicity Analysis

First, we find the first derivative $f'(x)$ using the quotient rule: $f'(x) = \frac{(2x - 3)(x^2 + 2x + 4) - (x^2 - 3x - 6)(2x + 2)}{(x^2 + 2x + 4)^2}$

Expanding the terms in the numerator: $(2x - 3)(x^2 + 2x + 4) = 2x^3 + 4x^2 + 8x - 3x^2 - 6x - 12 = 2x^3 + x^2 + 2x - 12$ $(x^2 - 3x - 6)(2x + 2) = 2x^3 + 2x^2 - 6x^2 - 6x - 12x - 12 = 2x^3 - 4x^2 - 18x - 12$

Subtracting the two expressions gives the numerator: $\text{Numerator} = (2x^3 + x^2 + 2x - 12) - (2x^3 - 4x^2 - 18x - 12) = 5x^2 + 20x = 5x(x + 4)$

Since the denominator $(x^2 + 2x + 4)^2 > 0$ for all real $x$ (as the discriminant of $x^2 + 2x + 4$ is $2^2 - 4(1)(4) = -12 < 0$), the sign of $f'(x)$ is determined solely by $5x(x + 4)$:

• $f'(x) > 0$ for $x \in (-\infty, -4) \cup (0, \infty) \implies f(x)$ is strictly increasing on $(-\infty, -4)$ and $(0, \infty)$.
• $f'(x) < 0$ for $x \in (-4, 0) \implies f(x)$ is strictly decreasing on $(-4, 0)$.

Now, evaluating statement (A) and (B):

• Option A: The interval $(-2, -1) \subset (-4, 0)$, where $f'(x) < 0$. Therefore, $f$ is decreasing in $(-2, -1)$. (TRUE)
• Option B: The interval $(1, 2) \subset (0, \infty)$, where $f'(x) > 0$. Therefore, $f$ is increasing in $(1, 2)$. (TRUE)

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Step 2: Range and Surjectivity (Onto/Into)

To find the extrema and range of $f(x)$:

• The critical points of $f(x)$ occur at $x = -4$ and $x = 0$.
• Local maximum value at $x = -4$: $f(-4) = \frac{(-4)^2 - 3(-4) - 6}{(-4)^2 + 2(-4) + 4} = \frac{16 + 12 - 6}{16 - 8 + 4} = \frac{22}{12} = \frac{11}{6}$
• Local minimum value at $x = 0$: $f(0) = \frac{0^2 - 3(0) - 6}{0^2 + 2(0) + 4} = \frac{-6}{4} = -\frac{3}{2}$
• As $x \to \pm\infty$: $\lim_{x \to \pm\infty} f(x) = \lim_{x \to \pm\infty} \frac{x^2 - 3x - 6}{x^2 + 2x + 4} = 1$

Since $f(x)$ is continuous everywhere on $\mathbb{R}$, its range is given by: $\text{Range of } f = \left[-\frac{3}{2}, \frac{11}{6}\right]$

Now, evaluating statement (C) and (D):

• Option C: Since the range $\left[-\frac{3}{2}, \frac{11}{6}\right] \neq \mathbb{R}$ (the codomain), $f$ is not onto (it is into). (FALSE)
• Option D: The range of $f$ is $\left[-\frac{3}{2}, \frac{11}{6}\right]$, which is not $\left[-\frac{3}{2}, 2\right]$. (FALSE)

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Conclusion

The correct options are A and B.