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

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$A, B$

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(D) Given $f(x) = \frac{x^2-3x-6}{x^2+2x+4}$.First,find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{(x^2+2x+4)(2x-3) - (x^2-3x-6)(2x+2)}{(x^2+2x+4)^2}$$f'(x) = \frac{(2x^3 - 3x^2 + 4x^2 - 6x + 8x - 12) - (2x^3 + 2x^2 - 6x^2 - 6x - 12x - 12)}{(x^2+2x+4)^2}$$f'(x) = \frac{2x^3 + x^2 + 2x - 12 - (2x^3 - 4x^2 - 18x - 12)}{(x^2+2x+4)^2}$$f'(x) = \frac{5x^2 + 20x}{(x^2+2x+4)^2} = \frac{5x(x+4)}{(x^2+2x+4)^2}$.Critical points are $x = 0$ and $x = -4$.For $x \in (-4, 0)$,$f'(x) For $x \in (1, 2)$,$f'(x) > 0$,so $f$ is increasing. Thus,$(B)$ is $TRUE$.To find the range,evaluate $f(0) = -\frac{6}{4} = -\frac{3}{2}$ and $f(-4) = \frac{16+12-6}{16-8+4} = \frac{22}{12} = \frac{11}{6}$.As $x \rightarrow \pm \infty$,$f(x) \rightarrow 1$.The range is $[-\frac{3}{2}, \frac{11}{6}]$.Since the range is not $R$,$f$ is not onto. Thus,$(C)$ and $(D)$ are $FALSE$.

Let $f: R \rightarrow R$ be defined by $f(x) = \frac{x^2-3x-6}{x^2+2x+4}$. Then which of the following statements is (are) $TRUE$?
$(A)$ $f$ is decreasing in the interval $(-2, -1)$
$(B)$ $f$ is increasing in the interval $(1, 2)$
$(C)$ $f$ is onto
$(D)$ Range of $f$ is $[-\frac{3}{2}, 2]$

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Let $f: R \rightarrow R$ be defined by $f(x) = \frac{x^2-3x-6}{x^2+2x+4}$. Then which of the following statements is (are) $TRUE$?$(A)$ $f$ is decreasing in the interval $(-2, -1)$$(B)$ $f$ is increasing in the interval $(1, 2)$$(C)$ $f$ is onto$(D)$ Range of $f$ is $[-\frac{3}{2}, 2]$