Question

Let $f:R\to R$ be defined by $f\left(x\right)=\frac{x^2-3x-6}{x^2+2x+4}$. Then which of the following statements is(are) TRUE?

• A. $f$ is decreasing in the interval $(-2,-1)$
• B. Range of $f$ is $[-\frac{3}{2},2]$
• C. $f$ is onto
• D. $f$ is increasing in the interval $(1,2)$

Answer 1

Answer: (A), (D)

$f$ is increasing in the interval $(1,2)$

$f\left(x\right)=\frac{x^2-3x-6}{x^2+2x+4}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(x^2+2x+4)(2x-3)-(x^2-3x-6)(2x+2)}{(x^2+2x+4{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{5x(x+4)}{(x^2+2x+4{)}^2}$
Sign scheme of $f^{\prime }\left(x\right):$
Hence, $f$ is decreasing in the interval $(-2,-1)$

Hence, $f$ is increasing in the interval $(1,2)$

$f\left(x\right)$ has local maximum at $x=-4$ and minimum at $x=0$
$f(-4)=\frac{11}{6},f\left(0\right)=-\frac{3}{2}$ and
$\underset{x\to \pm \infty }{lim}f\left(x\right)=1$
Hence, range of $f\left(x\right)$ is $[-\frac{3}{2},\frac{11}{6}]$

Hence, $f$ is not onto
As we can see in the figure: