Let $f:\mathbb{R} \to \mathbb{R}$ be defined by $$f(x) = \frac{{{x^2} - 3x - 6}}{{{x^2} + 2x + 4}}$$ Then which of the following statements is (are) TRUE?
$f(0) = \frac {-3}{2}$, 0 being the point of local minima.
(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 $\left[ { - \frac{3}{2},2} \right]$
(B) f is increasing in the interval (1, 2)
(C) f is onto
(D) Range of f is $\left[ { - \frac{3}{2},2} \right]$
Solution
We have, $f'(x) = \frac{{({x^2} + 2x + 4).(2x - 3) - ({x^2} - 3x - 6).(2x + 2)}}{{{{({x^2} + 2x + 4)}^2}}}$
$ \Rightarrow f'(x) = \frac{{5x(x + 4)}}{{{{({x^2} + 2x + 4)}^2}}}$
Sign-scheme for f'(x) will be decided by $x.(x+4)$. -4 & 0 are the points of local extremum. The figure below shows increasing/decreasing etc. details.
$f(0) = \frac {-3}{2}$, 0 being the point of local minima.
$f(-4)=\frac {11}{6}$, -4 being the point of local maxima.
Range of function is $\left[ { - \frac{3}{2},\frac{{11}}{6}} \right]$
Options (A) & (B) fall within the values shown in the above figure making them correct. Other options are incorrect.