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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

Let $f:\mathbb{R} \to \mathbb{R}$ be defined by ....

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?

(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 $\left[ { - \frac{3}{2},2} \right]$

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.