JEE Advanced 2021 — Paper 1 — Question 24

Compute $f(x) = \frac{x^2 - 3x - 6}{x^2 + 2x + 4}$.

Differentiate using quotient rule:

$$f'(x) = \frac{(2x-3)(x^2+2x+4) - (x^2-3x-6)(2x+2)}{(x^2+2x+4)^2}.$$

Simplify numerator to $5x^2 + 20x = 5x(x+4)$.

Thus $f'(x) = \frac{5x(x+4)}{(x^2+2x+4)^2}$. Denominator is always positive, so sign of $f'$ is sign of $x(x+4)$.

Critical points: $x=-4$ and $x=0$. Sign analysis: $f'>0$ on $(-\infty,-4)$, $f'<0$ on $(-4,0)$, $f'>0$ on $(0,\infty)$. Thus $f$ is decreasing on $(-4,0)$. Interval $(-2,-1)$ is subset of $(-4,0)$, so $f$ is decreasing there (option A true). Interval $(1,2)$ is subset of $(0,\infty)$, so $f$ is increasing there (option B true). As $x \to \pm\infty$, $f(x) \to 1$. Minimum at $x=0$: $f(0) = -3/2$.

Maximum at $x=-4$: $f(-4) = 11/6$.

So range of $f$ is $[-3/2, 11/6]$.

Since $2$ is not in the range, option D false. $f$ is not onto because codomain $\mathbb{R}$ but range is a bounded interval (option C false). Thus only options A and B are true.