JEE Advanced 2021 · Paper 1, Mathematics question 12

Derivative.
$$f'(x)=\frac{(2x-3)(x^2+2x+4)-(x^2-3x-6)(2x+2)}{(x^2+2x+4)^2}=\frac{5x(x+4)}{(x^2+2x+4)^2}.$$
So $f$ increases on $(-\infty,-4)$ and $(0,\infty)$, and decreases on $(-4,0)$.

Extremes. $f(0)=-\frac32$ is a local minimum, $f(-4)=\frac{11}6$ is a local maximum, and $f\to1$ at $\pm\infty$. So the range is $\left[-\frac32,\frac{11}6\right]$.

Check each option.

• (A) $f$ decreases on $(-2,-1)$. True.
• (B) $f$ increases on $(1,2)$. True.
• (C) $f$ is not onto. False.
• (D) The range is not $\left[-\frac32,2\right]$. False.

(A), (B)