JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Application of Derivatives

$f^{\prime }(x)=\frac{(x^2+2x+4)(2x-3)-(x^2-3x-6)(2x+2)}{(x^2+2x+4{)}^2}=\frac{5x(x+4)}{(x^2+2x+4{)}^2}$. The sign of $f^{\prime }(x)$ is determined by $5x(x+4)$: negative for $x\in (-4,0)$ and positive for $x\in (-\infty ,-4)\cup (0,\infty )$. Therefore, $f$ is decreasing on $(-4,0)$, which includes $(-2,-1)$, and increasing on $(0,\infty )$, which includes $(1,2)$. So (A) and (B) are true. Computing key values: $f(-4)=\frac{11}{6}$, $f(0)=-\frac{3}{2}$, and ${\lim }_{x\to \pm \infty }f(x)=1$. The range is $\left[-\frac{3}{2},\frac{11}{6}\right]$, so $f$ is not onto and (C) is false. The range is not $\left[-\frac{3}{2},2\right]$, so (D) is false.