Let f: R → R be defined by f(x) = x ^ 2 - 3x - 6 x ^ 2 + 2x + 4...

f\left(x\right) = \dfrac{{x}^{2} - 3x - 6}{{x}^{2} + 2x + 4}$</p><p>${f}^{'}\left(x\right) = \dfrac{\left({x}^{2} + 2x + 4\right)\left(2x - 3\right) - \left({x}^{2} - 3x - 6\right)\left(2x + 2\right)}{{\left({x}^{2} + 2x + 4\right)}^{2}}$</p><p>$= \dfrac{5{x}^{2} + 20x}{{\left({x}^{2} + 2x + 4\right)}^{2}} = \dfrac{5x ( x + 4 )}{{\left({x}^{2} + 2x + 4\right)}^{2}}$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/0de6bb8c-70ea-48e9-a43f-3b29dca08fa0-image.png" style="width: 140px; height: 58px;" /></p><p>Hence, it is increasing in$( - \infty , - 4 ] \cup [ 0, \infty )$</p><p> </p><p>Now as$x \to - \infty$then$f\left(x\right) = 1,$also,$x \to \infty$then$f\left(x\right) = 1$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/5db986cf-14a3-460c-b7d6-d0a97afbd93f-image.png" style="width: 318px; height: 250px;" /></p><p>$f\left(- 4\right) = \dfrac{16 + 12 - 6}{16 - 8 + 4} = \dfrac{22}{12} = \dfrac{11}{6}$</p><p>$f\left(0\right) = \dfrac{- 3}{2}$</p><p>Hence, range is \left[\dfrac{- 3}{2}, \dfrac{11}{6}\right]$, and the function is into.