JEE Main 202331 Jan, Evening shiftMathematicsMediumSets, Relations and Functions168× since 2002Let f:R−{2,6}→Rf: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}f:R−{2,6}→R be real valued function defined as f(x)=x2+2x+1x2−8x+12f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}f(x)=x2−8x+12x2+2x+1. Then range of fff isAOption A. (−∞,−214]∪[1,∞)\left(-\infty,-\frac{21}{4}\right] \cup[1, \infty)(−∞,−421]∪[1,∞)BOption B. (−∞,−214)∪(0,∞)\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty)(−∞,−421)∪(0,∞)COption C. (−∞,−214]∪[0,∞)\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)(−∞,−421]∪[0,∞)DOption D. (−∞,−214]∪[214,∞)\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)(−∞,−421]∪[421,∞)CheckSkip · show solution