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Let f:R-2,6→ R be real valued function defined as f(x)= x+2x+1 x^2-8x+12. Then range of f is

JEE Main 2023 — Mathematics Algebra

2023mcqhard

Let f:R−2,6→Rf:R-{2,6}\rightarrow Rf:R−2,6→R be real valued function defined as f(x)=x+2x+1x2−8x+12f(x)=\frac{x+2x+1}{{x}^{2}-8x+12}f(x)=x2−8x+12x+2x+1​. Then range of fff is

Official previous-year question

Held on 31 Jan 2023 · Verified 6 Jul 2026.

Options

  1. A

    (−∞,−214]∪[214,∞)(-\infty ,-\frac{21}{4}]\cup [\frac{21}{4},\infty )(−∞,−421​]∪[421​,∞)

  2. B

    (−∞,−214]∪[0,∞)(-\infty ,-\frac{21}{4}]\cup [0,\infty )(−∞,−421​]∪[0,∞)

  3. C

    (−∞,−214)∪(0,∞)(-\infty ,-\frac{21}{4})\cup (0,\infty )(−∞,−421​)∪(0,∞)

  4. D

    (−∞,−214]∪[1,∞)(-\infty ,-\frac{21}{4}]\cup [1,\infty )(−∞,−421​]∪[1,∞)

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