ExamBro
ExamBro
JEE Mains · Maths · STD 11- 2. Relation and Function

माना \(\mathrm{f}: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}\) एक वास्तविक मान फलन है, जो \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^2+2 \mathrm{x}+1}{\mathrm{x}^2-8 \mathrm{x}+12}\) द्वारा परिभाषित है तो \(\mathrm{f}\) का परिसर है

  1. A \(\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)\)
  2. B \(\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty)\)
  3. C \(\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)\)
  4. D \(\left(-\infty,-\frac{21}{4}\right] \cup[1, \infty)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)\)

Step-by-step Solution

Detailed explanation

Let \(y=\frac{x^2+2 x+1}{x^2-8 x+12}\) By cross multiplying \(y x^2-8 x y+12 y-x^2-2 x-1=0\) \(x^2(y-1)-x(8 y+2)+(12 y-1)=0\) Case \(1, y \neq 1\) \(D \geq 0\) \(\Rightarrow(8 y+2)^2-4(y-1)(12 y-1) \geq 0\) \(\Rightarrow y(4 y+21) \geq 0\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app