JEE Mains · Maths · STD 11- 2. Relation and Function
Let \(f: R -\{2,6\} \rightarrow R\) be real valued function defined as \(f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}\). Then range of \(f\) is
- A \(\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)\)
- B \(\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty)\)
- C \(\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)\)
- D \(\left(-\infty,-\frac{21}{4}\right] \cup[1, \infty)\)
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 \(1, y \neq 1\) \(D \geq 0\) \(\Rightarrow(8 y+2)^2-4(y-1)(12 y-1) \geq 0\) \(\Rightarrow y(4 y+21) \geq 0\)…
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