TS EAMCET · Maths · Quadratic Equation
If \(\frac{x^2+a x+3}{x^2+x+1}\) takes real values for all real values of \(x\) then \(a\) lies in the interval
- A \((-2-\sqrt{11}, \sqrt{11}-2)\)
- B \((4,3)\)
- C \((-2+\sqrt{2}, 2+\sqrt{2})\)
- D Not possible for y to have real values for all real values of x.
Answer & Solution
Correct Answer
(D) Not possible for y to have real values for all real values of x.
Step-by-step Solution
Detailed explanation
Let \(\frac{x^2+a x+3}{x^2+x+1}=y\), where \(y \in \mathbf{R}\) Now, \((y-1) x^2+(y-a) x+(y-3)=0\)…
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