AP EAMCET · Maths · Quadratic Equation
For real values of \(x\) and \(a\), if the expression \(\frac{x+a}{2 x^2-3 x+1}\) assumes all real values, then
- A \(a \lt -1\) or \(a\gt-\frac{1}{2}\)
- B \(-1 \lt a \lt -\frac{1}{2}\)
- C \(\frac{1}{2} \lt a \lt 1\)
- D \(a \lt \frac{1}{2}\) or \(a\gt1\)
Answer & Solution
Correct Answer
(B) \(-1 \lt a \lt -\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Let } \frac{x+a}{2 x^2-3 x+1}=y, \text { where } y \in \mathrm{R} \\ \Rightarrow & x+a=2 y x^2-3 y x+y\end{aligned}\) \(\begin{aligned}& \Rightarrow 2 y x^2-(3 y+1) x+y-a=0 \\ & (3 y+1)^2-4.2 y(y-a) \geq 0\end{aligned}\)…
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