AP EAMCET · Maths · Quadratic Equation
If \(x \in R\) and \(1 \leq \frac{3 x^2-7 x+8}{x^2+1} \leq 2\), then the minimum and maximum values of \(x\) are respectively.
- A 1,2
- B 5,12
- C 6,10
- D 1, 6
Answer & Solution
Correct Answer
(D) 1, 6
Step-by-step Solution
Detailed explanation
Given, \(1 \leq \frac{3 x^2-7 x+8}{x^2+1} \leq 2 \quad\left[\because x^2+1 \geq 0\right]\) Now, \(\quad 1 \leq \frac{x^2-7 x+8}{x^2+1}\) \(0 \leq 2 x^2-7 x+7\) \(\therefore \quad f(x)=2 x^2-7 x+7\) \(\Delta=(7)^2-4(2)(0)\) \(\Delta 0\) \(\Rightarrow \quad x \in R\) and…
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