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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.

  1. A 1,2
  2. B 5,12
  3. C 6,10
  4. D 1, 6
Verified Solution

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…