If x∈R and 1≤3x2−7x+8x2+1≤2, then the minimum values of x are reapectively

Given 1≤3x2−7x+8x2+1≤2Let 1≤3x2−7x+8x2+1⇒ x2+1≤3x2−7x+8⇒ 2x2−7x+7≥0and 3x2−7x+8x2+1≤2⇒ 3x2−7x+8≤2x2+2⇒ x2−7x+6≤0⇒ (x−1)(x−6)≤0∴ 1≤x≤6

If x∈R and 1≤3x2−7x+8x2+1≤2, then the minimum values of x are reapectively

Given 1≤3x2−7x+8x2+1≤2Let 1≤3x2−7x+8x2+1⇒ x2+1≤3x2−7x+8⇒ 2x2−7x+7≥0and 3x2−7x+8x2+1≤2⇒ 3x2−7x+8≤2x2+2⇒ x2−7x+6≤0⇒ (x−1)(x−6)≤0∴ 1≤x≤6

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$I f x \in R a n d 1 \leq \frac{3 x^{2} - 7 x + 8}{x^{2} + 1} \leq 2, t h e n t h e \min i m u m v a l u e s o f x a r e r e a p e c t i v e l y$

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1, 6

1, 2

6, 10

5, 12

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n 1 \leq \frac{3 x^{2} - 7 x + 8}{x^{2} + 1} \leq 2 \\ L e t 1 \leq \frac{3 x^{2} - 7 x + 8}{x^{2} + 1} \Rightarrow x^{2} + 1 \leq 3 x^{2} - 7 x + 8 \\ \Rightarrow 2 x^{2} - 7 x + 7 \geq 0 \\ a n d \frac{3 x^{2} - 7 x + 8}{x^{2} + 1} \leq 2 \\ \Rightarrow 3 x^{2} - 7 x + 8 \leq 2 x^{2} + 2 \\ \Rightarrow x^{2} - 7 x + 6 \leq 0 \\ \Rightarrow (x - 1) (x - 6) \leq 0 \\ ∴ 1 \leq x \leq 6 \end{array}$