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.

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Aicharya · Accepted Answer

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 $\frac{3 x^2-7 x+8}{x^2+1} \leq 2$ $\Rightarrow \quad x^2-7 x+6 \leq 0$ $\Rightarrow \quad(x-1)(x-6) \leq 0 \Rightarrow x \in[1,6]$ Hence, minimum and maximum values are 1 and 6.

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 \(x\)…

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