Let p, q be real numbers. If alpha is a root | vedclass

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(D) Given that $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$,we have $\alpha^{2}+3 p^{2} \alpha+5 q^{2}=0$. Since $\beta$ is a root of $x^{2}+9 p^

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$\alpha < \gamma < \beta$

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(D) Given that $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$,we have $\alpha^{2}+3 p^{2} \alpha+5 q^{2}=0$. Since $\beta$ is a root of $x^{2}+9 p^{2} x+15 q^{2}=0$,we have $\beta^{2}+9 p^{2} \beta+15 q^{2}=0$. Let $f(x)=x^{2}+6 p^{2} x+10 q^{2}$. Evaluating $f(\alpha)$: $f(\alpha)=\alpha^{2}+6 p^{2} \alpha+10 q^{2} = (\alpha^{2}+3 p^{2} \alpha+5 q^{2}) + 3 p^{2} \alpha + 5 q^{2} = 0 + 3 p^{2} \alpha + 5 q^{2}$. Since $\alpha > 0$ and $p, q$ are real numbers (assuming $p, q eq 0$ for non-trivial roots),$f(\alpha) > 0$. Evaluating $f(\beta)$: $f(\beta)=\beta^{2}+6 p^{2} \beta+10 q^{2} = (\beta^{2}+9 p^{2} \beta+15 q^{2}) - (3 p^{2} \beta + 5 q^{2}) = 0 - (3 p^{2} \beta + 5 q^{2})$. Since $\beta > \alpha > 0$,$f(\beta) Since $f(x)$ is a continuous polynomial function and $f(\alpha) > 0$ while $f(\beta) < 0$,by the Intermediate Value Theorem,there must exist a root $\gamma$ such that $\alpha < \gamma < \beta$.

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$,$\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$,$\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

Let $p, q$ be real numbers. If $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$,$\beta$ is a root of $x^{2}+9 p^{2} x+15 q^{2}=0$ and $0 < \alpha < \beta$,then the equation $x^{2}+6 p^{2} x+10 q^{2}=0$ has a root $\gamma$ that always satisfies:

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