Let A, G and H be the arithmetic mean,geometr | vedclass

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(B) Given that $A, G, H$ are the arithmetic,geometric,and harmonic means of two distinct positive real numbers,we know that $A > G > H > 0$ and $AH =

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$0 < \alpha < 1$

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(B) Given that $A, G, H$ are the arithmetic,geometric,and harmonic means of two distinct positive real numbers,we know that $A > G > H > 0$ and $AH = G^2$.The given quadratic equation is $A(G-H) x^2 + G(H-A) x + H(A-G) = 0$.Let $f(x) = A(G-H) x^2 + G(H-A) x + H(A-G)$.Evaluating $f(1) = A(G-H) + G(H-A) + H(A-G) = AG - AH + GH - GA + HA - HG = 0$.Since $f(1) = 0$,$x = 1$ is one root of the equation.Let the roots be $\alpha$ and $\beta = 1$. The product of the roots is given by $\alpha \cdot \beta = \frac{H(A-G)}{A(G-H)}$.Since $\beta = 1$,we have $\alpha = \frac{H(A-G)}{A(G-H)}$.Using $AH = G^2$,we substitute $H = \frac{G^2}{A}$:$\alpha = \frac{\frac{G^2}{A}(A-G)}{A(G-\frac{G^2}{A})} = \frac{\frac{G^2}{A}(A-G)}{A(\frac{AG-G^2}{A})} = \frac{G^2(A-G)}{A(AG-G^2)} = \frac{G^2(A-G)}{AG(A-G)} = \frac{G}{A}$.Since $A > G > 0$,it follows that $0 < \frac{G}{A} < 1$. Thus,$0 < \alpha < 1$.

Let $A, G$ and $H$ be the arithmetic mean,geometric mean and harmonic mean,respectively,of two distinct positive real numbers. If $\alpha$ is the smallest of the two roots of the equation $A(G-H) x^2 + G(H-A) x + H(A-G) = 0$,then:

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