Question

Let $A,G$ and $H$ be the arithmetic mean, geometric mean and harmonic mean, respetively 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

• A. $-2<\alpha <-1$
• B. $0<\alpha <1$
• C. $-1<\alpha <0$
• D. $1<\alpha <2$

Answer 1

The correct option is B $0<\alpha <1$

Given :
$A(G–H)x^2+G(H–A)x+H(A–G)=0$
Putting $x=1$ in the L.H.S of the given equation
$\Rightarrow A(G–H)1^2+G(H–A)1+H(A–G)=0$
$x=1$ is a root of given equation.
So the other root is $\alpha$,
Product of roots $\Rightarrow \alpha \times 1=\frac{H(A-G)}{A(G-H)}$
$\Rightarrow \alpha =\frac{HA-HG}{A(G-H)}=\frac{G^2-HG}{A(G-H)}=\frac{G(G-H)}{A(G-H)}\Rightarrow \alpha =\frac{G}{A}\therefore \alpha <1$
$[\because A.M>G.M]$