Let A,G and H be the arithmetic mean, geometric mean and harmonic mean, respectively of two distinct positive real numbers. If α is the smallest of the two roots of the equation A(G−H)x2+G(H−A)x+H(A−G)=0, then
A
−2<α<−1
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B
0<α<1
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C
−1<α<0
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D
1<α<2
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Solution
The correct option is A0<α<1 Consider two positive real numbers a and b.
Then, A=a+b2, G=√ab and H=2aba+b.
From that we can conclude that AH=G2
Now consider the given equation,
One of the root is 1.
Given the other roots is α.
Product of roots =α=H(A−G)A(G−H).
Now substitute H=G2A in the equation of α.
⇒α=G(A−G)A(A−G)
⇒α=GA
We know that A>G>H (Because the numbers are distinct and positive)