For 0 < c < b < a, let (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0

Question

For \(0<\mathrm{c}<\mathrm{b}<\mathrm{a}\), let \((\mathrm{a}+\mathrm{b}-2 \mathrm{c}) \mathrm{x}^{2}+(\mathrm{b}+\mathrm{c}-2 \mathrm{a}) \mathrm{x}+(\mathrm c+\mathrm a-2 \mathrm b)=0\) and \(\alpha \neq 1\) be one of its root. Then, among the two statements

(I) If \(\alpha \in(-1,0)\), then \(\mathrm{b}\) cannot be the geometric mean of \(\mathrm{a}\) and \(\mathrm{c}\)

(II) If \(\alpha \in(0,1)\), then $b$ may be the geometric mean of \(\mathrm{a}\) and \(\mathrm{c}\)

(1) Both (I) and (II) are true

(2) Neither (I) nor (II) is true

(3) Only (II) is true

(4) Only (I) is true

Answer 1

Correct option is (1) Both (I) and (II) are true

\(\mathrm {f(x)}=(\mathrm{a}+\mathrm{b}-2 \mathrm{c}) \mathrm{x}^{2}+(\mathrm{b}+\mathrm{c}-2 \mathrm{a}) \mathrm{x}+(\mathrm c+\mathrm a-2 \mathrm b)=0\)

\(\mathrm{f}(\mathrm{x})=\mathrm{a}+\mathrm{b}-2 \mathrm{c}+\mathrm{b}+\mathrm{c}-2 \mathrm{a}+\mathrm{c}+\mathrm{a}-2 \mathrm{~b}=0\)

\(\mathrm{f}(1)=0\)

\(\therefore \alpha \cdot 1=\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}}\)

\(\alpha=\frac{\mathrm c+\mathrm a-2\mathrm b}{\mathrm a+\mathrm b-2 \mathrm c}\)

If, \(-1<\alpha<0\)

\(-1<\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}}<0\)

\(\mathrm{b}+\mathrm{c}<2 \mathrm{a}\text{ and }\mathrm{b}>\frac{\mathrm{a}+\mathrm{c}}{2}\)

therefore, \( \mathrm{b}\) cannot be G.M. between \(\mathrm{a}\) and \(\mathrm{c}\).

If, \(0<\alpha<1\)

\(0<\frac{\mathrm c+\mathrm a-2 \mathrm b}{\mathrm a+\mathrm b-2 \mathrm c}<1\)

\(\mathrm{b}>\mathrm{c}\text{ and }\mathrm{b}<\frac{\mathrm{a}+\mathrm{c}}{2}\)

Therefore, \( \mathrm{b}\) may be the G.M. between \(\mathrm{a}\) and \(\mathrm{c}\).