- Home
- Standard 11
- Mathematics
For $0<\mathrm{c}<\mathrm{b}<\mathrm{a}$, let $(\mathrm{a}+\mathrm{b}-2 \mathrm{c}) \mathrm{x}^2+(\mathrm{b}+\mathrm{c}-2 \mathrm{a}) \mathrm{x}$ $+(c+a-2 b)=0$ 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 $\mathrm{b}$ may be the geometric mean of $a$ and $c$
Both $(I)$ and $(II) $are true
Neither $(I)$ nor $(II)$ is true
Only $(II)$ is true
Only $(I)$ is true
Solution
$\mathrm{f}(\mathrm{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}) $
$ \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}} $
$ \text { 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, $0<\alpha<1$
$0<\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}}<1$
$\mathrm{b}>\mathrm{c}$ and $\mathrm{b}<\frac{\mathrm{a}+\mathrm{c}}{2}$
Therefore, $\mathrm{b}$ may be the $G.M.$ between $\mathrm{a}$ and $\mathrm{c}$.
