Question

Let $a,b,c$ be three distinct numbers, such that $a+b+c=xS$ and $a^2+b^2+c^2=S^2$. If $a,b,c$ are in G.P., then which of the following is/are correct ?

• A. $\frac{1}{3}<x^2<3$
• B. $\frac{1}{3}<x^2<1$
• C. $1<x^2<3$
• D. $\frac{1}{3}\le x^2\le 3$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{1}{3}<x^2<1$
C $1<x^2<3$

Let $b=ar,c=a{r}^2,r\ne \pm 1$ as numbers are distinct
$a^2+a^2{r}^2+a^2{r}^4=S^2...\left(1\right)$
$a+ar+a{r}^2=xS...\left(2\right)$
Squaring both sides
$a^2(1+r+r^2{)}^2=x^2{S}^2...(3)$

Equation $\left(1\right)$ divided by equation $\left(3\right)$
$\Rightarrow \frac{a^2(1+r^2+r^4)}{a^2(1+r+r^2{)}^2}=\frac{S^2}{x^2{S}^2}$
$\Rightarrow \frac{(1+r^2{)}^2-r^2}{(1+r+r^2{)}^2}=\frac{1}{x^2}$
$\Rightarrow \frac{1+r^2-r^2}{1+r+r^2}=\frac{1}{x^2}$
$\Rightarrow x^2=\frac{r+\frac{1}{r}+1}{r+\frac{1}{r}-1}$

Putting
$y=r+\frac{1}{r};|y|>2$
$\Rightarrow x^2=\frac{y+1}{y-1}$
$\Rightarrow x^2=1+\frac{2}{y-1}$
Now, when
$y>2\Rightarrow y-1>1\Rightarrow 0<\frac{2}{y-1}<2\Rightarrow 1<1+\frac{2}{y-1}<3\Rightarrow 1<x^2<3$

When,
$y<-2\Rightarrow y-1<-3\Rightarrow 0>\frac{2}{y-1}>-\frac{2}{3}\Rightarrow 1>1+\frac{2}{y-1}>\frac{1}{3}\Rightarrow 1>x^2>\frac{1}{3}$
$\therefore x^2\in (\frac{1}{3},1)\cup (1,3)$


Alternate:
$y=\frac{x^2+1}{x^2-1}(\text{where}{x}^2\ne 1)...\left(4\right)$
$\because r\ne \pm 1,\therefore |y|=|r+\frac{1}{r}|>2$
$\Rightarrow \left|\frac{x^2+1}{x^2-1}\right|>2$
$\Rightarrow |x^2+1|>2|x^2-1|$
$\Rightarrow (x^2+1{)}^2-[2(x^2-1){]}^2>0$
$\Rightarrow \{(x^2+1)+\left[2\right(x^2-1\left)\right]\}\{(x^2+1)-\left[2\right(x^2-1\left)\right]\}>0$
$\Rightarrow (3x^2-1)(3-x^2)>0$
$\Rightarrow \frac{1}{3}<x^2<3$
From eqn$\left(4\right)$, $x^2\ne 1$
$\therefore x^2\in (\frac{1}{3},1)\cup (1,3)$