JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Quadratic Equations and Inequalities

Since the coefficients of the quadratic sum to zero ($(a+b-2c)+(b+c-2a)+(c+a-2b)=0$), $x=1$ is always a root. By Vieta's formulas, the product of the two roots equals the other root $\alpha$ (since one root is $1$):


Since $0<c<b<a$, the denominator $D=a+b-2c=(a-c)+(b-c)>0$. Let $N=c+a-2b$ be the numerator.

Statement (I): If $\alpha \in (-1,0)$, then in particular $N<0$ (from $\alpha <0$), i.e. $c+a<2b$, so $b>\frac{a+c}{2}$. Since $a\ne c$, the AM–GM inequality is strict: $\frac{a+c}{2}>\sqrt{ac}$. So $b>\sqrt{ac}$ strictly, and $b$ cannot equal the geometric mean of $a,c$. Statement (I) is true.

Statement (II): Take $c=1$, $a=t^2$ for any $t>1$, and set $b=t=\sqrt{ac}$ (the geometric mean), which satisfies $c<b<a$ for $t>1$. Then


As $t$ ranges over $(1,\infty )$, $\alpha =\frac{t-1}{t+2}$ ranges over $(0,1)$. So whenever $b$ equals the geometric mean of $a$ and $c$, $\alpha$ automatically lies in $(0,1)$ — confirming $b$ may indeed be the geometric mean when $\alpha \in (0,1)$. Statement (II) is true.

So both (I) and (II) are true.