Quadratic Equations
Quadratic with Parameter — Geometric Mean Condition
nta_pyq_2024_jan
Grade 11
Question:
For $0<c<b<a$, let $(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0$ and $\alpha\neq1$ be one of its roots. Then, among the two statements: (I) If $\alpha\in(-1,0)$, then $b$ cannot be the geometric mean of $a$ and $c$. (II) If $\alpha\in(0,1)$, then $b$ 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
Step-by-Step Solution
Key Concept: $f(1)=0$ so $x=1$ is always a root. The other root is $\alpha=\frac{c+a-2b}{a+b-2c}$. For $\alpha\in(-1,0)$: conditions imply $b>(a+c)/2$ (AM), so $b$ can't be GM$=\sqrt{ac}<(a+c)/2$. For $\alpha\in(0,1)$: $b>c$ and $b<(a+c)/2$, which is compatible with $b=\sqrt{ac}$.
Both statements are true.
Correct Answer: 1