1. Given quadratic and one obvious root

The equation is

Let

Then

So $x=1$ is a root.

Since one root is given to be $\alpha\ne 1$, the two roots are $1$ and $\alpha$.

2. Find the other root $\alpha$ using product of roots

For the quadratic $Ax^2+Bx+C=0$, product of roots is

Hence

Therefore

Now use the condition $0<c<b<a.$

Observe:

• $a+b-2c>0$ because $a>c$ and $b>c$.
• So the sign of $\alpha$ is the sign of $a+c-2b$.

3. Analyze statement (I): If $\alpha\in(-1,0)$, then $b$ cannot be the geometric mean of $a$ and $c$

If $b$ is the geometric mean of $a$ and $c$, then $b^2=ac.$ Write

because $0<c<b<a$.

Then

Since $r>1$,

and also clearly

So if $b$ is geometric mean, then necessarily $\alpha\in(0,1),$ not in $(-1,0)$.

Hence, if $\alpha\in(-1,0)$, then $b$ cannot be the geometric mean.

So (I) is true.

4. Analyze statement (II): If $\alpha\in(0,1)$, then $b$ may be the geometric mean of $a$ and $c$

We only need to check whether it is possible.

From the calculation above, if we choose $b$ as geometric mean, i.e. $a=cr^2,\ b=cr \quad (r>1),$ then

For every $r>1$,

Thus there do exist values with $b$ as geometric mean and $\alpha\in(0,1)$.

For example, take $c=1, b=2, a=4$. Then $b^2=ac=4$, so $b$ is geometric mean. Also

Thus statement (II) is also true.

So (II) is true.

5. Conclusion

Both statements are true.

Therefore the correct option is

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They match.