Step-by-step Derivation

1. Analyze the given quadratic equations

We are given two quadratic equations: Equation 1: $x^2 + 2px + q = 0$, with roots $\alpha, \beta$. From Vieta's formulas:

• Sum of roots: $\alpha + \beta = -2p$ (i)
• Product of roots: $\alpha \beta = q$ (ii) The discriminant is $D_1 = (2p)^2 - 4q = 4(p^2 - q)$.

Equation 2: $ax^2 + 2bx + c = 0$, with roots $\alpha, 1/\beta$. From Vieta's formulas:

• Sum of roots: $\alpha + \frac{1}{\beta} = -\frac{2b}{a}$ (iii)
• Product of roots: $\frac{\alpha}{\beta} = \frac{c}{a}$ (iv) The discriminant is $D_2 = (2b)^2 - 4ac = 4(b^2 - ac)$.

We are given that $a,b,c,p,q$ are real numbers. This implies that the roots of each equation are either both real or a complex conjugate pair.

2. Analyze the condition on $\beta$

We are given $\beta^2 \in \{-1, 0, 1\}$. For the roots of the second equation to be defined, we must have $1/\beta$ defined, which means $\beta \ne 0$. Thus, $\beta^2 \ne 0$. So the condition simplifies to $\beta^2 \in \{-1, 1\}$.

3. Case 1: $\beta^2 = -1$

This implies $\beta$ is a purely imaginary number, i.e., $\beta = i$ or $\beta = -i$. Since the coefficients $p, q$ of the first equation, $x^2+2px+q=0$, are real, its complex roots must occur in a conjugate pair. Therefore, $\alpha = \bar{\beta}$.

• If $\beta = i$, then $\alpha = -i$.
• If $\beta = -i$, then $\alpha = i$. In both cases, the roots of the first equation are $\{i, -i\}$. This equation is $(x-i)(x+i) = x^2+1=0$. Comparing with $x^2+2px+q=0$, we get $2p=0 \implies p=0$ and $q=1$.

Now, let's find the roots of the second equation, $ax^2+2bx+c=0$. The roots are $\alpha$ and $1/\beta$.

• If $\beta=i$ and $\alpha=-i$, the roots are $\{-i, 1/i\} = \{-i, -i\}$.
• If $\beta=-i$ and $\alpha=i$, the roots are $\{i, 1/(-i)\} = \{i, i\}$. In either scenario, the two roots are identical and non-real. However, the coefficients $a,b,c$ of the second equation are real. A quadratic equation with real coefficients must have roots that are a conjugate pair if they are complex. Let a root be $z$. Then $\bar{z}$ must also be a root. If the roots are $\{-i, -i\}$, the conjugate of the root $-i$ is $i$. So $i$ must be a root, which is not the case unless $-i=i$, which is false. This means that a quadratic equation with real coefficients cannot have roots $\{-i, -i\}$ (or $\{i, i\}$). Therefore, the case $\beta^2 = -1$ is impossible under the given conditions.

4. Case 2: $\beta^2 = 1$

This implies $\beta = 1$ or $\beta = -1$. In this case, $\beta$ is real. Since one root ($\eta$) of the first equation ($x^2+2px+q=0$ with real coefficients) is real, the other root $\alpha$ must also be real. Also, if $\beta^2=1$, then $1/\beta = \beta$. So, the roots of the second equation are $\alpha$ and $1/\beta = \beta$. This means both equations have the same set of roots, $\{\alpha, \beta\}$. Since the two quadratic equations have the same roots, their coefficients must be proportional: $\frac{a}{1} = \frac{2b}{2p} = \frac{c}{q}$ This gives us $b = pa$ and $c = qa$.

5. Evaluate STATEMENT - 1

STATEMENT - 1: $(p^2 - q)(b^2 - ac) \ge 0$.

Since the only possible scenario is $\beta^2=1$, the roots $\alpha, \beta$ must be real. For the first equation with real roots, its discriminant must be non-negative. $D_1 = 4(p^2 - q) \ge 0 \implies p^2 - q \ge 0$. Similarly, for the second equation with real roots $\alpha, \beta$, its discriminant must also be non-negative. $D_2 = 4(b^2 - ac) \ge 0 \implies b^2 - ac \ge 0$. The product of two non-negative numbers is non-negative. Therefore, $(p^2 - q)(b^2 - ac) \ge 0$.

Alternatively, using the relations from Case 2: $b=pa$ and $c=qa$. $b^2 - ac = (pa)^2 - a(qa) = a^2p^2 - a^2q = a^2(p^2 - q)$. So, $(p^2 - q)(b^2 - ac) = (p^2 - q) \cdot a^2(p^2 - q) = a^2(p^2 - q)^2$. Since $a$ and $(p^2-q)$ are real, $a^2 \ge 0$ and $(p^2 - q)^2 \ge 0$. Thus, $a^2(p^2 - q)^2 \ge 0$. So, STATEMENT - 1 is True.

6. Evaluate STATEMENT - 2

STATEMENT - 2: $b \ne pa$ or $c \ne qa$.

We have established that the only possible case under the problem's constraints is $\beta^2=1$. In this case, we proved that $b=pa$ and $c=qa$. This means the statement "$b \ne pa$ or $c \ne qa$" is false, because the negation, "$b = pa$ and $c = qa$", is true. So, STATEMENT - 2 is False.

7. Conclusion

STATEMENT - 1 is True and STATEMENT - 2 is False. This corresponds to option C.

The stored answer is B, which implies both statements are true. This contradicts our rigorous analysis. The contradiction arises because the premises for the case $\beta^2 = -1$ are inconsistent. When a problem's premises for a specific case lead to a contradiction, that case is considered impossible and is excluded from consideration. The statements are then evaluated based on the remaining possible cases. In this problem, the only possible case is $\beta^2=1$, which makes Statement-2 false. Therefore, the stored answer appears to be incorrect.