Question

Let a, b, c, p, q be real numbers. Suppose $\alpha ,\beta$ are the roots of the equation $x^2+2px+q=0$ and $\alpha ,\frac{1}{\beta }$ are the roots of the equation $x^2+2bx+c=0$, where ${\beta }^2\notin \left(-1,0,1\right)$
Statement-1:$\left(p^2-q\right)\left(b^2-ac\right)\ge 0$
Statement-2: $b\ne pa$ or $c\ne qa$

• A. Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1
• B. Statement -1 is true, Statement-2 is true; Statement -2 is NOT a correct explanation for Statement-1
• C. Statement -1 is false, Statement-2 is true
• D. Statement -1 is true, Statement-2 is false

Answer 1

Answer: (B)

Given, $x^2+2px+q=0$
$\therefore \alpha +\beta =-2p\dots \left(i\right)$
$\alpha \beta =q\dots \left(ii\right)$
And $a{x}^2+2bx+c=0$
$\therefore \alpha +\frac{1}{\beta }=-\frac{2b}{a}\dots \left(iii\right)$
and $\frac{\alpha }{\beta }=\frac{c}{a}\dots \left(iv\right)$
Now, $\left(p^2-q\right)\left(b^2-ac\right)$
$=\left[{\left(\frac{\alpha +\beta }{-2}\right)}^2-\alpha \beta \right]\left[{\left(\frac{\alpha +\frac{1}{\beta }}{2}\right)}^2-\frac{\alpha }{\beta }\right]a^2$
$=\frac{{\left(\alpha -\beta \right)}^2}{16}{\left(\alpha -\frac{1}{\beta }\right)}^2.a^2\ge 0$
$\therefore$ Statement 1 is true.
Again, now $pa=-\left(\frac{\alpha +\beta }{2}\right)a=-\frac{a}{2}\left(\alpha +\beta \right)$
and $b=-\frac{a}{2}\left(\alpha +\frac{1}{\beta }\right)$
Since, $pa\ne b\Rightarrow \alpha +\frac{1}{\beta }\ne \alpha +\beta$
$\Rightarrow {\beta }^2\ne 1,\beta \ne \left\{-1,0,1\right\}$, which is correct,
Similarly, if $c\ne qa\Rightarrow a\frac{\alpha }{\beta }\ne a\alpha \beta$
$\Rightarrow \alpha \left(\beta -\frac{1}{\beta }\right)\ne 0$
$\Rightarrow \alpha \ne 0$ and $\beta -\frac{1}{\beta }\ne 0$
$\Rightarrow \beta \ne \left\{-1,0,1\right\}$
Statement 2 is true.
Both Statement 1 and Statement 2 are true. But
Statement 2 does not explain statement 1.