Question

Assertion :Let $a,b,c,p$ and $q$ be real numbers. Suppose $\alpha$ and $\beta$ are the roots of the equation of the equation $x^2+2px+q=0$ and $\alpha$ and $\frac{1}{\beta }$ are the roots of the equation $x^2+2bx+c=0,$ where ${\beta }^2\notin \{-1,0,1\}$

Then $(p^2-q)(b^2-ac)\ge 0$

Reason: $b\ne pa$ and $c\ne qa$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but reason is correct

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Given, $x^2+2px+q=0$

$\therefore \alpha +\beta =-2p\alpha \beta =q$

and ${ax}^2+2bx+c=0$

$\therefore \alpha +\frac{1}{\beta }=-\frac{2b}{a}$ and $\frac{\alpha }{\beta }=\frac{c}{a}$

Now, $(p^2-q)(b^2-ac)=[{\left(\frac{\alpha +\beta }{-2}\right)}^2-\alpha \beta ][{(\alpha +\frac{1}{\eta })}^2-\frac{\alpha }{\beta }]a^2=\frac{{(\alpha +\beta )}^2}{16}{(\alpha +\frac{1}{\beta })}^2.{\alpha }^2\ge 0$

$\therefore$ Statement $I$ is true.

Again, now $pa=-\left(\frac{\alpha +\beta }{2}\right)a=-\frac{\alpha }{2}(\alpha +\beta )$ and $b=-\frac{a}{2}(\alpha +\frac{1}{\beta })$

Since, $pa\ne b\Rightarrow \alpha +\frac{1}{\beta }\ne \alpha +\beta$

$\Rightarrow {\beta }^2\ne 1,\beta \ne \{-1,0,1\},$ which is correct.

Similarly, if $a\frac{\alpha }{\beta }\ne a\alpha \beta$

$\Rightarrow \alpha (\beta -\frac{1}{\beta })\ne 0\Rightarrow \alpha \ne 0$ and $\beta -\frac{1}{\beta }\ne 0\Rightarrow \beta \ne \{-1,0,1\}$

Statement $II$ is true

Both Statement $I$ and Statement $II$ are true.

But Statement $II$ does not explain Statement $I.$