Question

Let $\alpha ,\beta$ be the non-zero real roots of the quadratic equation $a_1{x}^2+b_{1}x+c_1=0$ and $\gamma ,\delta$ are the non-zero real roots of the quadratic equation $a_2{x}^2+b_{2}x+c_2=0$. Suppose $D_1\equiv b_1{}^2-4a_1{c}_1$ and $D_2\equiv b_2^2-4a_2{c}_2$.
Statement-1: If $\alpha ,\beta ,\gamma ,\delta$ are in H.P. then $D_1:D_2=c_1{}^2:c_2{}^2$.
Statement-2: If $\alpha ,\beta ,\gamma ,\delta$ are in A.P. then $D_1:D_2=a_1{}^2:a_2{}^2$.

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true. . .

Answer 1

Answer: (B)

Statement-1: $\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{\delta }$ are in A.P. (As $\alpha ,\beta ,\gamma ,\delta$ are in H.P.)
[11th, 12-09-2010, P-2]
$\Rightarrow \frac{1}{\beta }-\frac{1}{\alpha }=\frac{1}{\delta }-\frac{1}{\gamma }\Rightarrow \frac{(\alpha -\beta {)}^2}{{\alpha }^2{\beta }^2}=\frac{(\gamma -\delta {)}^2}{{\gamma }^2{\delta }^2}\Rightarrow \frac{(\alpha +\beta {)}^2-4\alpha \beta }{{\alpha }^2{\beta }^2}=\frac{(\gamma +\delta {)}^2-4\gamma \delta }{{\gamma }^2{\delta }^2}$
$\Rightarrow \frac{b_1^2-4a_1{c}_1}{c_1^2}=\frac{b_2^2-4a_2{c}_2}{c_2^2}\Rightarrow \frac{D_1^2}{D_2^2}=\frac{c_1^2}{c_2^2}$.
Statement-2: We have
$\beta -\alpha =\delta -\gamma (\text{ As }\alpha ,\beta ,\gamma ,\delta \text{ are in A.P.) }$
$\Rightarrow (\alpha +\beta {)}^2-4\alpha \beta =(\gamma +\delta {)}^2-4\gamma \delta \Rightarrow \frac{b_1^2}{a_1^2}-\frac{4c_1}{a_1}=\frac{b_2^2}{a_2^2}-\frac{4c_2}{a_2}$
$\Rightarrow \frac{b_1^2-4a_1{c}_1}{a_1^2}=\frac{b_2^2-4a_2{c}_2}{a_2^2}\Rightarrow \frac{D_1^2}{D_2^2}=\frac{a_1^2}{a_2}$
Clearly both Statement -1 and Statement- 2 are true.