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 $D_1 \equiv b_1{ }^2-4 a_1 c_1$ and $D_2 \equiv b_2^2-4 a_2 c_2$.
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_1^2-4 a_1 c_1}{c_1^2}=\frac{b_2^2-4 a_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{4 c_1}{a_1}=\frac{b_2^2}{a_2^2}-\frac{4 c_2}{a_2} $
$\Rightarrow \frac{b_1^2-4 a_1 c_1}{a_1^2}=\frac{b_2^2-4 a_2 c_2}{a_2^2} \Rightarrow \frac{D_1^2}{D_2^2}=\frac{a_1^2}{a_2}$
Clearly both Statement -1 and Statement- 2 are true.