Question

Let $\alpha_1, \beta_1\left(\alpha_1<\beta_1\right)$ be the roots of the equation $x^2-b x+1=0$ and $\alpha_2, \beta_2\left(\alpha_2<\beta_2\right)$ be the roots of the equation $ax ^2+ bx +2=0$ such that $\frac{\alpha_1+\alpha_2}{\alpha_1 \alpha_2}=\frac{\beta_1+\beta_2}{\beta_1 \beta_2}=1$.

List I		List II
P	The value of a equals	1	-4
Q	The value of $b$ equals	2	1
R	The value of $\left(\beta_1+2 \alpha_2\right)$ equals	3	3
S	The value of $\left(\alpha_1-2 \alpha_2\right)$ equals	4	4

• A. P-1, Q-4, R-3, S-2
• B. P-2, Q-3, R-1, S-4
• C. P-1, Q-1, R-3, S-4
• D. P-1, Q-3, R-2, S-1

Answer 1

Answer: (A)

Given, $\frac{1}{\alpha_1}+\frac{1}{\alpha_2}=1$ and $\frac{1}{\beta_1}+\frac{1}{\beta_2}=1$
$\text { Add } \Rightarrow\left(\frac{1}{\alpha_1}+\frac{1}{\alpha_2}\right)+\left(\frac{1}{\beta_1}+\frac{1}{\beta_2}\right)=2 \Rightarrow\left(\frac{1}{\alpha_1}+\frac{1}{\beta_1}\right)+\left(\frac{1}{\alpha_2}+\frac{1}{\beta_2}\right)=2 \Rightarrow \frac{\alpha_1+\beta_1}{\alpha_1 \beta_1}+\frac{\alpha_2+\beta_2}{\alpha_2 \beta_2}=2 $
$\Rightarrow \frac{b}{1}+\frac{-b / a}{2 / a}=2 \Rightarrow b-\frac{b}{2}=2 \Rightarrow b=4$
Hence, roots of the first equation are $2+\sqrt{3}, 2-\sqrt{3}$
$\therefore \alpha_1=2-\sqrt{3} \text { and } \beta_1=2+\sqrt{3}$
Also second equation is $ax ^2+4 x +2=0$.
Now, $\frac{1}{\alpha_1}+\frac{1}{\alpha_2}=1$
$\therefore \frac{1}{\alpha_2}=1-\frac{1}{\alpha_1} \Rightarrow \frac{1}{\alpha_2}=1-(2+\sqrt{3})=-1-\sqrt{3} \Rightarrow \alpha_2=\frac{-1}{1+\sqrt{3}}$
Put $\alpha_2$ in second equation we get $a =-4 . $