Question

Given that $\alpha ,\gamma$ are the roots of the equation $A{x}^2-4x+1=0and\beta ,\delta$ are the roots of the equation $B{x}^2-6x+1=0$, then the values of A and B respectively such that$\alpha ,\beta ,\gamma and\delta$ are in HP:

• A. -5, 9
• B. $\frac{3}{2},5$
• C. 3, 8
• D. None of these

Answer 1

The correct option is C 3, 8

Let us consider choice (a). When we put the values of A and B respectively, we get the values of $\alpha ,\beta ,\gamma and\delta$ as $-1,\frac{1}{3},\frac{1}{5},\frac{1}{3}$ which are not in HP. So this option is not correct.
Now for our convenience we consider choice (c), then by substituting the values of A and B, we get the values of $\alpha ,\beta ,\gamma and\delta$ as $1,\frac{1}{2},\frac{1}{3}and\frac{1}{4}$ which are in HP. Hence this could be the correct choice.
Alternatively:
$\begin{array}{ccc}A{x}^2-4x+1=0& \dots \left(I\right)& \\ & \alpha +\gamma =\frac{4}{A}& \dots \left(1\right)\\ & \alpha \gamma =\frac{1}{A}& \dots \left(2\right)\\ and& B{x}^2-6x+1=0& \dots \left(ii\right)\\ & \beta +\delta =\frac{6}{B}& \dots \left(3\right)\\ \beta \delta =\frac{1}{B}& \dots \left(4\right)& \end{array}$
Since it is given that $\alpha ,\beta ,\gamma ,\delta$ are in HP.
$\begin{array}{cc}\therefore & \beta =\frac{2\alpha \gamma }{\alpha +\gamma }=\frac{1}{2}\\ and& \gamma =\frac{2\beta \delta }{\beta +\delta }=\frac{1}{3}\end{array}$
Again since $\beta$ and $\gamma$ are the roots of the given equation hence they must satisfy the equation. So
$\begin{array}{cc}& B{\beta }^2-6\beta +1=0\\ and& A{\gamma }^2-4\gamma +1=0\\ \Rightarrow & B=8andA=3\end{array}$
Hence option (c) is correct