Question

Let ${\alpha }_1$ and ${\alpha }_2$ be the roots of the equation $x^2-4x+P_1=0$, and ${\alpha }_3$ and ${\alpha }_4$ be the roots of the equation $x^2-36x+P_2=0$. If ${\alpha }_1<{\alpha }_2<{\alpha }_3<{\alpha }_4$ and ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ are in G.P., then the product $P_1{P}_2$ equals

• A. 81
• B. 243
• C. 729
• D. 27

Answer 1

The correct option is C 729

Given ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ are in G.P
let ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ be $\frac{a}{r^3},\frac{a}{r},ar,a{r}^3$ respectively.
Then,
$\frac{a}{r^3}+\frac{a}{r}=4ar+a{r}^3=36$
Solving the above two equations
$\frac{a(r+r^3)}{a(\frac{1}{r^3}+\frac{1}{r})}=\frac{36}{4}\Rightarrow r^3\times r=9\Rightarrow r=\sqrt{3}\Rightarrow a=\frac{36}{r+r^3}=3\sqrt{3}$
$\Rightarrow P_1{P}_2={\alpha }_1\times {\alpha }_2\times {\alpha }_3\times {\alpha }_4=\frac{a}{r^3}\times \frac{a}{r}\times ar\times a{r}^3=a^4={\left(3\sqrt{3}\right)}^4=729$
Hence, option 'C' is correct.