Question

Suppose $E_1,E_2,E_3$ be three mutually exclusive events such that $P\left(E_i\right)=p_i$, for $i=1,2,3$.

If $p_1,p_2,p_3$ are the roots of $27x^3-27x^2+ax-1=0$, then value of $a$ is,

• A. $9$
• B. $6$
• C. $3$
• D. $0$

Answer 1

Answer: (A)

$9$

As $p_1,p_2,p_3$ are roots of $27x^3-27x^2+ax-1=0$

Then $s_1=p_1+p_2+p_3=-(-\frac{27}{27})=1$ ...$\left(1\right)$

$s_2=p_1{p}_2+p_2{p}_3+p_1{p}_3=\frac{a}{27}$ ...$\left(2\right)$

And $s_3=p_1{p}_2{p}_3=-\left(\frac{-1}{27}\right)=\frac{1}{27}$ ....$\left(3\right)$

Now applying $A.M\ge G.M.$ on $p_1,p_2,p_3$

we get $\frac{p_1+p_2+p_3}{3}\ge {\left(p_1{p}_2{p}_3\right)}^{\frac{1}{3}}$

From $\left(1\right)$ and $\left(3\right)$

$\frac{1}{3}\ge {\left(\frac{1}{27}\right)}^{\frac{1}{3}}\Rightarrow p_1=p_2=p_3=\frac{1}{3}$ $\because A.M.=G.M.henceallvaluesmustbeequal$

From $\left(2\right)$ $p_1{p}_2+p_2{p}_3+p_1{p}_3=\frac{a}{27}$

$\Rightarrow \frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{1}{3}=\frac{a}{27}$

$\Rightarrow a=9$