Question

Two sets A and B are given as
$A=\{x|x\text{is an integer root of the equation}{x}^5-6x^4+11x^3-6x^2=0\}$
$B=\{x|x\text{is a real root of the equation}a{x}^5+2a{x}^3+2b{x}^2+b=0,a,b\in R\text{such that the given equation have}\text{maximum number of real roots.}\}$
Then, which of the following is correct?

• A. $n\left(A\right)>n\left(B\right)$
• B. $n\left(A\right)=n\left(B\right)$
• C. $n\left(A\right)<n\left(B\right)$
• D. cannot comment because $n\left(A\right)$ varries as the values of $a$ and $b$ vary

Answer 1

The correct option is A $n\left(A\right)>n\left(B\right)$

Given,
$\text{Set}A=\{x|x\text{is an integer root of the equation}{x}^5-6x^4+11x^3-6x^2=0\}$
$x^5-6x^4+11x^3-6x^2=0$
$\Rightarrow x^2(x^3-6x^2+11x-6)=0$
$\Rightarrow x^2\left(x^2\right(x-1)-5x(x-1)+6(x-1\left)\right)=0$
$\Rightarrow x^2(x-1)(x^2-5x+6)=0$
$\Rightarrow x^2(x-1)(x-2)(x-3)=0$
$A=\{0,1,2,3\}$
$\therefore n\left(A\right)=4$

$\text{Set}B=\{x|x\text{is a real root of the equation}a{x}^5+2a{x}^3+2b{x}^2+b=0,a,b\in R\text{such that the given equation have}\text{maximum number of real roots.}\}$

Case 1: $ab>0$
$f\left(x\right)=a{x}^5$+$2a{x}^3$+$2b{x}^2$+$b$
Using Descartes' rule of Signs
$f\left(x\right)$ = No sign change $\Rightarrow$ zero positive roots
$f(-x)=-a{x}^5$+$-2a{x}^3$+$2b{x}^2$+$b$
$\Rightarrow$ one sign change
It can have maximum one real root.
As it is odd polynomials, it must have atleast one real root.
$\therefore$ It has exactly one root.
Case 2: $ab<0$
Proceeding in the same way as Case 1.
Set B contains only one real element
$\therefore n\left(B\right)=1$
Hence, $n\left(A\right)>n\left(B\right)$