Question

The equation $p{x}^3-(p+1)x^2+px=0$ has real roots and $p$ is any positive integer, then the equation has

• A. exactly one root less than $1$
• B. more than one root $\ge 1$
• C. all roots greater than $1$
• D. none of these

Answer 1

Answer: (A), (B)

exactly one root less than $1$
more than one root $\ge 1$

Let $\alpha ,\beta ,\gamma$ be the roots of $p{x}^3-(p+1)x^2+px=0$
$\Rightarrow p{x}^3-(p+1)x^2+px=0$

$\Rightarrow x(p{x}^2-(p+1)x+p)=0$
So, $\alpha$ is root of $x=0$
And $\beta ,\gamma$ are roots of $f\left(x\right)=p{x}^2-(p+1)x+p$
Now as $\beta +\gamma =\frac{(p+1)}{p}>0$
$\beta \gamma =1>0$
And
$f\left(1\right)=p{.1}^2-(p+1)1+p\ge 0$
Hence, only one root is less than $1$.
And more than one root is $\ge$ $1$
Hence, options 'A' and 'B' are correct.