Question

If $f\left(x\right)={\prod }_{k=1}^{999}\left(x^2-47x+k\right)$ , then product of all real roots of $f\left(x\right)=0$ is

• A. $550!$
• B. $551!$
• C. $552!$
• D. $999!$

Answer 1

Answer: (C)

Consider $x^2-47x+k=0$
For real roots, $47^2-4k\ge 0\Rightarrow k\le 552$
$\therefore k=1,2,3\dots .,552$
Product of real roots $=1\times 2\times 3\times 4\times \dots ..\times 552=552!$