Question

Let $f(x)=(x-1)(x-2)(x-3)\dots (x-n),n\in N$ and $f^{\prime }(n)=5040$, then the value of $n$ is

Answer 1

Answer: 8

$\ln (f(x))=\ln (x-1)+\ln (x-2)+\dots .+\ln (x-n)$
$\Rightarrow f^{\prime }(x)=f(x)\left[\frac{1}{x-1}+\frac{1}{x-2}+\dots +\frac{1}{x-n}\right]$
$\Rightarrow f^{\prime }(x)=(x-2)(x-3)(x-n)+(x-1)(x-3)\dots (x-n)+\dots .+$
$(x-1)(x-2)\dots (x-(n-1))$
$\Rightarrow f^{\prime }(n)=(n-1)(n-2)(n-3)\cdot 3\cdot 2\cdot 1($ all other factors except the last vanishes when $x=n$ )
$\Rightarrow 5040=(n-1)!$
$\Rightarrow n=8$