Question

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

Answer 1

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