JEE maths#6

$\begin{aligned} f(x) & = (x-1)^{100}(x-2)^{2(99)}(x-3)^{3(98)} \cdots (x-100)^{100} \\ & = \prod_{n=1}^{100} (x-n)^{n(101-n)} \\ f'(x) & = f(x) \sum_{n=1}^{100} \frac {n(101-n)}{x-n} & \small \color{#3D99F6} \text{See note.} \\ \frac {f'(x)}{f(x)} & = \sum_{n=1}^{100} \frac {n(101-n)}{x-n} \\ \implies k & = \sum_{n=1}^{100} \frac {n(101-n)}{101-n} = \sum_{n=1}^{100} n = \frac {100(101)}2 = 5050 \end{aligned}$

$\implies \dfrac k{50}-97 = \dfrac {5050}{50} - 97 = \boxed{4}$

Note:

$\small \begin{aligned} f'(x) & = {\color{#3D99F6}100(x-1)^{99}}(x-2)^{2(99)}(x-3)^{3(98)} \cdots (x-100)^{100} \\ & \quad + (x-1)^{100}{\color{#3D99F6}(2(99))(x-2)^{2(99)-1}}(x-3)^{3(98)} \cdots (x-100)^{100} \\ & \quad + (x-1)^{100}(x-2)^{2(99)}{\color{#3D99F6}(3(98))(x-3)^{3(98)-1}} \cdots (x-100)^{100} \\ & \quad + \cdots + (x-1)^{100}(x-2)^{2(99)}(x-3)^{3(98)} \cdots {\color{#3D99F6}100(x-100)^{99}} \\ & = {\color{#3D99F6}\frac{100}{x-1}(x-1)^{100}}(x-2)^{2(99)}(x-3)^{3(98)} \cdots (x-100)^{100} \\ & \quad + {\color{#3D99F6}\frac{2(99)}{x-2}}(x-1)^{100}{\color{#3D99F6}(x-2)^{2(99)}}(x-3)^{3(98)} \cdots (x-100)^{100} \\ & \quad + {\color{#3D99F6}\frac{3(98)}{x-3}}(x-1)^{100}(x-2)^{2(99)}{\color{#3D99F6}(x-3)^{3(98)}} \cdots (x-100)^{100} \\ & \quad + \cdots + {\color{#3D99F6}\frac{100}{x-100}}(x-1)^{100}(x-2)^{2(99)}(x-3)^{3(98)} \cdots {\color{#3D99F6}(x-100)^{100}} \\ & = {\color{#3D99F6}\frac{100}{x-1}}f(x) + {\color{#3D99F6}\frac{2(99)}{x-2}}f(x) + {\color{#3D99F6}\frac{3(98)}{x-3}}f(x) + \cdots + {\color{#3D99F6}\frac{100}{x-100}}f(x) \\ & = f(x) \sum_{n=1}^{100} \frac {n(101-n)}{x-n} \end{aligned}$