Let f(x) = (x+1)(x+2)...(x+100) and g(x) = f(x).f''(x) - ( f'(x))^2 . Let n be number of real roots of g(x)=0, then : (a.) n2 (c.) n100

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right).....\left(\mathrm{x}+10\right) \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)...\left(\mathrm{x}+9\right)+\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)...\left(\mathrm{x}+8\right)\left(\mathrm{x}+10\right)+...+\left(\mathrm{x}+2\right).....\left(\mathrm{x}+10\right) \\ \frac{\mathrm{f}\text{'}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}=\frac{1}{\mathrm{x}+10}+\frac{1}{\mathrm{x}+9}+....+\frac{1}{\mathrm{x}+1}=\sum _{\mathrm{r}=1}^{10} \frac{1}{\mathrm{x}+\mathrm{r}} \\ \Rightarrow \frac{\mathrm{f}\text{'}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}= \sum _{\mathrm{r}=1}^{10} \frac{1}{\mathrm{x}+\mathrm{r}} \\ \mathrm{Differentiate} \mathrm{both} \mathrm{sides} \\ \Rightarrow {\left[\frac{\mathrm{f}\text{'}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\right]}^{\text{'}} =-\sum _{\mathrm{r}=1}^{10} \frac{1}{{\left(\mathrm{x}+\mathrm{r}\right)}^2} \\ \mathrm{Now} \\ \mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)-{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2={\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2{\left[\frac{\mathrm{f}\text{'}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\right]}^{\text{'}} \\ =-{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2\sum _{\mathrm{r}=1}^{10} \frac{1}{{\left(\mathrm{x}+\mathrm{r}\right)}^2} \\ \mathrm{g}\left(\mathrm{x}\right)=0 \\ \mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)-{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2=0 \\ \mathrm{If} \mathrm{we} \mathrm{represent} \mathrm{roots} \mathrm{as} {\mathrm{a}}_{\mathrm{i}} \end{gathered}$$

Hence n=0

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