Which of the following functions is not p.d.f. of a continuous random variable X?

Since \(P(X=1)=P(X=2)=\ldots \ldots \ldots=P(X=n)\)
and \(P(X=1)+P(X=2)+\ldots \ldots+P(X=n)=1\),
We get \(P(X=1)=P(X=2)=P(X=n)=\frac{1}{n}\)

\(\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{1}{\mathrm{n}}+\frac{2}{\mathrm{n}}+\frac{3}{\mathrm{n}}+\ldots \ldots+\frac{\mathrm{n}}{\mathrm{n}}=\frac{1}{\mathrm{n}}(1+2+3+\) \(\ldots \ldots \ldots \mathrm{n})=\frac{1}{\mathrm{n}} \frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{\mathrm{n}+1}{2}\)
\(\sum x_{1}{ }^{2} p_{1} =\frac{1^{2}}{n}+\frac{2^{2}}{n}+\frac{3^{2}}{n}+\ldots \ldots+\frac{n^{2}}{n} \)
\( =\frac{1}{n}\left(1^{2}+2^{2}+3^{2}+\ldots+n^{2}\right)=\) \(\frac{1}{n} \frac{n(n+1)(2 n+1)}{6}-\frac{(n+1)(2 n+1)}{6}\)
Given \(E(X)=V(X)\)
\(\sum x_{i} p_{i}=\sum x_{i}^{2} p_{i}-\left[\sum x_{i} p_{i}\right]^{2} \)
\( \frac{n+1}{2}=\frac{(n+1)(2 n+1)}{6}-\left(\frac{n+1}{2}\right)^{2} \)
\( \frac{n+1}{2}=(n+1)\left(\frac{2 n+1}{6}-\frac{n+1}{4}\right) \Rightarrow \frac{1}{2}=\frac{8 n+4-6 n-6}{24} \)
\( 12=2 n-2 \Rightarrow n=7\)