By Binomial Theorem,
\(\underset{r=0} {\overset{n}∑} \,{ ^nC_r} a^{n-r} b^r= (a+b)^n\)
By putting \(b=3\) and \(a = 1\)in the above eqution, we obtain
\(\underset{r=0} {\overset{n}∑} \,{ ^nC_r} (1)^{n-r} (3)^r= (1+3)^n\)
⇒ \(\underset{r=0} {\overset{n}∑} { 3^r }\,{ ^nC_r} = 4^n\)
Hence, proved.
\(f(x) = \begin{cases} x^2, & \quad 0≤x≤3\\ 3x, & \quad 3≤x≤10 \end{cases}\)
The relation g is defined by
\(g(x) = \begin{cases} x^2, & \quad 0≤x≤2\\ 3x, & \quad 2≤x≤10 \end{cases}\)
Show that f is a function and g is not a function.