By splitting \(1.1\) and then applying Binomial Theorem, the first few terms of \((1.1)^{10000}\) can be obtained as
\((1.1) ^{10000} = (1+0.1)^{ 10000}\)
=\(^{10000}C_0 + ^{10000}C_1(1.1)\,+\,\text{Other positive terms}\)
=\(1+10000×1.1+ \text{Other positive terms }\)
=\(1+11000+\text{Other positive terms}\)
\(> 1000\)
\(\text{Hence,}\, (1.1)^{10000} >1000\).
\(f(x) = \begin{cases} x^2, & \quad 0≤x≤3\\ 3x, & \quad 3≤x≤10 \end{cases}\)
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.