Question

The relation \(f\) is defined by \(f(x)=\left\{\begin{matrix} x^{2}, &0\leq x\leq 3\\ 3x, &3\leq x\leq 10 \end{matrix}\right.\).The relation \(g\) is defined by \(g(x)=\left\{\begin{matrix} x^{2}, &0\leq x\leq 2\\ 3x, &2\leq x\leq 10 \end{matrix}\right.\).Show that \(f\) is a function \(g\) is not a function.

Answer 1

Step 1: Given \(f(x)=\left\{\begin{matrix} x^{2}, &0\leq x\leq 3\\ 3x, &3\leq x\leq 10 \end{matrix}\right.\) and \(g(x)=\left\{\begin{matrix} x^{2}, &0\leq x\leq 2\\ 3x, &2\leq x\leq 10 \end{matrix}\right.\)

Step 2: Check relation \(f\) has two relations at \(x=3\) given.
As we can see, f(x) has two relations at x=3
Now, check given relation is function or not.
Put \(x=3\)
If \(f(x)=x^{2}\)
\(f(3)=9\) and if \(f(x)=3x\)
\(f(3)=9\) i.e., at \(x=3, f(x)=9\) (unique image)
So, relation has unique image in \([0,10]\)
Hence the given relation is a function.

Step 3: Check relation g is a function or not
g(x) also has two relations at x=2
Now, check given relation is function or not.
at \(x=2\)
if \(f(x)=x^2\)
\(f(2)=4\)
and if \(f(x)=3x\)
\(f(2)= 3 \times 2=6\)
So, relation \(g(x)\) has two values at \(x=2\) i.e ,\(4\) and \(6\). Hence the given relation is not a function.