Explanation:
Exp: (D) : Given,
\(f(x)=7-3 x\)
For one-one function
\(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\)
\(7-3 \mathrm{x}_1=7-3 \mathrm{x}_2\)
\(\mathrm{x}_1=\mathrm{x}_2\)
Thus, \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) or
\(\mathrm{x}_1=\mathrm{x}_2, \forall \mathrm{x}_1, \mathrm{x}_2 \in \mathrm{R}\)
Hence, \(\mathrm{f}(\mathrm{x})\) is a one-one function
For onto function -
Let, \(\quad y=7-3 x\)
\(y-7=-3 x\)
\(x=\left[-\left(\frac{y-7}{3}\right)\right]\)
Clearly, \(\mathrm{x}\) is real number .
Thus, for each \(y \in R\) (codomain), there exists \(x \in R\) (domain) such that
\(f(x)=f\left[-\left(\frac{y-7}{3}\right)\right]\)
\(=7-3\left[-\left(\frac{y-7}{3}\right)\right]\)
\(7+3 \times\left(\frac{y-7}{3}\right)\)
\(=7+y-7\)
\(=y\)
Therefore, \(f(x)\) is an onto function.
Hence, \(f(x)\) is one-one and onto.