Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be defined as \(f(x) = 100x + 1\), where \(\mathrm{R}\) is a set of real numbers, then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → f is both one-one and onto
One-one: suppose \(f(x_1) = f(x_2)\). Then
\(100x_1 + 1 = 100x_2 + 1\)
\(x_1 = x_2\)
So f is one-one.
Onto: take any real y and solve \(f(x) = y\):
\(\displaystyle x = \frac{y - 1}{100}\)
This x is a real number and \(f(x) = (y - 1) + 1 = y\), so every real y is an image. Hence f is onto.