Let a and b be real numbers and let \(f : R \to R\) be defined by \(f(x) = ax + b\), where R is the set of real numbers. Then f is one-one and onto if and only if
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(a \ne 0\)
Let \(a \ne 0\).
One-one: if \(f(p) = f(q)\), then
\(ap + b = aq + b\)
\(a(p - q) = 0\)
\(p = q\)
Onto: for any real y, the real number below satisfies \(f(x) = y\):
\( \displaystyle x = \frac{y - b}{a} \)
Let \(a = 0\). Then \(f(x) = b\) for every x, a constant function, which is neither one-one nor onto.
So f is one-one and onto exactly when \(a \ne 0\).
Why the other options are incorrect
Option (2) \(b \ne 0\) — Incorrect. With \(a = 0\), \(b = 1\), f is the constant function 1, which is not one-one.
Option (3) \(a \gt 0\) — Incorrect. \(f(x) = -2x\) has \(a \lt 0\) and is still one-one and onto.
Option (4) \(a = b\) — Incorrect. With \(a = b = 0\), f is the constant function 0.