If fand gare both one-to-one, then the sum function f+gis also one-to-one.✔False True

False.

Step-by-step explanation:

1. Definition of one-to-one function: A function $f$ is one-to-one (injective) if $f(a) = f(b)$ implies $a = b$.

2. Given: $f$ and $g$ are both one-to-one functions.

3. To Prove/Disprove: The sum function $(f+g)(x) = f(x) + g(x)$ is one-to-one.

4. Counterexample: Consider $f(x) = x$ and $g(x) = -x$, both of which are one-to-one functions.

5. Sum function: $(f+g)(x) = f(x) + g(x) = x + (-x) = 0$.

6. Analysis: The sum function $(f+g)(x) = 0$ for all $x$, which is not one-to-one because different values of $x$ (e.g., $x = 1$ and $x = 2$) map to the same value (0).

7. Conclusion: The sum function $f+g$ is not necessarily one-to-one even if $f$ and $g$ are both one-to-one. Therefore, the statement is false.