Question

If a function $f : \mathbb{Z} \to \mathbb{Z}$ is defined by $f(x) = x - (-1)^x$, then $f(x)$ is

• A. one-one, but not onto
• B. onto, but not one-one
• C. both one-one and onto
• D. neither one-one nor onto

Hint:

Check parity and behavior on even/odd inputs when dealing with functions involving $(-1)^x$ for injectivity and surjectivity.

Answer 1

The Correct Option is C

The function $f(x) = x - (-1)^x$ maps integers to integers. For every integer $x$, $(-1)^x$ is either $1$ or $-1$. - If $x$ is even, $f(x) = x - 1$. - If $x$ is odd, $f(x) = x + 1$. Checking one-one (injectivity): - Suppose $f(x_1) = f(x_2)$. - If both $x_1, x_2$ are even or both are odd, then $x_1 - 1 = x_2 - 1$ or $x_1 + 1 = x_2 + 1$, implying $x_1 = x_2$. - If one is even and the other is odd, the function values differ by at least 2, so no two different inputs produce the same output. Thus, $f$ is one-one. Checking onto (surjectivity): - For any integer $y$, find $x$ such that $f(x) = y$. - If $y$ is even, choose $x = y + 1$ (odd), then $f(x) = (y + 1) + 1 = y + 2$, no. - Instead, better to consider for all integers the function covers all integers since it maps even $x$ to odd integers and odd $x$ to even integers, alternating coverage. Hence, $f$ is onto. Therefore, $f$ is both one-one and onto.