Question-1

function

Let \(f\) and \(g\) be two real valued functions such that \(g(x) = f(x) - 2\). Select all true options.

If \(f\) is odd, then \(g\) is odd.
If \(f\) is even, then \(g\) is even.
If \(g\) is odd, then \(f\) is odd.
If \(g\) is even, then \(f\) is even.

Hint

An odd function is one that satisfies \(f(-x) = -f(x)\) for all \(x \in \mathbb{R}\).
An even function is one that satisfies \(f(-x) = f(x)\) for all \(x \in \mathbb{R}\).

Answer

If \(f\) is odd, then \(g\) is odd.
If \(f\) is even, then \(g\) is even.
If \(g\) is odd, then \(f\) is odd.
If \(g\) is even, then \(f\) is even.

Solution

If \(f\) is odd, then \(f(-x) = -f(x)\). Then: \[ \begin{aligned} g(-x) &= f(-x) - 2\\ &= -f(x) - 2\\ &= -[f(x) + 2] \end{aligned} \] This doesn’t allow us to conclude anything about the nature of \(g\). A simple counter-example will show that if \(f\) is odd, \(g\) need not be odd. For instance, if \(f(x)= x\), we have \(g(x) = x - 2\), which is neither odd nor even.

If \(f\) is even, then \(f(-x) = f(x)\). Then: \[ \begin{aligned} g(-x) &= f(-x) - 2\\ &= f(x) - 2\\ &= g(x) \end{aligned} \] Therefore, if \(f\) is even, then \(g\) is also even. A similar argument holds for the case when \(g\) is even.