Question

Consider the function \(f:\mathbb{R}\to\mathbb{R}\) defined by

\(f(x)=\begin{cases}\left(2-\sin\left(\frac{1}{x}\right)\right)|x|, & x\ne 0\\ 0, & x=0\end{cases}\)

. Then \(f\) is :

• A. not monotonic on \((-\infty,0)\) and \((0,\infty)\)
• B. monotonic on \((0,\infty)\) only
• C. monotonic on \((-\infty,0)\) only
• D. monotonic on \((-\infty,0)\cup(0,\infty)\)

Hint:

For \(x\gt 0\), \(f'(x)=2-\sin\frac{1}{x}+\frac{1}{x}\cos\frac{1}{x}\), whose sign changes infinitely often near 0; also \(f(-x)=f(x)\), so the same holds for \(x\lt 0\).

Answer 1

The Correct Option is A

For \(x\gt 0\): \(f(x)=\left(2-\sin\frac{1}{x}\right)x\), so \(f'(x)=2-\sin\frac{1}{x}+x\left(-\cos\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)=2-\sin\frac{1}{x}+\frac{1}{x}\cos\frac{1}{x}\).
At \(x=\frac{1}{2n\pi}\): \(\sin=0,\ \cos=1\), so \(f'=2+2n\pi\gt 0\).
At \(x=\frac{1}{(2n+1)\pi}\): \(\sin=0,\ \cos=-1\), so \(f'=2-(2n+1)\pi\lt 0\) for \(n\ge 1\).
Both sequences tend to 0, so in every interval \((0,\varepsilon)\) the derivative takes both signs: \(f\) is not monotonic on \((0,\infty)\).
Also \(f(-x)=f(x)\) (even function), so on \((-\infty,0)\) the function is a mirror image and is also not monotonic.
Answer: (1) not monotonic on \((-\infty,0)\) and \((0,\infty)\)