Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined by $f(x) = x + |x|\cos x$. Then which of the following statements is TRUE?
A) $f$ is one-one but NOT onto
B) $f$ is onto but NOT one-one
C) $f$ is BOTH one-one and onto
D) $f$ is NEITHER one-one NOR onto
Step-by-Step Solution
Key Concept: For $x\geq 0$: $f(x)=x(1+\cos x)\geq 0$, and $f(x)=0$ at $x=\pi,3\pi,\ldots$ — so $f$ is many-one. For $x\geq 0$, $f(x)\in[0,\infty)$ as $x\to\infty$. For $x<0$: $f(x)=x(1-\cos x)\leq 0$, reaching $(-\infty,0)$. Range $=\mathbb{R}$, so $f$ is onto.
$f(x)=0$ at $x=0$ and $x=\pi,3\pi,\ldots$ — multiple preimages of $0$, so not injective. Since $f(x)\to+\infty$ for $x\to+\infty$ and $f(x)\to-\infty$ for $x\to-\infty$, by IVT it is surjective. Answer: B.
Correct Answer: B