Discussion
Let \(f, g : \mathbf{R} \to \mathbf{R}\) be defined by \(f(x) = \sin\dfrac{1}{x}\) and \(g(x) = x\sin\dfrac{1}{x}\).
Then :
Answer: D
Neither \(f\) nor \(g\) is of bounded variation.
\(\sin(1/x)\) oscillates between -1 and 1 infinitely often as \(x \to 0\), so near 0 the variation adds up to 2 on each oscillation and is infinite.
\(x\sin(1/x)\) is continuous but still of unbounded variation near 0. Taking the points \(x_{k} = 1/((k + \frac{1}{2})\pi)\), where \(|g(x_{k})| = x_{k}\), the variation is at least about \(\sum 2x_{k}\), a multiple of the harmonic series, which diverges.
Since both functions have infinite variation on every interval around 0, every option claiming bounded variation for either of them is wrong.
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