设函数 ${f_1}\left( x \right) = {x^2} $，$ {f_2}\left( x \right) = 2\left( {x - {x^2}} \right) $，$ {f_3}\left( x \right) = \dfrac{1}{3}\left| {\sin 2{\mathrm \pi} x} \right|$，${a_i} = \dfrac{i}{99}$，$i = 0,1,2, \cdots ,99$．记 ${I_k} = \left| {{f_k}\left( {a_1} \right) - {f_k}\left( {a_0} \right)} \right| + \left| {{f_k}\left( {a_2} \right) - {f_k}\left( {a_1} \right)} \right| + \cdots + \left| {{f_k}\left( {{a_{99}}} \right) - {f_k}\left( {{a_{98}}} \right)} \right|$，$k = 1,2,3$，则 \((\qquad)\)
【难度】
【出处】
无
【标注】
【答案】
B
【解析】
按条件列出 $I_k$，再比较即可．注意及时利用平方差公式、等差数列的求和公式、三角函数的公式等进行变形化简．由\[\left| \left( \dfrac{ i }{99} \right)^2 - {\left(\dfrac{ i - 1}{99}\right)^2} \right| = \dfrac{1}{99} \cdot \dfrac{2 i - 1}{99}\]得\[\begin{split}{I_1} & = \dfrac{1}{99}\left(\dfrac{1}{99} + \dfrac{3}{99} + \dfrac{5}{99} + \cdots + \dfrac{2 \times 99 - 1}{99}\right) \\& \overset{\left[a\right]}= \dfrac{1}{99} \cdot \dfrac{{{{\left(99\right)}^2}}}{99} \\& = 1,\end{split}\]（推导中用到：[a]）由\[2 \left| \dfrac{ i }{99} - \dfrac{ i - 1}{99} - \left( \dfrac{i}{99} \right)^2 + {\left(\dfrac{i - 1}{99}\right)^2} \right| = \dfrac{2}{99} \left|\dfrac{99 - \left(2i - 1\right)}{99} \right|,\]得\[\begin{split}{I_2} &= 2 \times \dfrac{1}{99} \times 2 \times \dfrac{50\left(98 + 0\right)}{2 \times 99} \\&\overset{\left[b\right]} = \dfrac{98}{99} \cdot \dfrac{100}{99} < 1 ,\\ {I_3} & = \dfrac{1}{3}\left( \left| \left|\sin \dfrac{2{\mathrm \pi} }{99} \right| - \left|\sin \dfrac{0\times{2{\mathrm \pi} }}{99} \right| \right| + \cdots + \left| \left|\sin \dfrac{{99\times{2{\mathrm \pi} }}}{99} \right| - \left|\sin \dfrac{{98\times{2{\mathrm \pi} }}}{99} \right| \right|\right) \\& \overset{\left[c\right]}= \dfrac{1}{3}\left(2\sin \dfrac{50{\mathrm \pi} }{99} - 2\sin \dfrac{148{\mathrm \pi} }{99}\right) > 1,\end{split}\]（推导中用到：[b][c]）故 ${I_2} < {I_1} < {I_3}$．
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