f(x) = \(\int_0^1\left|t x-t^2\right| d t, 0 \leq x \leq 1\) అయితే, f(x) యొక్క గరిష్ట విలువ

1
\(\frac{1}{3}+\frac{1}{3 \sqrt{2}}\)
2
\(\frac{1}{3}\)
3
\(\frac{1}{6}\)
4
\(\frac{1}{2}\)

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