The total number of distinct x in [0, 1] for | vedclass

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(A) Let $f(x) = \int_0^x \frac{t^2}{1+t^4} dt - 2x + 1$ for $x \in [0, 1]$.Taking the derivative,$f'(x) = \frac{x^2}{1+x^4} - 2$.Since $x^4 + 1 \geq 2

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(A) Let $f(x) = \int_0^x \frac{t^2}{1+t^4} dt - 2x + 1$ for $x \in [0, 1]$.Taking the derivative,$f'(x) = \frac{x^2}{1+x^4} - 2$.Since $x^4 + 1 \geq 2x^2$ by $AM$-$GM$ inequality,we have $\frac{x^2}{1+x^4} \leq \frac{1}{2}$.Thus,$f'(x) = \frac{x^2}{1+x^4} - 2 \leq \frac{1}{2} - 2 = -\frac{3}{2} Since $f'(x) We evaluate the endpoints:$f(0) = \int_0^0 \frac{t^2}{1+t^4} dt - 2(0) + 1 = 1$.$f(1) = \int_0^1 \frac{t^2}{1+t^4} dt - 2(1) + 1 = \int_0^1 \frac{t^2}{1+t^4} dt - 1$.Since $\frac{t^2}{1+t^4} Since $f(0) > 0$ and $f(1) < 0$ and $f(x)$ is continuous and strictly decreasing,by the Intermediate Value Theorem,there exists exactly one root $x \in [0, 1]$ such that $f(x) = 0$.

The total number of distinct $x \in [0, 1]$ for which $\int_0^x \frac{t^2}{1+t^4} dt = 2x - 1$ is

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