The total number of distinct x ∈ [0, 1] for which, ∫ _ 0 ^ x t ^ 2...

Let $f\left(x\right) = \left({∫}_{0}^{x}\dfrac{{t}^{2}}{1 + {t}^{4}}dt\right) - 2x + 1$
$⇒$ ${f}^{′}$\left(x\right) = \dfrac{{x}^{2}}{1 + {x}^{4}} - 2$<br />As$\dfrac{1 + {x}^{4}}{{x}^{2}} \ge 2 ⇒ \dfrac{{x}^{2}}{1 + {x}^{4}} \le \dfrac{1}{2}$<br />$⇒ {f}^{′}\left(x\right) \le - \dfrac{3}{2}$<br />$⇒ f\left(x\right)$is continuous and decreasing<br />$f\left(0\right) = 1$and$f\left(1\right) = \left({∫}_{0}^{1}\dfrac{{t}^{2}}{1 + {t}^{4}}dt\right) - 1 \le \dfrac{-1}{2}$<br />By intermediate value theorem (IVT),$f\left(x\right) = 0$possesses exactly one solution in [0, 1]. As$f ( 0 ) f ( 1 ) < 0