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$Consider\quad f\left( x \right) =\quad \int _{ 0 }^{ x }{ \frac { { t }^{ 2 } }{ 1+{ t }^{ 4 } } dx } -2x+1\\ First\quad Notice\quad that\quad \quad \frac { { x }^{ 2 } }{ 1+{ x }^{ 4 } } =\frac { 1 }{ { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } } \quad and\quad Hence\quad by\quad using\quad AM-GM\\ we\quad can\quad say\quad { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \ge 2\quad or\quad \frac { { x }^{ 2 } }{ 1+{ x }^{ 4 } } \le \frac { 1 }{ 2 } .................(i)\\ also\quad \int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 } }{ 1+{ x }^{ 4 } } dx } \le \int _{ 0 }^{ 1 }{ \frac { 1 }{ 2 } dx } \quad =>\int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 } }{ 1+{ x }^{ 4 } } dx } \le \frac { 1 }{ 2 } ..............(ii)\\ Now\quad f^{ ' }\left( x \right) =\frac { { x }^{ 2 } }{ 1+{ x }^{ 4 } } -2\quad hence\quad f^{ ' }\left( x \right) \le 0\quad from\quad (i)\\ So\quad f\left( x \right) \quad is\quad a\quad montonically\quad decreasing\quad function\quad \forall x\in \Re \\ Now\quad f\left( 0 \right) =1\quad and\quad f\left( 1 \right) <0\quad using\quad (ii)\\ Hence\quad there\quad is\quad only\quad one\quad zero\quad of\quad f\left( x \right) \quad in\quad \left[ 0,1 \right]$