A) 0
B) 1
C) 2
D) 3
Correct Answer: B
Solution :
| Let \[f(x)=\int_{0}^{x}{\frac{{{t}^{2}}}{1+{{t}^{4}}}dt\,-2x+1}\] |
| \[\Rightarrow \]\[f'(x)=\frac{{{x}^{2}}}{1+{{x}^{4}}}-2<0\forall x\in [0,1]\] |
| \[\therefore \]f is decreasing on [0,1] |
| Also \[f(0)=1\] |
| and \[f(1)=\int_{0}^{1}{\frac{{{t}^{2}}}{1+{{t}^{4}}}}dt-1\] |
| For \[0\le t\le 1\Rightarrow 0\le \frac{{{t}^{2}}}{1+{{t}^{4}}}<\frac{1}{2}\] |
| \[\therefore \int_{0}^{1}{\frac{{{t}^{2}}}{1+{{t}^{4}}}dt<\frac{1}{2}\Rightarrow f(1)<0}\] |
| \[\therefore \]\[f(x)\]crosses \[x-\operatorname{axis}\]exactly once in \[[0,1]\] |
| \[\therefore \]\[f(x)=0\]has exactly one root in \[[0,1]\] |
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