A) 1
B) 2
C) 3
D) 4
Correct Answer: A
Solution :
[a]Let \[f(x)=\int\limits_{0}^{x}{\frac{{{t}^{2}}}{1+{{t}^{4}}}dt-2x+1}\] |
\[\Rightarrow \]\[f'(x)=\frac{{{x}^{2}}}{1+{{x}^{4}}}-2\] as \[\frac{1+{{x}^{4}}}{{{x}^{2}}}\ge 2\] |
\[\therefore \]\[\frac{{{x}^{2}}}{1+{{x}^{4}}}\le \frac{1}{2}\] |
\[\Rightarrow \]\[f'(x)\le -\frac{3}{2}\] |
\[\Rightarrow \]\[f\,(x)\]is continuous and decreasing. |
\[\therefore \,\,f(0)=1\]and \[f(1)=\int\limits_{0}^{1}{\frac{{{t}^{2}}}{1+{{t}^{4}}}dt-2\le -\frac{3}{2}}\] |
By intermediate value theorem \[f\,(x)=0\]possesses exactly one solution in [0, 1]. |
You need to login to perform this action.
You will be redirected in
3 sec