A) \[0\]
B) \[1\]
C) \[\frac{1}{2}\]
D) \[-\frac{1}{2}\]
Correct Answer: A
Solution :
[a]We have, \[f(x)=\int_{0}^{x}{\sin ({{t}^{2}}-t+x)dt}\] |
\[\Rightarrow \]\[f'(x)=\sin ({{x}^{2}}-x+2)+\int_{0}^{x}{\cos ({{t}^{2}}-t+x)dt}\] |
\[\Rightarrow \]\[f''(x)=2x\cos {{x}^{2}}+\cos ({{x}^{2}})-\int_{0}^{x}{\sin ({{t}^{2}}-t+x)dt}\] |
\[\Rightarrow \]\[f''(x)+f(x)=(2x+1){{\cos }^{2}}x\] |
\[\Rightarrow \]\[f''\left( -\frac{1}{2} \right)+f\left( \frac{-1}{2} \right)=\left( 2\left( \frac{-1}{2} \right)+1 \right)\cos \left( \frac{1}{4} \right)=0\] |
You need to login to perform this action.
You will be redirected in
3 sec