A) 1
B) \[\frac{1}{2}\]
C) \[-\frac{1}{2}\]
D) 0
Correct Answer: A
Solution :
| \[f\,(x)=\int\limits_{1}^{x}{\frac{x\,f\,(x)+1}{{{x}^{2}}}}\,\,dx\] |
| \[f\,(x)=\frac{x\,f\,(x)+1}{{{x}^{2}}}=\frac{1}{x}\,f\,(x)+\frac{1}{{{x}^{2}}}\] |
| i.e. \[f'\,(x)-\frac{1}{x}f\,(x)=\frac{1}{{{x}^{2}}}\] |
| Which is linear differential equation on solving it, we get |
| \[\therefore \] \[\frac{1}{x}f\,(x)=\int{\frac{1}{{{x}^{3}}}}\,\,dx+c=-\frac{1}{2{{x}^{2}}}+c\] |
| i.e. \[f\,(x)=-\frac{1}{2x}+cx\] |
| \[\frac{3}{4}=f\,(2)=-\frac{1}{4}+2c\] \[\Rightarrow \] \[c=\frac{1}{2}\] |
| \[f'\,(x)=-\frac{1}{2{{x}^{2}}}+c\] |
| \[\therefore \] \[f'\,(1)=\frac{1}{2}+\frac{1}{2}=1\] |
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