A) exists and equals\[\frac{4}{7}.\]
B) exists and equals 4.
C) does not exist.
D) exists and equals 0.
Correct Answer: B
Solution :
\[f'(x)=7-\frac{3}{4}.\frac{f(x)}{x},x>0\] |
\[\therefore \] \[f'(x)+\frac{3}{4x}f(x)=7\] (Linear) |
\[f(x).{{e}^{\int{\frac{3}{4x}dx}}}=\int{7.{{e}^{\int{\frac{3}{4x}dx}}}}+c\] |
\[f(x).{{x}^{3/4}}=\int{7.{{x}^{3/4}}}+c\] |
\[=7\frac{{{x}^{7/4}}}{\frac{7}{4}}+c\] |
\[\therefore \] \[f(x)=4x+c{{x}^{-3/4}}\] |
\[\therefore \] \[f\left( \frac{1}{x} \right)=\frac{4}{x}+c{{x}^{3/4}}\] |
\[\therefore \] \[\underset{x\to {{0}^{+}}}{\mathop{Lt}}\,xf\left( \frac{1}{x} \right)=\underset{x\to {{0}^{+}}}{\mathop{Lt}}\,4+c{{x}^{7/4}}=4.\] |
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