A) does not exist
B) exists and equals \[\frac{4}{7}\]
C) exists and equals 4
D) exists and equals 0
Correct Answer: C
Solution :
\[f'(x)=7-\frac{3}{4}\frac{f(x)}{x}\,(x\,\,>\,\,0)\] Given\[f(1)\ne 4\], \[\underset{x\to 0+}{\mathop{\lim }}\,x\,+\,\left( \frac{1}{x} \right)\,\,=\,\,7\] \[\Rightarrow \,\,\,\,\frac{dy}{dx}+\frac{3}{4}\frac{y}{x}\,\,=\,\,7\] (linear diff. equation) \[I.F.\,\,=\,\,{{e}^{\int{\frac{3}{4x}dx}}}\,=\,{{e}^{\frac{3}{4}\,\log \left| x \right|}}\,=\,{{x}^{3/4}}\] \[y{{x}^{3/4}}\,=\,\int{7.{{x}^{3/4}}dx}\] \[\Rightarrow \,\,\,\,y{{x}^{3/4}}\,=\,\,\,7\frac{{{x}^{7/4}}}{(7/4)}\,\,+\,\,C\] \[\Rightarrow \,\,\,f\left( x \right)=y=4x+C{{x}^{-}}{{^{3}}^{/}}^{4}\] \[\Rightarrow \,\,\,\,f\left( \frac{1}{x} \right)\,=\,\frac{4}{x}\,+\,C{{x}^{3/4}}\] \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\,xf\left( \frac{1}{x} \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,(4+C.{{x}^{7/4}})\,\,=\,\,4\]
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