A) \[\frac{13}{6}\]
B) \[\frac{23}{18}\]
C) \[\frac{25}{9}\]
D) \[\frac{31}{18}\]
Correct Answer: D
Solution :
Differentiate w.r.t. t \[\underset{t\to x}{\mathop{lim}}\,\frac{2t\,f(x)-{{x}^{2}}f'(t)}{1}=1\] \[\Rightarrow \]\[2xf(x)-{{x}^{2}}f'(x)=1\] \[f'(x)=\frac{2x\,f(x)-1}{{{x}^{2}}}\] \[\frac{dy}{dx}=\frac{2y}{x}-\frac{1}{{{x}^{2}}}\] \[I.F.={{e}^{-\int_{{}}^{{}}{\frac{2}{x}dx}}}\] \[={{e}^{-2\ell nx}}=\frac{1}{{{x}^{2}}}\] \[y\left( \frac{1}{{{x}^{2}}} \right)=\int_{{}}^{{}}{-\frac{1}{{{x}^{4}}}dx}\] \[\frac{y}{{{x}^{2}}}=\frac{1}{3{{x}^{3}}}+c\] At \[x=1,y=1\]\[\Rightarrow \]\[c=\frac{2}{3}\] \[f(x)=\frac{1}{3x}+\frac{2{{x}^{2}}}{3}\] \[f\left( \frac{3}{2} \right)=\frac{31}{18}\]
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