A) \[\frac{2}{7}\]
B) \[\frac{1}{2}\]
C) \[2\]
D) \[\frac{7}{2}\]
Correct Answer: B
Solution :
\[3f(x)-2f\left( \frac{1}{x} \right)=x\] ... (i) Let\[\frac{1}{x}=y\], then\[3f\left( \frac{1}{y} \right)-2f(y)=\frac{1}{y}\] \[\Rightarrow \] \[-2f(y)+3f\left( \frac{1}{y} \right)=\frac{1}{y}\] \[\Rightarrow \] \[-2f(x)+3f\left( \frac{1}{x} \right)=\frac{1}{x}\] ? (ii) On multiplying Eq. (i) by 3 and Eq. (ii) by 2 and then adding them, we get \[9f(x)-6f\left( \frac{1}{x} \right)-4f(x)+6f\left( \frac{1}{x} \right)=3x+\frac{2}{x}\] \[\Rightarrow \] \[5f(x)=3x+\frac{2}{x}\] \[\Rightarrow \] \[f(x)=\frac{1}{5}\left[ 3x+\frac{2}{x} \right]\] \[\therefore \] \[f'(x)=\frac{1}{5}\left[ 3-\frac{2}{{{x}^{2}}} \right]\] Then, \[f'(2)=\frac{1}{5}\left[ 3-\frac{2}{4} \right]=\frac{1}{2}\]
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