A) 4
B) 3
C) 2
D) 1
Correct Answer: B
Solution :
\[\because \,\,f(x)-f\left( \frac{x}{7} \right)=\frac{x}{7}\] \[\Rightarrow \] \[f\left( \frac{x}{7} \right)-f\left( \frac{x}{{{7}^{2}}} \right)=\frac{x}{{{7}^{2}}}\] \[\Rightarrow \] \[f\left( \frac{x}{{{7}^{2}}} \right)-f\left( \frac{x}{{{7}^{3}}} \right)=\frac{x}{{{7}^{3}}}\] \[f\left( \frac{x}{{{7}^{n-1}}} \right)-f\left( \frac{x}{{{7}^{n}}} \right)=\frac{x}{{{7}^{n}}}\] Adding, we get \[f(x)-f\left( \frac{x}{{{7}^{n}}} \right)=\frac{x}{7}\,\left( 1+\frac{1}{7}+\frac{1}{{{7}^{2}}}+.....+\frac{1}{{{7}^{n-1}}} \right)\]\[\Rightarrow \,f(x)-f\left( \frac{x}{{{7}^{n}}} \right)=\frac{x}{6}\left( 1-\frac{1}{{{7}^{n}}} \right)\] Taking limit \[n\to \infty \] \[f(x)-f(0)=\frac{x}{6}\] \[\Rightarrow \,f(x)=1+\frac{x}{6}\] \[\therefore \] Required area \[=\frac{1}{2}\times 6\times 1=3\]You need to login to perform this action.
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