question_answer

Let \[f(x)\] be a continuous function such that \[f(0)=1\] and \[f(x)-f\left( \frac{x}{7} \right)=\frac{x}{7}\forall x\in R,\] then area bounded by the curve \[y=f(x)\] and the co-ordinate axes is

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 \[=\frac{1}{2}\times 6\times 1=3\]