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Q. Let $g: R \rightarrow R$ be a continuous function such that $g(x+1)=\frac{1}{3} g(x)$ for all $x \in R$ and let $\alpha_{\mathrm{n}}=\int\limits_0^{\mathrm{n}} \mathrm{g}(\mathrm{x}) \mathrm{dx}$ for all integer $\mathrm{n} \geqslant 1$. Then $\displaystyle\lim _{n \rightarrow \infty} \alpha_n$

JEE AdvancedJEE Advanced 2019

Solution:

$ \mathrm{g}(\mathrm{x}+1)=\frac{1}{3} \mathrm{~g}(\mathrm{x}) $
$\mathrm{g}(\mathrm{x}+2)=\frac{1}{3} \mathrm{~g}(\mathrm{x}+1)=\frac{1}{9} \mathrm{~g}(\mathrm{x})$
Now,
$ \displaystyle\lim _{\mathrm{n} \rightarrow \infty} \alpha_{\mathrm{n}}=\displaystyle\lim _{\mathrm{n} \rightarrow \infty}\left(\int\limits_0^1 \mathrm{~g}(\mathrm{x}) \mathrm{dx}+\int\limits_1^2 \mathrm{~g}(\mathrm{x}) \mathrm{dx}+\int\limits_2^3 \mathrm{~g}(\mathrm{x}) \mathrm{dx}+\ldots \ldots\right) $
$ \int\limits_2^3 \mathrm{~g}(\mathrm{x}) \mathrm{dx} ; \mathrm{x}=\mathrm{t}+2 ; \int\limits_0^1 \mathrm{~g}(\mathrm{t}+2) \mathrm{dt}=\frac{1}{9} \int_0^1 \mathrm{~g}(\mathrm{x}) \mathrm{dx} $
$\text { i.e. } \displaystyle\lim _{\mathrm{n} \rightarrow \infty} \alpha_{\mathrm{n}}=\int\limits_0^1 \mathrm{~g}(\mathrm{x}) \mathrm{dx}\left(1+\frac{1}{3}+\frac{1}{9}+\ldots \infty\right)$
$ =\frac{3}{2} \int\limits_0^1 \mathrm{~g}(\mathrm{x}) \mathrm{dx}$