Question

Let $g:R\to R$ be a continuous function such that $g(x+1)=\frac{1}{3}g(x)$ for all $x\in R$ and let ${\alpha }_{\mathrm{n}}=\underset{0}{\overset{\mathrm{n}}{\int }}\mathrm{g}(\mathrm{x})\mathrm{dx}$ for all integer $\mathrm{n}⩾1$. Then $\underset{n\to \infty }{\lim }{\alpha }_n$

• A. Exist
• B. Does not exist
• C. is equal to $\frac{3}{2}\underset{0}{\overset{1}{\int }}\mathrm{g}(\mathrm{x})\mathrm{dx}$
• D. not equal to $\frac{3}{2}\underset{0}{\overset{1}{\int }}\mathrm{g}(\mathrm{x})\mathrm{dx}$

Answer 1

Answer: (A), (C)

$\mathrm{g}(\mathrm{x}+1)=\frac{1}{3}\mathrm{g}(\mathrm{x})$
$\mathrm{g}(\mathrm{x}+2)=\frac{1}{3}\mathrm{g}(\mathrm{x}+1)=\frac{1}{9}\mathrm{g}(\mathrm{x})$
Now,
$\underset{\mathrm{n}\to \infty }{\lim }{\alpha }_{\mathrm{n}}=\underset{\mathrm{n}\to \infty }{\lim }\left(\underset{0}{\overset{1}{\int }}\mathrm{g}(\mathrm{x})\mathrm{dx}+\underset{1}{\overset{2}{\int }}\mathrm{g}(\mathrm{x})\mathrm{dx}+\underset{2}{\overset{3}{\int }}\mathrm{g}(\mathrm{x})\mathrm{dx}+\dots \dots \right)$
$\underset{2}{\overset{3}{\int }}\mathrm{g}(\mathrm{x})\mathrm{dx};\mathrm{x}=\mathrm{t}+2;\underset{0}{\overset{1}{\int }}\mathrm{g}(\mathrm{t}+2)\mathrm{dt}=\frac{1}{9}{\int }_0^1\mathrm{g}(\mathrm{x})\mathrm{dx}$
$\text{ i.e. }\underset{\mathrm{n}\to \infty }{\lim }{\alpha }_{\mathrm{n}}=\underset{0}{\overset{1}{\int }}\mathrm{g}(\mathrm{x})\mathrm{dx}\left(1+\frac{1}{3}+\frac{1}{9}+\dots \infty \right)$
$=\frac{3}{2}\underset{0}{\overset{1}{\int }}\mathrm{g}(\mathrm{x})\mathrm{dx}$