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(D) For $x 
eq 0$,the function is given by the sum of a geometric series: $f(x) = |x|^\alpha \sum \limits_{n=0}^{\infty} \left(\frac{1}{1+x^2}ight)

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more than $4$ elements

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(D) For $x 
eq 0$,the function is given by the sum of a geometric series: $f(x) = |x|^\alpha \sum \limits_{n=0}^{\infty} \left(\frac{1}{1+x^2}ight)^n$. Since $|\frac{1}{1+x^2}| Thus,$f(x) = |x|^\alpha \cdot \frac{1+x^2}{x^2} = |x|^\alpha \cdot |x|^{-2} (1+x^2) = |x|^{\alpha-2} (1+x^2)$. For $f$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 0$. $\lim_{x 	o 0} |x|^{\alpha-2} (1+x^2) = 0$ holds if and only if $\alpha - 2 > 0$,which means $\alpha > 2$. The set of such real numbers $\alpha$ is the interval $(2, \infty)$. Since this interval contains infinitely many real numbers,the set has more than $4$ elements.

Let $R$ be the set of all real numbers and $\alpha \in R$ be positive. Define a function $f: R \rightarrow R$ by $f(0)=0$ and $f(x)=|x|^\alpha \sum \limits_{n=0}^{\infty}\left(1+x^2\right)^{-n}$,for $x \neq 0$. Then the set of real numbers $\alpha$ for which $f$ is continuous at $x = 0$ has

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