For every real number x,let f(x) = frac{x}{1! | vedclass

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(B) Given,$f(x) = \sum_{n=1}^{\infty} \frac{2^n - 1}{n!} x^n$. We can rewrite this as $f(x) = \sum_{n=1}^{\infty} \frac{(2x)^n}{n!} - \sum_{n=1}^{\inf

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exactly one real solution

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(B) Given,$f(x) = \sum_{n=1}^{\infty} \frac{2^n - 1}{n!} x^n$. We can rewrite this as $f(x) = \sum_{n=1}^{\infty} \frac{(2x)^n}{n!} - \sum_{n=1}^{\infty} \frac{x^n}{n!}$. Since $e^y = 1 + \sum_{n=1}^{\infty} \frac{y^n}{n!}$,we have $\sum_{n=1}^{\infty} \frac{y^n}{n!} = e^y - 1$. Substituting this,we get $f(x) = (e^{2x} - 1) - (e^x - 1) = e^{2x} - e^x$. Setting $f(x) = 0$,we have $e^{2x} - e^x = 0$,which implies $e^x(e^x - 1) = 0$. Since $e^x > 0$ for all real $x$,we must have $e^x - 1 = 0$,which gives $e^x = 1$. Thus,$x = 0$ is the only real solution.

For every real number $x$,let $f(x) = \frac{x}{1!} + \frac{3}{2!} x^2 + \frac{7}{3!} x^3 + \frac{15}{4!} x^4 + \dots$. Then the equation $f(x) = 0$ has

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