Let f : R to R be defined as f(x) = e^{x^2} + | vedclass

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(D) Given the function $f(x) = e^{x^2} + \cos x$. Since $f(-x) = e^{(-x)^2} + \cos(-x) = e^{x^2} + \cos x = f(x)$,the function is an even function. Fo

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many-one into

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(D) Given the function $f(x) = e^{x^2} + \cos x$. Since $f(-x) = e^{(-x)^2} + \cos(-x) = e^{x^2} + \cos x = f(x)$,the function is an even function. For any even function defined on $R$,$f(x) = f(-x)$ for all $x 
eq 0$,which implies the function is many-one. Next,we check the range of the function. As $x 	o \infty$,$f(x) 	o \infty$. At $x = 0$,$f(0) = e^0 + \cos(0) = 1 + 1 = 2$. Since $e^{x^2} \ge 1$ and $\cos x \ge -1$,the minimum value of $f(x)$ is $2$ at $x=0$. Because the range of $f(x)$ is $[2, \infty)$,which is a proper subset of the codomain $R$,the function is into. Therefore,$f$ is many-one into.

Let $f : R \to R$ be defined as $f(x) = e^{x^2} + \cos x$. Then $f$ is:

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