The function f: R rightarrow R defined by f(x | vedclass

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(D) Step $1$: Check for one-one property. $f(x) = e^x + e^{-x}$. Since $f(-x) = e^{-x} + e^{-(-x)} = e^{-x} + e^x = f(x)$, the function is an even fun

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not bijective

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(D) Step $1$: Check for one-one property. $f(x) = e^x + e^{-x}$. Since $f(-x) = e^{-x} + e^{-(-x)} = e^{-x} + e^x = f(x)$, the function is an even function. For an even function, $f(x_1) = f(x_2)$ does not imply $x_1 = x_2$ (e.g., $f(1) = f(-1)$). Therefore, the function is not one-one. Step $2$: Check for onto property. We know that $e^x > 0$ for all $x \in R$. By the Arithmetic Mean-Geometric Mean inequality, $\frac{e^x + e^{-x}}{2} \geq \sqrt{e^x \cdot e^{-x}} = 1$, which implies $e^x + e^{-x} \geq 2$. Thus, the range of $f(x)$ is $[2, \infty)$. Since the codomain is $R$ and the range is $[2, \infty)$, the range $
eq$ codomain. Therefore, the function is not onto. Conclusion: Since the function is neither one-one nor onto, it is not bijective.

The function $f: R \rightarrow R$ defined by $f(x) = e^x + e^{-x}$ is

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