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66. The function [tex]f: \mathbb{R} \rightarrow \mathbb{R}[/tex] defined by [tex]f(x) = e^x + e^{-x}[/tex] is:

A. one-one
B. onto
C. bijective
D. not bijective

Answer :

To determine whether the function [tex]f: \mathbb{R} \rightarrow \mathbb{R}[/tex] defined by [tex]f(x) = e^x + e^{-x}[/tex] is one-one, onto, bijective, or not bijective, let's analyze its properties.

  1. One-One (Injective):

    A function is one-one if different inputs produce different outputs. To check if [tex]f[/tex] is one-one, we assume:

    [tex]f(x_1) = f(x_2)[/tex]

    This implies:

    [tex]e^{x_1} + e^{-x_1} = e^{x_2} + e^{-x_2}[/tex]

    Rearranging and considering symmetry and properties of the exponential function, the only solution over the reals where this is always true is [tex]x_1 = x_2[/tex]. Thus, [tex]f(x)[/tex] is one-one.

  2. Onto (Surjective):

    A function is onto if for every element [tex]y[/tex] in the codomain [tex]\mathbb{R}[/tex], there exists some [tex]x[/tex] in the domain such that [tex]f(x) = y[/tex]. However, [tex]e^x + e^{-x}[/tex] approaches infinity as [tex]x[/tex] approaches both [tex]+\infty[/tex] and [tex]-\infty[/tex]. The minimum value occurs when the derivative [tex]f'(x) = e^x - e^{-x}[/tex] is zero, i.e., at [tex]x = 0[/tex] with [tex]f(0) = 2[/tex]. Since [tex]f(x)[/tex] has a minimum of 2 and extends to infinity, it is not surjective as it cannot produce values less than 2.

  3. Bijective:

    A function is bijective if it is both one-one and onto. Since [tex]f(x)[/tex] is not onto, [tex]f(x)[/tex] is also not bijective.

Conclusion:

The correct choice is (D) not bijective.

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