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To determine which of these functions are even, let's recall the definition of an even function. A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex] for every [tex]\( x \)[/tex] in the domain of [tex]\( f \)[/tex].
Let’s examine each function step by step to see if it satisfies the condition of being even.
### A. [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex]
To determine if [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] is even, we need to check [tex]\( f(-\alpha) \)[/tex]:
[tex]\[ f(-\alpha) = 1 + \sec(-\alpha) \][/tex]
Since [tex]\(\sec(-\alpha) = \sec(\alpha)\)[/tex] (because the secant function is even), we have:
[tex]\[ f(-\alpha) = 1 + \sec(\alpha) = f(\alpha) \][/tex]
Thus, [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] is even.
### B. [tex]\( f(x) = x \cos(x) \)[/tex]
To determine if [tex]\( f(x) = x \cos(x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (-x) \cos(-x) \][/tex]
Since [tex]\(\cos(-x) = \cos(x)\)[/tex] (because the cosine function is even), we have:
[tex]\[ f(-x) = (-x) \cos(x) = -x \cos(x) \][/tex]
Since [tex]\( f(-x) = -x \cos(x) \neq f(x) = x \cos(x) \)[/tex], the function [tex]\( f(x) = x \cos(x) \)[/tex] is not even.
### C. [tex]\( f(x) = \csc(x^2) \)[/tex]
To determine if [tex]\( f(x) = \csc(x^2) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \csc((-x)^2) = \csc(x^2) \][/tex]
Since [tex]\( f(-x) = \csc(x^2) = f(x) \)[/tex], the function [tex]\( f(x) = \csc(x^2) \)[/tex] is even.
### D. [tex]\( f(x) = \sin(2x) \)[/tex]
To determine if [tex]\( f(x) = \sin(2x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \sin(2(-x)) = \sin(-2x) \][/tex]
Since [tex]\(\sin(-2x) = -\sin(2x)\)[/tex] (because the sine function is odd), we have:
[tex]\[ f(-x) = -\sin(2x) \neq f(x) = \sin(2x) \][/tex]
Thus, the function [tex]\( f(x) = \sin(2x) \)[/tex] is not even.
### E. [tex]\( f(x) = \cos(2x) \)[/tex]
To determine if [tex]\( f(x) = \cos(2x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \cos(2(-x)) = \cos(-2x) \][/tex]
Since [tex]\(\cos(-2x) = \cos(2x)\)[/tex] (because the cosine function is even), we have:
[tex]\[ f(-x) = \cos(2x) = f(x) \][/tex]
Thus, the function [tex]\( f(x) = \cos(2x) \)[/tex] is even.
### F. [tex]\( f(t) = \sec^2(t) - 1 \)[/tex]
To determine if [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] is even, we need to check [tex]\( f(-t) \)[/tex]:
[tex]\[ f(-t) = \sec^2(-t) - 1 \][/tex]
Since [tex]\(\sec(-t) = \sec(t)\)[/tex] (because the secant function is even), we have:
[tex]\[ \sec^2(-t) = \sec^2(t) \][/tex]
[tex]\[ f(-t) = \sec^2(t) - 1 = f(t) \][/tex]
Thus, the function [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] is even.
### Conclusion:
The even functions are:
- [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] (A)
- [tex]\( f(x) = \csc(x^2) \)[/tex] (C)
- [tex]\( f(x) = \cos(2x) \)[/tex] (E)
- [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] (F)
Therefore, the even functions are [tex]\(\boxed{A, C, E, F}\)[/tex].
Let’s examine each function step by step to see if it satisfies the condition of being even.
### A. [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex]
To determine if [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] is even, we need to check [tex]\( f(-\alpha) \)[/tex]:
[tex]\[ f(-\alpha) = 1 + \sec(-\alpha) \][/tex]
Since [tex]\(\sec(-\alpha) = \sec(\alpha)\)[/tex] (because the secant function is even), we have:
[tex]\[ f(-\alpha) = 1 + \sec(\alpha) = f(\alpha) \][/tex]
Thus, [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] is even.
### B. [tex]\( f(x) = x \cos(x) \)[/tex]
To determine if [tex]\( f(x) = x \cos(x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (-x) \cos(-x) \][/tex]
Since [tex]\(\cos(-x) = \cos(x)\)[/tex] (because the cosine function is even), we have:
[tex]\[ f(-x) = (-x) \cos(x) = -x \cos(x) \][/tex]
Since [tex]\( f(-x) = -x \cos(x) \neq f(x) = x \cos(x) \)[/tex], the function [tex]\( f(x) = x \cos(x) \)[/tex] is not even.
### C. [tex]\( f(x) = \csc(x^2) \)[/tex]
To determine if [tex]\( f(x) = \csc(x^2) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \csc((-x)^2) = \csc(x^2) \][/tex]
Since [tex]\( f(-x) = \csc(x^2) = f(x) \)[/tex], the function [tex]\( f(x) = \csc(x^2) \)[/tex] is even.
### D. [tex]\( f(x) = \sin(2x) \)[/tex]
To determine if [tex]\( f(x) = \sin(2x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \sin(2(-x)) = \sin(-2x) \][/tex]
Since [tex]\(\sin(-2x) = -\sin(2x)\)[/tex] (because the sine function is odd), we have:
[tex]\[ f(-x) = -\sin(2x) \neq f(x) = \sin(2x) \][/tex]
Thus, the function [tex]\( f(x) = \sin(2x) \)[/tex] is not even.
### E. [tex]\( f(x) = \cos(2x) \)[/tex]
To determine if [tex]\( f(x) = \cos(2x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \cos(2(-x)) = \cos(-2x) \][/tex]
Since [tex]\(\cos(-2x) = \cos(2x)\)[/tex] (because the cosine function is even), we have:
[tex]\[ f(-x) = \cos(2x) = f(x) \][/tex]
Thus, the function [tex]\( f(x) = \cos(2x) \)[/tex] is even.
### F. [tex]\( f(t) = \sec^2(t) - 1 \)[/tex]
To determine if [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] is even, we need to check [tex]\( f(-t) \)[/tex]:
[tex]\[ f(-t) = \sec^2(-t) - 1 \][/tex]
Since [tex]\(\sec(-t) = \sec(t)\)[/tex] (because the secant function is even), we have:
[tex]\[ \sec^2(-t) = \sec^2(t) \][/tex]
[tex]\[ f(-t) = \sec^2(t) - 1 = f(t) \][/tex]
Thus, the function [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] is even.
### Conclusion:
The even functions are:
- [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] (A)
- [tex]\( f(x) = \csc(x^2) \)[/tex] (C)
- [tex]\( f(x) = \cos(2x) \)[/tex] (E)
- [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] (F)
Therefore, the even functions are [tex]\(\boxed{A, C, E, F}\)[/tex].
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