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21 May, 21:37

Which function has an asymptote at x=pi/2?

a) y = sec x

b) y = cot x

c) y = csc x

d) y = cos x

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Answers (1)
  1. 22 May, 00:10
    0
    Step-by-step explanation:

    Note that y = sec x is the same as y = 1 / cos x, and that cos x is 0 at π/2. Division by zero is not defined. Thus, x cannot = π/2, and there is a vertical asymptote there.

    Thus, the first answer choice, y = sec x, is the correct one.
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