Which of the following functions have a vertical asymptote for values of theta such that cos theta=1

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Heather W.

Algebra

6 months, 1 week ago

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Related Question

Which of the following functions has two horizontal asymptotes (A) $\mathrm{y}=\frac{|\mathrm{x}|}{\mathrm{x}+1}$ (B) $\mathrm{y}=\frac{2 \mathrm{x}}{\sqrt{\mathrm{x}^{2}+1}}$ (C) $y=\frac{\sin x}{x^{2}+1}$ (D) $y=\cot ^{-1}(2 x+1)$

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Video Transcript

in discussion. We need to remind that which of the following has two horizontal assembles. The given are the four options. So we know for the horizontal symptoms. Ah The limit that is limit for the functionary limit extends to plus minus in finite for the function F. Of X is equal to eight then Y equal to a. Is a. I seem to if there are two different values then it will have are two different horizontal assim toads. So what you call to asia horizontal assembled. So let us check for the given four options. Which one has the horizontal the same toad. Now for the option A The given limit is as limit extends to ah in fa night that we will calculate for plus infinity and minus infinity. So for plus infinity the function given as it is um Margiela's X over X plus one For which we will guide it as uh as it is plus and 30. So this will be in the numerator. This will become excellently and therefore this will become one over ah one, 1/1 plus one by infinite. That will become one only. So this will be one And that is the limit for negative infinite. That is extends to minus infinity. This will become more or less eggs over X plus one as it is negative. So model S X will result in two minus X. Therefore this will become minus 1/1 plus um one by infrared. So that will become -1. So we got the two values that is vehicle savannah and vehicles to -1 so that it will have to swim towards the option is correct. Option is correct. Now for the second option there is option B. We have the function that is um two x over square root X squared plus one. And therefore we will check for the two limits that is plus infinity and minus infinity. So limit Exchange two plus in finite over that is actually square over Scare Root Access Square Plus one on dividing by access card on the numerator and denominator boat. Um So this is two x not x square here. So two X now dividing by X on the numerator and denominator we will get it as limit. Exchange two in Finite. So you plus in finite this will become too over scared route here. This will become divided by X. In the numerator. Inside this square that will become X square. So x squared y X square will be one plus one by X square. Now substituting this limit we will get it as two. Similarly for the minus infinite, this will become and it extends to minus infinity. This will become That is -2. So as dividing by minus X in the numerator and denominator that will become So two x over X square root X squared plus one. This limit will be called to minus two. So this will also have to ascend towards that is to horizontal A symptom. So option B is also correct. Now we will check for the function in the option C. Now it is given as a limit, that is sine X over X squared plus one. So we will check for the limit extends to plus infinity for sin X over X squared plus one. Now on substitution we will get um sign infinite over infinite. We know that some infinite will will be ranging between zero and one, but in the denominator is infinite, so this will become um zero, Similarly for minus infrared, this will become zero extends to minus infinity, syntax over X squared plus one. This will also be zero as something up over infinite will be you know now therefore in the options, see there are no there are no two sim towards there is horizontal symptom a symptom So uh option C is incorrect now in the option D and the option D the given function is is contingent universe two X plus one. So limit X turns to plus infinity this will become contingent universe um two x plus one on substituting plus infinity at the place of X will get contingent universe infinite plus in fact that will be at zero, we know that Contingent Universe, zero is infrared. So this will become zero for the negative um there is limited extents to minus infinity contingent universe two x plus one on substitution access minus. In fact this will become contingent universe minus infrared. We know that contingent is negative in that is for pipe as minus infinite for pie. So this will become bye. So this also has two values, so there will be two horizontal ascent roads. Therefore, we can write that option D is also correct. Option D is also correct. Now, we from the our solution, we conclude that option A B and D are correct. So A B and D are correct. I hope all of you got discussion. Thank you.

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