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asked in Calculus by Boss (8.6k points)   | 58 views
is it option A?
Yes but how?

1 Answer

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I think this should be the logic.

Let f(x) = $\frac{\cos ^{m}x}{x^{n}}$

Value of x Value of n
<1 <1
<1 >=1
>=1 <1
>=1 >=1

 

Now, we focus for first 2 cases (i.e. x<1 since x has more values less than 1  rather than greter than 1)

 

Now, if we consider n>=1, then consider what will be instantaneous value of f(x).

e.g n=100 and x=0.02

$\therefore$ cosmx<1 but xn$\rightarrow$0

So, instantaneous value is very high and thus, integration of f(x) will never converge.

 

Now, if we consider n<1, then consider what will be instantaneous value of f(x).

e.g  n=0.9  x=0.02

$\therefore$  instantaneous value of f(x) will be finite. (just try taking an example)

So, this will converge.

 

So, my answer was just a calculated guess. I dont know if there is any solution by using the series expansion of cos(x).

 

 

answered by Veteran (15k points)  

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