Let f ( x ) = 1 + e 1 / x 1 . If
P V ∫ − 1 1 d x d f ( x ) d x = λ + 1 + e 1 − e ,
where λ is some constant, what is the value of λ ?
Note : The " P V " before the integral indicates the Cauchy principal value .
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Nice problem and solution. I almost chose 0 before recognizing the discontinuity. This may be a dumb question, but why is the title "Smart Fish"?
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Problems, I believe, are like fish. You can keep catching a lot of them, but however good you think you are, there are fish smart enough to bait you.
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Hahaha. Yes, so true. :)
You should make this question sounds more rewarding in order to make people wanting to ensure for a correct strive.
You caught me, ya smart fish :) Nice problem!
Ahhhhh you got me! I love the problem, and this was a pretty smart fish.
I hate you! :D
It's a very fun problem and great solution! However, it is important to indicate that the problem is asking for the Cauchy principal value (otherwise, the integral is undefined). I've edited the problem statement to reflect this.
Nice title :)
Never met with this. But it was a mistake of 1 + e 1 - 1 − e 1 1 being found and thought happy that ignored a check required. Didn't expect a genuine trap but thought for catching careless workings or doing differentiation and integration. Rushed because a lack of luck for correct answer.
Now I remember and I shall be careful. The thing was I didn't agree with the question of two images of seeing a mirror for finding only one image as a fault. Nevertheless, this man scored very fast by answering only not so many questions.
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Note that the function and its derivative are discontinuous at x = 0 . This means that we are forced to evaluate our integral using improper bounds. i.e:
∫ − 1 1 d x d ( f ( x ) ) d x = ∫ − 1 0 d x d ( f ( x ) ) d x + ∫ 0 1 d x d ( f ( x ) ) d x
= x → 0 − lim [ f ( x ) − f ( − 1 ) ] + y → 0 + lim [ f ( 1 ) − f ( y ) ]
= 1 + 1 + e 1 − e
⇒ λ = 1
Note:
lim x → 0 − f ( x ) = 1 , lim y → 0 + f ( y ) = 0