Smart Fish

Calculus Level 4

Let f ( x ) = 1 1 + e 1 / x \displaystyle f(x) = \frac 1{1+e^{1/x}} . If

P V 1 1 d d x f ( x ) d x = λ + 1 e 1 + e , PV\ \int_{-1}^{1} \frac{d}{dx} f(x) \ dx = \lambda + \frac{1-e}{1+e},

where λ \lambda is some constant, what is the value of λ ? \lambda?

Note : The " P V PV " before the integral indicates the Cauchy principal value .

0 1 e \frac {1} {e} 1 2 e e None of the above

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1 solution

Note that the function and its derivative are discontinuous at x = 0 x=0 . This means that we are forced to evaluate our integral using improper bounds. i.e:

1 1 d ( f ( x ) ) d x d x = 1 0 d ( f ( x ) ) d x d x + 0 1 d ( f ( x ) ) d x d x \int _{ -1 }^{ 1 }{ \frac { d(f\left( x \right) ) }{ dx } } dx=\int _{ -1 }^{ 0 }{ \frac { d(f\left( x \right) ) }{ dx } } dx+\int _{ 0 }^{ 1 }{ \frac { d(f\left( x \right) ) }{ dx } } dx

= lim x 0 [ f ( x ) f ( 1 ) ] + lim y 0 + [ f ( 1 ) f ( y ) ] =\lim _{ x\rightarrow { 0 }^{ - } }{ [f(x)-f(-1)] } +\quad \lim _{ y\rightarrow { 0 }^{ + } }{ [f(1)-f(y)] }

= 1 + 1 e 1 + e =1+\frac { 1-e }{ 1+e }

λ = 1 \Rightarrow \boxed{\lambda=1}

Note:

lim x 0 f ( x ) = 1 , lim y 0 + f ( y ) = 0 \lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =1,\lim _{ y\rightarrow { 0 }^{ + } }{ f(y) } =0

Nice problem and solution. I almost chose 0 0 before recognizing the discontinuity. This may be a dumb question, but why is the title "Smart Fish"?

Brian Charlesworth - 6 years, 1 month ago

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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.

Raghav Vaidyanathan - 6 years, 1 month ago

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Hahaha. Yes, so true. :)

Brian Charlesworth - 6 years, 1 month ago

You should make this question sounds more rewarding in order to make people wanting to ensure for a correct strive.

Lu Chee Ket - 5 years, 7 months ago

You caught me, ya smart fish :) Nice problem!

Jake Lai - 6 years, 1 month ago

Ahhhhh you got me! I love the problem, and this was a pretty smart fish.

Seth Lovelace - 6 years, 1 month ago

I hate you! :D

Vincent Miller Moral - 6 years, 1 month ago

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.

Andrew Hayes Staff - 4 years, 4 months ago

Nice title :)

Deepak Kumar - 6 years, 1 month ago

Never met with this. But it was a mistake of 1 1 + e \displaystyle \frac{1}{1 + e} - 1 1 1 e \displaystyle \frac {1}{1 - \frac {1}{e}} 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.

Lu Chee Ket - 5 years, 7 months ago

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