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Differential Calculus

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7 Mar 2009 10:00:33 IST

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This problem appeared in Leningrad Mathematical Olympiad 1988

Let $f: \mathbb{R}\to\mathbb{R}$ be a continuous function satisfying $f(x)\ f(f(x))=1$ for all $x\in\mathbb{R}$ . If $f(1000)=999$ , find $f(500)$ .

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akshay A NEW BEGINNING...

Blazing goIITian

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7 Mar 2009 10:38:49 IST

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my ans 2/1000.

Ankit

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7 Mar 2009 10:44:18 IST

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edited.......................

first put x=1000

f(1000)* f(f(1000) )=1

999* f(999)=1

therefore

f(999)=1/999- (i)

Now we know that our function takes values 999 (at x=1000) and [ 1/999] (at x=999), and therefore, since it is given to be continuous, we can be sure that it takes all values between these two, and that includes x=500.

Hence, there exists a such that f(a)=500. Substituting x=a in the equation (i)

f(500)= 1/ 500

second method

from f(x)*f (f(x)) =1

Now we know that f is continoous, so it takes all possible values from 1/999 to 999.

Also 500

Hence f(500)=1/500

Anant Kumar

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7 Mar 2009 10:49:52 IST

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Akshay, would you care to explain your answer.

@Falling up, do you think that f(x)=1/x is contiunous throughout the real numbers (as is required by the problem)?

Ankit

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7 Mar 2009 10:55:21 IST

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I left this

since f is continuous it can have all values from 1/999 to 999

Anant Kumar

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7 Mar 2009 11:06:57 IST

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If it does take all values from 1/500 to 500, then what. And how does that help? Further, do you think that the given data f(1000)=500 is redundant? At least from your solution, it seems so.

Ankit

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7 Mar 2009 11:16:30 IST

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akshay A NEW BEGINNING...

Blazing goIITian

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7 Mar 2009 11:23:14 IST

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again getting same ans2/1000 = 1/500

given f(x) * f(f(x)) = 1

and f(1000) =999

hence f(1000) * f( f(1000)) = 1999* f(999) = 1

hence f(999) = 1/999

now again using given equation

we have

f(999) * f(f(999)) = 1

1/999 * f(1/999) = 1

hence we have f(1/999) = 999

hence we can conclude that f(x) = (1/x)

hence f(500) = 1/500

Sir,, plss correct if something wrong.( i am still giving it a try will post if will a find a more suitable method)

Hari Shankar

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7 Mar 2009 11:24:23 IST

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@fallingup thats brilliant!!!!!! just for my sake can you make one more post here so that I can rate you again .

Ankit Rana

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7 Mar 2009 11:26:16 IST

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given that f(x) . f[f(x] = 1

and f(1000) = 999

hence for x = 1000 we have f(1000) . f [ f(1000) ] = 1

i.e. f(1000) f(999) = 1

hence f (999) = 1/f(1000) = 1/999

now since f is a continous function

it must attain all values between 1/999 and 999 in the interval [999,1000] since f(999) = 1/999 and f(1000) = 999

hence for some x between 999 and 1000 the function will attain 500 such that f(x) = 500

hence we have for that x f(x) . f[ f(x)] = 500 f(500) = 1

hence f(500) = 1/500

here u need not know the function but u can definitely say that f(500) = 1/500 since we are using the property of continous functions.

akshay A NEW BEGINNING...

Blazing goIITian

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7 Mar 2009 11:26:52 IST

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am i right?

Hari Shankar

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7 Mar 2009 11:27:42 IST

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@akshay, f(x) = 1/x if x is an image under this function. What falling up has done is to prove that 500 is an image in this function using continuity of x and the intermediate value theorem.

Great thinking!

Ankit Rana

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7 Mar 2009 11:30:11 IST

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oops im too slow in typing in the editor the solution was already posted!!!

akshay A NEW BEGINNING...

Blazing goIITian

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7 Mar 2009 11:33:56 IST

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@ankit.. no matters... i know u have also solved it.so rating u.

Anant Kumar

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7 Mar 2009 11:47:48 IST

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That was (ultimately) what I was looking for: the use of intermediate value property of continuous functions. And Falling up, you did a nice job.

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