A calculus problem by Archit Tripathi

Calculus Level 5

If f f is a differentiable function [other than ( f ( x ) = x ) (f(x) = x) ] which satisfies:

f ( x + f ( y ) + x f ( y ) ) = y + f ( x ) + y f ( x ) f(x+f(y)+xf(y)) = y+f(x)+yf(x) for all real x , y ≠ − 1 x, \ y \ne -1 , then find the value of 4034 [ 1 + f ( 2016 ) ] . 4034[1+f(2016)].


The answer is 2.

Archit Tripathi by Archit Tripathi , 23, India

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1 solution

Kushal Bose
Sep 13, 2016

In the above equation put x=0 and y=z we get:

f ( f ( z ) ) = z + ( z + 1 ) f ( 0 ) f(f(z))=z+(z+1)f(0)

In the question it is said that function is not defined for x = − 1 x=-1 .So it can be said that function has a denominator ( x + 1 ) (x+1)

So, the function may be of these types:

f ( x ) = a x x + 1 f(x)=\dfrac{ax}{x+1} or f ( x ) = a x − 1 x + 1 f(x)=a\frac{x-1}{x+1} or f ( x ) = a x + 1 f(x)=\dfrac{a}{x+1}

Putting these functions in the above equation we get the function as : f ( x ) = − x x + 1 f(x)=-\frac{x}{x+1}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I took composition and arrived at fof=f

satyam mani - 4 years, 9 months ago

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Can you post your solution

Kushal Bose - 4 years, 9 months ago

first of all, your assumption that the function has a denominator x+1 is not necessarily true as the function could be piecewise and still be continuous and diffrentiable. may be a trivial question, but why doesnt f(x) = x work here?? i could always define a function whose domain excludes one point. that way function doesnt exist at x=-1,neither does its derivative .

Rohith M.Athreya - 4 years, 8 months ago

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