Functions in two variables.

Algebra Level 5

Find sum of solutions of f ( 123 ) f(123) for defined equation f ( x f ( x ) + f ( y ) ) = y + f ( x ) 2 , x , y ∈ R f\left( xf\left( x \right) +f\left( y \right) \right) =y+{ f\left( x \right) }^{ 2 },\quad x,y\in R .


The answer is 0.

Shivamani Patil by shivamani patil , 21, India

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1 solution

Shivamani Patil
Aug 29, 2014

Okay i will write the solution:

let f ( x ) = 0 f(x)=0 then we have ( f ( f ( y ) ) = y \boxed{(f(f(y))=y} .

Now plug x = f ( x ) x=f(x) in original equation

Then we get f ( f ( x ) f ( f ( x ) ) + f ( y ) = y + f ( f ( x ) ) 2 f(f(x)f(f(x))+f(y)=y+{ f(f(x)) }^{ 2 }

which equals

f ( f ( x ) x + f ( y ) ) = y + x 2 f(f(x)x+f(y))=y+{ x }^{ 2 } ....Equation 1

Now notice that LHS of original equation and LHS of equation 1 are equal

Therefore we have

y + x 2 = y + f ( x ) 2 y+{ x }^{ 2 }=y+{f(x) }^{ 2 }

x 2 = f ( x ) 2 { x }^{ 2 }={f(x) }^{ 2 }

f ( x ) = x f(x)=x or f ( x ) = − x f(x)=-x

Therefore

( f ( 123 ) = 123 \boxed{(f(123)=123} or ( f ( 123 ) = − 123 \boxed{(f(123)=-123}

Therefore sum of solutions is 123 − 123 = 0 123-123=0

2 Helpful 0 Interesting 0 Brilliant 0 Confused

@shivamani patil But , first you need to show that there exists a solution for f ( x ) = 0 f(x) = 0

Ankit Kumar Jain - 4 years, 3 months ago

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