Functions 1

Algebra Level 3

Let f ( x ) + f ( 2 x ) + f ( 2 − x ) + f ( 1 + x ) = x f\left( x \right) +f\left( 2x \right) +f\left( 2-x \right) +f\left( 1+x \right) =x for all x ∈ R x\in \mathbb R .

Find the value of f ( 0 ) f(0) .


The answer is -0.25.

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Utkarsh Bansal by Utkarsh Bansal , 23, India

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3 solutions

Tarun Singh
Mar 18, 2015

p u t x = 0 f ( 0 ) + f ( 0 ) + f ( 2 ) + f ( 1 ) = 0 O R , − 2 f ( 0 ) = f ( 1 ) + f ( 2 ) p u t x = 1 f ( 1 ) + f ( 2 ) + f ( 1 ) + f ( 2 ) = 1 2 { f ( 1 ) + f ( 2 ) } = 1 2 { − 2 ( f ( 0 ) } = 1 f ( 0 ) = − 0.25 put\quad x=0\\ f\left( 0 \right) +f\left( 0 \right) +f\left( 2 \right) +f\left( 1 \right) =0\\ OR,\quad -2f(0)=f(1)+f(2)\\ put\quad x=1\\ f\left( 1 \right) +f\left( 2 \right) +f\left( 1 \right) +f\left( 2 \right) =1\\ 2\{ f(1)+f(2)\} =1\\ 2\{ -2(f(0)\} =1\\ f(0)=-0.25

16 Helpful 0 Interesting 0 Brilliant 0 Confused

@Tarun Singh Thanks for the solution.

Utkarsh Bansal - 6 years, 2 months ago

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easy!!!!!!!!!!!!!!!........................ UB

kritarth lohomi - 5 years, 10 months ago

Ok thank you very much

Eru Yuil - 5 years, 1 month ago

I entered the same answer it didn't accepted :(

Rishabh Sethi - 5 years, 10 months ago
Hero Miles
Mar 18, 2015

Let x = 0 x = 0

Then f ( 0 ) + f ( 0 ) + f ( 1 ) + f ( 2 ) = 0 f(0) + f(0) + f(1) + f(2) = 0

And 2 f ( 0 ) + f ( 1 ) + f ( 2 ) = 0 2f(0) + f(1) + f(2) = 0

And f ( 1 ) + f ( 2 ) = − 2 f ( 0 ) f(1) + f(2) = -2f(0)

Let x = 1 x = 1

Then f ( 1 ) + f ( 1 ) + f ( 2 ) + f ( 2 ) = 1 f(1) + f(1) + f(2) + f(2) = 1

And 2 f ( 1 ) + 2 f ( 2 ) = 1 2f(1) + 2f(2) = 1

And f ( 1 ) + f ( 2 ) = 1 2 f(1) + f(2) = \dfrac{1}{2}

Notice that f ( 1 ) + f ( 2 ) = − 2 f ( 0 ) = 1 2 f(1) + f(2) = -2f(0) = \dfrac{1}{2}

Set − 2 f ( 0 ) = 1 2 -2f(0) = \dfrac{1}{2}

Then f ( 0 ) = − 1 4 f(0) = -\dfrac{1}{4}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

@Bruce Preston Nice solution , upvoted

Utkarsh Bansal - 6 years, 2 months ago
Ahmed Wahid
May 4, 2015

f (0)=-0.25

0 Helpful 0 Interesting 0 Brilliant 0 Confused

But how? You must show!

Md Zuhair - 4 years, 3 months ago

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