KVPY SA Exam 2014 Question Paper

Question : 2 of 80

Marks: +1, -0

Let R be the set of all real numbers and let f be a function from R to R such that

f(x)+(x+1/2)f(1-x)=1 ,

for all x ∊ R . Then

2 f( 0 ) + 3 f( 1 )

is equal to

Solution:

\f(x)+(x+1/2)\f(1-x)=1

\f(1-x)+(1-x+1/2)

\f(1-(1-x))=1

\f(1-x)+(3/2-x ) \f ( x)=1

{1-\f(x)} /{x+1/2 }+(3/2-x ) \f ( x)=1

1 -\f(x) +(3/2x-x^2+3/4-x/2)

\f(x) =x+1/2

\f(x) (x-x^2-1/4)=x-1/2

\f(x) (4x - 4x^2- 1) = 4x - 2

\f(x) = { -2+4x} /{ 4x - 4x^2- 1 }

2\f(0) + 3\f(1) = 2({-2+0}/{0-0-1})

+3({-2+4}/{4-4-1})

=+4+{3(+2)}/{-1}=+4-6=-2

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