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 xelement ofR. Then 2f(0)+3f(1) is equal to
A. 2
B. 0
C. -2
D. -4

Asked by subhajit.690 | 17th Dec, 2016, 12:57: AM

Expert Answer:

begin mathsize 16px style Given space that space straight f left parenthesis straight x right parenthesis plus open parentheses straight x plus 1 half close parentheses straight f left parenthesis 1 minus straight x right parenthesis equals 1
When space straight x equals 0 comma
straight f left parenthesis 0 right parenthesis plus open parentheses 0 plus 1 half close parentheses straight f left parenthesis 1 minus 0 right parenthesis equals 1
rightwards double arrow straight f left parenthesis 0 right parenthesis plus 1 half space straight f left parenthesis 1 right parenthesis equals 1 space space..... left parenthesis straight i right parenthesis
When space straight x equals 1 comma
straight f left parenthesis 1 right parenthesis plus open parentheses 1 plus 1 half close parentheses straight f left parenthesis 1 minus 1 right parenthesis equals 1
rightwards double arrow straight f left parenthesis 1 right parenthesis plus 3 over 2 space straight f left parenthesis 0 right parenthesis equals 1 space space space...... left parenthesis ii right parenthesis
Multiply space left parenthesis straight i right parenthesis space by space 3 over 2.
We space get comma space 3 over 2 straight f left parenthesis 0 right parenthesis plus 3 over 4 space straight f left parenthesis 1 right parenthesis equals 3 over 2 space.... left parenthesis iii right parenthesis
Subtracting space left parenthesis ii right parenthesis space from space left parenthesis iii right parenthesis comma space we space get
negative 1 fourth space straight f left parenthesis 1 right parenthesis equals 1 half rightwards double arrow straight f left parenthesis 1 right parenthesis equals negative 2
So comma space substituting space straight f left parenthesis 1 right parenthesis equals negative 2 space in space left parenthesis straight i right parenthesis comma space we space get space rightwards double arrow straight f left parenthesis 0 right parenthesis plus 1 half space left parenthesis negative 2 right parenthesis equals 1 space rightwards double arrow straight f left parenthesis 0 right parenthesis minus 1 equals 1 rightwards double arrow straight f left parenthesis 0 right parenthesis equals 2
So comma space 2 straight f left parenthesis 0 right parenthesis plus 3 straight f left parenthesis 1 right parenthesis equals 2 left parenthesis 2 right parenthesis plus 3 left parenthesis negative 2 right parenthesis equals 4 minus 6 equals negative 2
Hence comma space the space correct space option space is space left parenthesis straight C right parenthesis. end style

Answered by Rebecca Fernandes | 17th Dec, 2016, 08:35: PM

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