Maths-
General
Easy

Question

If f(x) satisfies the relation 2f(x) +f(1-x) = x2 for all real x , then f(x) is

  1. fraction numerator 2 x minus 1 plus x squared over denominator 6 end fraction
  2. fraction numerator 2 x minus 1 plus x squared over denominator 3 end fraction
  3. fraction numerator 4 x minus 1 plus x squared over denominator 6 end fraction
  4. fraction numerator 4 x minus 1 plus x squared over denominator 3 end fraction

The correct answer is: fraction numerator 2 x minus 1 plus x squared over denominator 3 end fraction


    We have given that

    2f(x) +f(1-x) = x2     - - -- - - - - -(i)
    We have to find the value of f(x)
    By replacing x by (1-x) in equation (i)we get,

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

    2f(1-x) + f(x) = 1 + x2 – 2x         - - - - - -(ii)
    Multiplying the equation(i) by 2 we get,

    4f(x) + 2f(1-x) = 2x2                - - - - - - (iii)
    Subtracting equation (ii) from (iii)

    3f(x) = x2 + 2x -1
    So,

    f left parenthesis x right parenthesis equals fraction numerator x squared plus 2 x minus 1 over denominator 3 end fraction
    Therefore option (b) is correct.

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