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AP EAMCET2003MathematicsBinomial Theorem

If 0 < y < 2^ 1 / 3 and x (y^3-1 )=1 , then 2 x + 2 3 x^3 + 2 5 x^5 + is equal to:

Options

  1. A( y^3 2-y^3 )
  2. B( y^3 1-y^3 )
  3. C( 2 y^3 1-y^3 )
  4. D( y^3 1-2 y^3 )

Correct answer

A. ( y^3 2-y^3 )

Step-by-step solution

We have, x (y^3-1 )=1 x= 1 y^3-1 = 1 k (say) k= 1 x Then, 2 x + 2 3 x^3 + 2 5 x^5 + =2 k+ 2 3 k^3+ 2 5 k^5+ = 1+k 1-k = 1+y^3-1 1-y^3+1 = y^3 2-y^3

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