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If 0 < x < 1 and $$y = {1 \over 2}{x^2} + {2 \over 3}{x^3} + {3 \over 4}{x^4} + ....$$, then the value of e1 + y at $$x = {1 \over 2}$$ is :

JEE · Math · previous-year question

  1. A.$${1 \over 2}{e^2}$$correct
  2. B.2e
  3. C.$${1 \over 2}\sqrt e $$
  4. D.2e2

Answer

A. $${1 \over 2}{e^2}$$

Explanation

$$y = \left( {1 - {1 \over 2}} \right){x^2} + \left( {1 - {1 \over 3}} \right){x^3} + ....$$ $$ = ({x^2} + {x^3} + {x^4} + ......) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + ....} \right)$$ $$ = {{{x^2}} \over {1 - x}} + x - \left( {x + {{{x^2}} \over 2} + {{{x^3}} \over 3} + ....} \right)$$ $$ = {x \over {1 - x}} + \ln (1 - x)$$ $$x = {1 \over 2} \Rightarrow y = 1 - \ln 2$$ $${e^{1 + y}} = {e^{1 + 1 - \ln 2}}$$ $$ = {e^{2 - \ln 2}} = {{{e^2}} \over 2}$$

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