If 0<x<1 and y=12x2+23x3+34x4+⋯, then the value of e1+y at x=12 is
A
2e
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B
12e2
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C
2e2
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D
12√e
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Solution
The correct option is B12e2 y=12x2+23x3+34x4+⋯ =(1−12)x2+(1−13)x3+(1−14)x4+⋯ =(x2+x3+x4+⋯)+(−x22−x33−x44−⋯) =x21−x+x+(−x1−x22−x33−x44−⋯) y=x1−x+ln(1−x) y+1=11−x+ln(1−x) ey+1=e11−x+ln(1−x)=e11−x×eln(1−x)=(1−x)e11−x ∴ at x=12,y=12e2