यदि `y=(1+x)(1+x^(2))(1+x^(4)) . . .(1+x^(2n))`, तब `x=0` पर `(dy)/(dx)` का मान है

Options

A. `0`
B. `-1`
C. 1
D. इनमे से कोई नहीं

Answer

C

Explanation

`becausey=(1+x)(1+x^(2))`
`(1+x^(4)) . . . (1+x^(2n))`
`(1-x)y=(1-x^(2))(1+x^(2))`
`(1+x^(4)) . . .(1+x^(2n))`
`=(1-x^(4))(1+x^(4)) . . .(1+x^(2n))`
. . . . . . . . . .
`=(1-x^(2n))(1+x^(2n))=1-x^(2n+1)`
`:.y=(1-x^(2n+1))/((1-x))`
`rArr(dy)/(dx)=([(1-x)(-2^(n+2)*x^((2^(n+1-1)))-(1-x^(2^(n+1)))(-1))])/((1-x)^(2))`
`rArr" "((dy)/(dx))_(x=0)`
`=((1-0)(-2^(n+1)*0)-(1-0)(-1))/(1)=1`