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Question

If y = (1+x)(1+x2)(1+x4)....(1+x2n), then dydx at x = 0 is


A

0

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B

-1

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C

1

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D

2

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Solution

The correct option is C

1


Since,y = (1+x)(1+x2)(1+x4)....(1+x2n)

(1x)y = (1x2)(1+x2)(1+x4)....(1+x2n)

= (1x4)(1+x4)...(1+x2n)

.......................................

= (1x2n)(1+x2n) = 1x2n+1

y = 1x2n+1(1x)

dydx = (1x)(2n+1.x2n(1x2n+1)(1)))(1x)2

dydxx=0 = (10)(2n+1.0)(10)(1)1

= 1


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