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Question: If \(y = (1 + x)(1 + x^{2})(1 + x^{4}).......(1 + x^{2^{n}})\) then \(\frac{dy}{dx}\) at \(x = 0\) i...

If y=(1+x)(1+x2)(1+x4).......(1+x2n)y = (1 + x)(1 + x^{2})(1 + x^{4}).......(1 + x^{2^{n}}) then dydx\frac{dy}{dx} at x=0x = 0 is

A

1

B

– 1

C

0

D

None of these

Answer

1

Explanation

Solution

y=(1−x)(1+x)(1+x2).....(1+x2n)1−x=1−x2n+11−xy = \frac{(1 - x)(1 + x)(1 + x^{2}).....(1 + x^{2^{n}})}{1 - x} = \frac{1 - x^{2^{n + 1}}}{1 - x}

∴\therefore dydx=−2n+1.x2n+1−1(1−x)+1−x2n+1(1−x)2\frac{dy}{dx} = \frac{- 2^{n + 1}.x^{2^{n + 1} - 1}(1 - x) + 1 - x^{2^{n + 1}}}{(1 - x)^{2}}

∴\therefore At x=0x = 0, dydx=−2n+10.1+1−012=1\frac{dy}{dx} = \frac{- 2^{n + 1}0.1 + 1 - 0}{1^{2}} = 1.