Question

If $y=(1+x)(1+x^2)(1+x^4)\dots \dots (1+x^{2^n})$, then
​$\frac{dy}{dx}$ at x=0 is

• A. -1
• B. 1
• C. none of these

Answer 1

The correct option is B 1

Since, $y=(1+x)(1+x^2)(1+x^4)\dots \dots (1+x^{2^n}),(1-x)y=(1-x^2)(1+x^2)(1+x^4)\dots \dots (1+x^{2^n})$
$=(1-x^4)(1+x^4)\dots (1+x^{2^n})\dots ldots=(1-x^{2^n})(1+x^{2^n}=1-x^{2^{n+1}}$
$\therefore y=\frac{1-x^{2^{n+1}}}{(1-x)}$
$\because \frac{dy}{dx}=\frac{(1-x)(-2^{n+1}.x^{2^{n+1}})-(-1-2^{n+1})(-1)}{(1-x{)}^2}$
$\therefore \frac{dy}{dx}{|}_{x=0}=\frac{(1-x)(-2^{n+1}.0)-(1-0)(-1)}{(1-x{)}^2}$
$=1$