Question

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

• A. 0
• B. -1
• C. 1
• D. 2

Answer 1

The correct option is C

1

Since,$y$ = $(1+x)(1+x^2)(1+x^4)....(1+x^{2n})$

$(1-x)y$ = $(1-x^2)(1+x^2)(1+x^4)....(1+x^{2n})$

= $(1-x^4)(1+x^4)...(1+x^{2n})$

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

= $(1-x^{2n})(1+x^{2n})$ = $1-x^{2n+1}$

$\therefore y$ = $\frac{1-x^{2n+1}}{(1-x)}$

$\because \frac{dy}{dx}$ = $\frac{(1-x)(-2n+1.x^{2n}-(1-x^{2n+1}\left)\right(-1\left)\right))}{(1-x{)}^2}$

$\therefore \frac{dy}{dx}{|}_{x=0}$ = $\frac{(1-0)(-2^{n+1}.0)-(1-0)(-1)}{1}$

= 1