Q43 / 160 · AP EAMCET 2011 · Mathematics
If $y=(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{2 n}\right)$, then $\left(\frac{d y}{d x}\right)_{x=0}$ is equal to
- A0
- B$\frac{1}{2}$
- Correct answerC1
- D2
Solution
Given
$\begin{aligned}
y & =\frac{(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{2 n}\right)}{(1-x)} \times(1-x) \\
& =\frac{\left(1-x^2\right)\left(1+x^2\right)\left(\left(1+x^4\right) \ldots\left(1+x^{2 n}\right)\right.}{1-x} \\
\Rightarrow y & =\frac{\left[1-\left(x^{2 n}\right)^2\right]}{1-x}=\frac{1-x^{4 n}}{1-x}
\end{aligned}$
$\begin{aligned} \frac{d y}{d x} & =\frac{(1-x)\left(-4 n x^{4 n-1}\right)-\left(1-x^{4 n}\right)(-1)}{(1-x)^2} \\ \left(\frac{d y}{d x}\right)_{x=0} & =\frac{1(-0)-(1-0)(-1)}{1^2} \\ & =\frac{1}{1}=1\end{aligned}$
Hence, the correct answer is option: 'C'.