Q45 / 90 · KVPY 2016 (SB/SX) · Mathematics
The remainder when the polynomial $1+x^{2}+x^{4}+x^{6}+\ldots+x^{22}$ is divided by $1+x+x^{2}+x^{3}+\ldots .+x^{11}$ is -
- A0
- B2
- C$1+x^{2}+x^{4}+\ldots+x^{10}$
- Correct answerD$2\left(1+x^{2}+x^{4}+\ldots+x^{10}\right)$
Solution
$$
\begin{array}{l}
P(x)=1+x^{2}+x^{4}+x^{6}+\ldots \ldots+x^{22}=\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{4}+x^{8}\right)\left(1-x^{4}+x^{8}\right) \\
Q(x)=1+x+x^{2}+x^{3}+\ldots \ldots+x^{11}=(1+x)\left(1+x^{2}\right)\left(1+x^{4}+x^{8}\right) \\
\frac{P(x)}{Q(x)}=\frac{\left(1+x^{4}\right)\left(1-x^{4}+x^{8}\right)}{(1+x)}=\frac{1-x^{4}+x^{8}+x^{4}-x^{8}+x^{12}}{1+x}=\frac{1+x^{12}}{1+x}
\end{array}
$$
Remainder. When $\left(1+x^{12}\right)$ is divided by $(1+x)$ is $=2$
Now remainder $\left.\mathrm{P}_{\Delta} \mathrm{x}\right)$ divided by $\mathrm{Q}(\mathrm{x})$
$$
\begin{array}{l}
=2\left(1+x^{2}\right)\left(1+x^{4}+x^{8}\right) \\
=2\left(1+x^{2}+\ldots \ldots+x^{10}\right)
\end{array}
$$
Hence, the correct answer is option: 'D'.