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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
$0$
$2$
$1+x^2+x^4+\ldots+x^{10}$
$2\left(1+x^2+x^4+\ldots+x^{10}\right)$
Solution
(d)
Let,$P(x)=1+x^2+x^4+x^6+\ldots+x^{22}$
$Q(x)=1+x+x^2+x^3+\ldots+x^{11}$
$P(x)=\frac{x^{24}-1}{x^2-1}$
$\left[\because a+a r+a r^2+\ldots+a r^{n-1}=\frac{a\left(r^n-1\right)}{r-1}\right]$
Similarly, $Q(x)=\frac{x^{12}-1}{x-1}$
$\therefore \quad \frac{P(x)}{Q(x)}=\left(\frac{x^{24}-1}{x^2-1}\right)\left(\frac{x-1}{x^{12}-1}\right)=\frac{x^{12}+1}{x+1}$
Now, $\left(1+x^{12}\right)$ is divided by $(x+1)$, then by remainder theorem remainder is $2$
Now, $P(x)$ is divided by $Q(x)$, then
remainder $=2\left(1+x^2\right)\left(1+x^4+x^8\right)$
$=2\left(1+x^2+x^4+\ldots+x^{10}\right)$
