The remainder when the polynomial $1 + x^{2} + x^{4} + x^{6} + x . . . . . + x^{22}$ is divided by $1 + x + x^{2} + x^{3} + . . . . + x^{11}$ is:

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 + x^{2} + x^{4} + . . . + x^{10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 (1 + x^{2} + x^{4} + . . . + x^{10})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0$
$1 + x + x^{2} + x^{3} + . . . . . + x^{n} = \frac{1 - x^{n}}{1 + x^{n}}$
$1 + x + x^{2} + x^{4} + . . . . . + x^{2 n} = \frac{1 - x^{n}}{1 - x^{2}}$
Hence
$\Rightarrow 1 + x^{2} + x^{4} + . . . . . + x^{2 n} = \frac{1 - x^{11}}{1 - x^{2}} . . . (i i i)$
$1 + x + x^{2} + x^{3} + . . . . . + x^{11} = \frac{1 - x^{11}}{1 + x} . . . (i v)$
$\frac{(i i i)}{(i v)} = \frac{1 - x^{11}}{1 - x^{2}} \times \frac{1 - x}{1 + x^{11}} = \frac{1 - x}{1 - x^{2}} = \frac{1 - x}{(1 - x) (1 + x)} = \frac{1}{1 + x}$

Suggest Corrections

Similar questions

1 + x^{2} + x^{4} + x^{6} + \dots . + x^{22}

1 + x + x^{2} + x^{3} + \dots . + x^{11}

1 + x^{2} + x^{4} + x^{6} + . . . . + x^{22}

1 + x + x^{2} + x^{3} + . . . . + x^{11}

1 + x^{2} + x^{4} + x^{6} + \dots + x^{34}

1 + x + x^{2} + x^{3} + \dots + x^{17}

1 + x^{2} + x^{4} + x^{6} + . . . . + x^{34}

1 + x + x^{2} + x^{3} + . . . . + x^{17}

q (x)

r (x)

f (x)

g (x)

p (x) = x^{6} + x^{4} - x^{2} - 1

g (x) = x^{3} - x^{2} + x - 1

Join BYJU'S Learning Program