Let Px be a polynomial of degree 2010- Suppose Pn-n1-n for all n-0-1-2-2010- Find P2012- correct answer - 5- wrong answer 0

Define Q(x)=(1+x)P(x) x Then Q(x) is a polynomial of degree 2011. Since Q(0)=Q(1)=Q(2)=⋯=Q(2010)=0, ∴ Q(x) can be written as Q(x)=Ax(x 1)(x 2)⋯(x 2010) 1=Q( 1)= ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Polynomial Functions

Let Px be a p...

Question

Let $P (x)$ be a polynomial of degree $2010$ . Suppose $P (n) = \frac{n}{1 + n}$ for all $n = 0, 1, 2, \dots, 2010.$ Find $P (2012) .$
(correct answer + 5, wrong answer 0)

Open in App

Solution

Define $Q (x) = (1 + x) P (x) - x$
Then $Q (x)$ is a polynomial of degree $2011$ .
Since $Q (0) = Q (1) = Q (2) = \dots = Q (2010) = 0,$
$∴ Q (x)$ can be written as
$Q (x) = A x (x - 1) (x - 2) \dots (x - 2010)$
$1 = Q (- 1) = A (- 1) (- 2) (- 3) \dots (- 2011) = - A \times (2011)!$
$\Rightarrow A = - \frac{1}{(2011)!}$

Now, $Q (2012) = A \times (2012)!$
$= - \frac{1}{(2011)!} \times (2012)!$
$= - 2012$
And $P (2012) = \frac{Q (2012) + 2012}{2013}$
$= \frac{- 2012 + 2012}{2013} = 0$

Suggest Corrections

Similar questions

Q. Let

P (x)

be a polynomial of degree

2010

. Suppose

P (n) = \frac{n}{1 + n}

for all

n = 0, 1, 2, \dots, 2010.

Find

P (2012) .

(correct answer + 5, wrong answer 0)

Q. If

(3 + x^{2008} + x^{2009})^{2010} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}

,
then the value of

a_{0} - \frac{1}{2} a_{1} - \frac{1}{2} a_{2} + a_{3} - \frac{1}{2} a_{4} - \frac{1}{2} a_{5} + a_{6} - \dots

upto

n

terms is

Q. Let

a_{1}, a_{2}, a_{3}, \dots

be a G.P. with

a_{1} = a

and common ratio

r,

where

a

and

r

positive integers, then the number of ordered pairs

(a, r)

such that

12 \sum r = 1 {log}_{8} a_{r} = 2010

is
(correct answer + 1, wrong answer - 0.25)

Q. Let

N = \frac{100!}{20^{m}} .

The smallest positive integer

m

for which

N

is not an integer is
(correct answer + 1, wrong answer - 0.25)

Q. If all non-negative real numbers

a_{1} \leq a_{2} \leq \dots \leq a_{n}

satisfying

n \sum i = 1 a_{i} = 96,

n \sum i = 1 a_{i}^{2} = 144

and

n \sum i = 1 a_{i}^{3} = 216,

then the value of

n

is
(correct answer + 1, wrong answer - 0.25)

Join BYJU'S Learning Program

Let P(x) be a polynomial of degree 2010. Suppose P(n)=n1+n for all n=0,1,2,…,2010. Find P(2012). (correct answer + 5, wrong answer 0)

Let $P (x)$ be a polynomial of degree $2010$ . Suppose $P (n) = \frac{n}{1 + n}$ for all $n = 0, 1, 2, \dots, 2010.$ Find $P (2012) .$
(correct answer + 5, wrong answer 0)