KVPY SX SB Exam 2015 Question Paper

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Question : 9 of 120

Marks: +1, -0

Let p(x) be a polynomial such that

p(x) – p'(x) = x^n

, where n is a positive integer. Then p(0) equals -

$n!$
$(n -1)!$
$1/n!$
$1/(n -1)!$

Solution:

Let

P(x) = a_0x^n + a_1x^{n-1} + a_2x^{n-2} +…+ a_n

P'(x) = na_0x^{n-1} + (n – 1)a_1x^{n-2} +…+ a^{n-1}

P(x)P'(x) = a_0x^n + (a_1 - na_0) x^{n-1}

+ (a_2 - (n -1)a_1) x^{n-2}+…+ (a_n - a_{n-1})

given

P(x) - P'(x) = a_0x^n

.
so that

\table a_1 - na_0 = 0,a_1/a_0=n;a_2 - (n - 1)a_1 = 0 ,a_2/a_1=(n-1);a_n - a_{n-1} = 0,a _n/a_{n-1}=1

P(0) =a_n (a_n/a_{n-1})(a_{n-1}/a_{n-2})...(a_1/a_0)

= 1 × 2 × 3 × …× n
= n!

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