Xavier Aptitude Test 2013 Paper

© examsiri.com

Question : 78 of 121

Marks: +1, -0

Consider a function f(x) = x

^{4}

+ x

^{3}

+ x

^{2}

+ x + 1, where x is a positive integer greater than 1. What will be the remainder if (x

^{5}

) is divided by
f(x)?

1
4
5
a monomial in x
a polynomial in x

Solution:

(c)

f(x)=x^{4}+x^{3}+x^{2}+x+1

f\ (x^{5}\ )=x^{20}+x^{15}+x^{10}+x^{5}+1

f (x^{5}\ )=x^{20}+1-1+x^{15}+1-1+x^{10}+1-1+x^{5}+1-1

=\ (x^{20}-1\ )+\ (x^{15}-1\ )+\ (x^{10}-1\ )+ (x^{5}-1\ )+5

All the terms of

f (\ x^{5}\ )

except 5 can be divisible by

\ (x^{5}-1\ )

as

\ (x^{5}-1\ )

can be factorised as

(x-1)\ (x^{4}+x^{3}+x^{2}+1\ )

∴

remainder

=5

© examsiri.com

Go to Question:

Prev Question

Next Question