KVPY 2019 SA Exam Previous Papers

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Question : 63 of 80

Marks: +1, -0

Let N be the set positive integers. For all n ∊ N, let

\f_n = (n + 1)^{1\/3} - n ^{1\/3}

and

A =\{ {n∊N:\f _(n +1)<1/{3(n+1) ^{2\/3}}<\f_n\}

Then

A = N
A is a finite set
the complement of A in N is nonempty but finite
A and its complement in N are both infinite

Solution:

\f_n = (n + 1)^{1\/3} - n ^{1\/3}

Rationalising

\f_n

get

\f_n=1/{(n+1) ^{2\/3}+(n ) ^{1\/3} (n+1) ^{1\/3}+n ^{2\/3}}>1/{3(n+1) ^{2\/3}}

similarty

\f_{n+1}=1/{(n+2) ^{2\/3}+(n+1 ) ^{1\/3} (n+2) ^{1\/3}+({n+1 }) ^{2\/3}}>1/{3(n+1) ^{2\/3}}

Hence

\f_{n+1}<1/{3(n+1) ^{2\/3}}<\f_n∀n∊N

Hence A = N

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