For every positive integer n , let n denote the integer closest to n , and let A_k=\n>0: n =k\ . The number of elements in A_49 is

By definition n =k means k is the integer closest to n , i.e. k- 12< n<k+ 12. Square (all quantities positive): (k- 12 )^2<n< (k+ 12 )^2, that is k^2-k+ 14<n<k^2+k+ 14. Count the integers n in this range. The integers strictly between k^2-k+ 14 and k^2+k+ 14 are k^2-k+1,\ k^2-k+2,\ ,\ k^2+k, and their number is (k^2+k)-(k^2-k+1)+1=2k . For k=49 : |A_49|=2 49=98 . Answer: B.

Inequalities Difficulty 4

TMUA practice: Inequalities, problem 5

A TMUA-calibre inequalities problem at difficulty 4 of 5, with the full worked solution.

The question

For every positive integer

n

, let

\langle n\rangle

denote the integer closest to

\sqrt{n}

, and let

A_{k}=\{n>0:\langle n\rangle=k\}

. The number of elements in

A_{49}

97

98

99

100

The correct answer is highlighted. B

Worked solution

By definition $\langle n\rangle=k$ means $k$ is the integer closest to $\sqrt{n}$ , i.e.

$k-\tfrac12<\sqrt{n}<k+\tfrac12.$

Square (all quantities positive):

\left(k-\tfrac12\right)^{2}<n<\left(k+\tfrac12\right)^{2}, \qquad\text{that is}\qquad k^{2}-k+\tfrac14<n<k^{2}+k+\tfrac14.

Count the integers $n$ in this range. The integers strictly between $k^{2}-k+\tfrac14$ and $k^{2}+k+\tfrac14$ are

$k^{2}-k+1,\ k^{2}-k+2,\ \ldots,\ k^{2}+k,$

and their number is $(k^{2}+k)-(k^{2}-k+1)+1=2k$ .

For $k=49$ : $|A_{49}|=2\cdot49=98$ .

Answer: B.