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Two alternative packages $A$ and $B$ are available for processing a database having $10k$ records. Package $A$ requires $0.0001n2$ time units and package $B$ requires $10 n \log 10n$ time units​ ​ to process $n$ records. What is the smallest value of $k$ for which package $B$ will be preferred over $A$?

  1. $12$
  2. $10$
  3. $6$
  4. $5$
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the smallest value of kk for which package BB will be preferred over A

B\leqslantA

$10nlog_{10}(n) ≤ 0.0001n^{2}$

Given$ n = 10^{k }$records. Therefore,

$10*10^{k}log_{10}(10^{k}) ≤ 0.0001(10^{k})^{2}$
$10^{k+1}*k ≤ 0.0001 × 10^{2k}$
$k ≤ 10^{2k−k−1−4}$
$k ≤ 10^{k−5}->(1)$

 

if $k=12$ put in equation 1

$12\leqslant10^{12-5}$

$12\leqslant10^{7}$ IT IS true

 

if $k=10$ put in equation 1

$10\leqslant10^{10-5}$

$10\leqslant10^{5}$ IT IS true

 

if $k=6$ put in equation 1

$6\leqslant10^{6-5}$

$6\leqslant10$ it  is true

 

if $k=5$ put in equation 1

$5\leqslant10^{5-5}$

$5\leqslant1$ it  is not  true

here  K=12,10,6 satisfied but in question asking minimum value which satisfied equation 1

 MINIMUM VALUE IS 6

Option C
Answer:

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