Question

Answer each part TRUE or FALSE

$$\begin{gathered} \mathbf{a}.\mathbf{n}=\mathbf{o}\left(\mathbf{2}\mathbf{n}\right). \\ \mathbf{b}. \mathbf{2}\mathbf{n}=\mathbf{o}\left({\mathbf{n}}^2\right).\mathbf{A} \\ \mathbf{c}\mathbf{.}\mathbf{2}{}^{\mathbf{n}}\mathbf{=}\mathbf{o}\left(3^n\right)\mathbf{.}\mathbf{A} \\ \mathbf{d}\mathbf{.}\mathbf{1}\mathbf{=}\mathbf{o}\left(n\right)\mathbf{.} \\ \mathbf{e}\mathbf{.}\mathbf{n}\mathbf{=}\mathbf{o}\left(\log n\right)\mathbf{.} \\ \mathbf{f}\mathbf{.}\mathbf{1}\mathbf{=}\mathbf{o}\left(1/n\right)\mathbf{.} \end{gathered}$$

Short answer

(a) $n=0\left(2n\right)$is true.

(b)$2n=O\left(n^2\right)$ true.

(c) $2^n=O\left(3^n\right)$is true.

(d) $1=O\left(n\right)$is true.

(e) $n=o(\log n)$is false.

(f) $1=o\left(\frac{1}{n}\right)$is false.

Step-by-step solution

Step 1:To Calculate the correct theory

Here, it shows each part calculation which show correct theory and formula base answer in linear time,quadratic time etc.

To check whether equation is True or False

a) TRUE

$2n$seems to be linear time, therefore n represents linear time as well, therefore$n=O\left(2n\right)$

b) TRUE
$n^2$Since has been quadratic whereas $2n$seems to be linear,$2n=O\left(n^2\right)$

c) TRUE
$2^n$ involves time that is exponentially andbecomes exponentially larger than $2^n$thus .$2^n=O\left(3^n\right)$

d) TRUE
$1$ represents continuous time (i.e., a moment), and therefore it is undeniably$1=O\left(n\right)$ where n is linear times.

e) FALSE
$n$is linear time$\log n$ Because it would be the logarithmic time, that's less than linear, $ncouldbeinBigOoflogn$.

f) FALSE
Important Note: $1=o\left(\frac{1}{n}\right)$s rather tough to obtain; it just implies that the algorithm will run quicker for a greater number of inputs; more tasks, less time; less tasks, more time.