**State TRUE or FALSE justifying your answer with proper reason. a. 2π^2 + 1 = π(π^2 ) b. π^2 (1 + βπ) = π(π^2 ) c. π^2 (1 + βπ) = π(π^2 log π) d. 3π^2 + βπ = π(π + πβπ + βπ) e. βπ log π = π(π)**

The **Answer to the Question**

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**Here's the Solution to this Question**

a.

True

$2π^2 + 1\le 3n^2$

b.

False

$\displaystyle \lim_{n\to \infin} \frac{π^2 (1 + \sqrtπ) }{n^2}=\infin$

c.

False

$\displaystyle \lim_{n\to \infin} \frac{π^2 (1 + \sqrtπ) }{n^2logn}=\infin$

d.

False

$\displaystyle \lim_{n\to \infin} \frac{ 3π^2 + \sqrtπ) }{π + π\sqrtπ + \sqrtπ}=\infin$

e.

True

$\sqrtπ log π\le n$