Question

a) Let Qi = {qi|x² + q²x +4 = 0, q is an element of N, x is aan element of R}.
Find ⋃∞ i=1 Qi and ⋂∞ i=1 Qi.

Answer 1

$Q_{i} = \{q^{*}i|x^{2} + q^{2}x + 4 = 0\mid q$ is an element of $N,x$ is an element of $R\}$

Then:

for $x^{2} + q^{2}x + 4 = 0$ :

Implicit plot:

The solution is:

$D = q^4 - 16$

$|q| < 2 \Rightarrow$ no solutions;

$|q| = 2 \Rightarrow x = -2$

$|q| > 2 \Rightarrow x = 0.5(-q^2 + (q^4 - 16)^{0.5})$ ; $x = 0.5(-q^2 - (q^4 - 16)^{0.5})$ .

Then we can see, that $q = \{2, 3, 4, 5, \ldots\}$ (as $q \in N$ ), and, respectively, $x$ takes real values. $\Rightarrow q = N \setminus \{1\}$

$Q_{1} = \{2,3,4,5,6,7,8,9,\ldots \}$

$Q_{2} = \{4,6,8,10,12,14,16,18,\ldots \}$

.

$Q_{k} = \{2k,3k,4k,5k,6k,7k,8k,\ldots \}$

From this we can assume:

$\cup \infty \mathrm{i} = 1\mathrm{Q_i} = \mathrm{Q_1} = \{\mathrm{q}|\mathrm{x}^2 +\mathrm{q}^2\mathrm{x} + 4 = 0\mid \mathrm{q}$ is an element of $\mathbf{N},\mathbf{x}$ is an element of $\mathbb{R}\}$

$\bigcap \infty \mathrm{i} = 1\mathrm{Q_i} = \varnothing$

www.AssignmentExpert.com