Question

Match the following columns:

Column I		Column II
(a)	$6.\overline{54}$ is ....... .	(p)	14
(b)	$\mathrm{\pi }$ is ...... .	(q)	6
(c)	The length of period of $\frac{1}{7}$=...... .	(r)	a rational number
(d)	If $x=\left(2-\sqrt{3}\right)$, then $\left(x^2+\frac{1}{x^2}\right)$....... .	(s)	an irrational number

(a) ........
(b) ........
(c) ........
(d) ........

Answer 1

(a) Because it is a non-terminating and repeating decimal, it is a rational number.

(b) $\mathrm{\pi }$ is an irrational number.

(c) $\frac{1}{7}=.142857142857$...
Hence, its period is 6.

(d)
$$\begin{gathered} x^2+\frac{1}{x^2} \\ ={\left(2-\sqrt{3}\right)}^2+\frac{1}{{\left(2-\sqrt{3}\right)}^2} \\ =\left(2^2+{\left(\sqrt{3}\right)}^3-2\times 2\times \sqrt{3}\right)+\frac{1}{\left(2^2+{\left(\sqrt{3}\right)}^3-2\times 2\times \sqrt{3}\right)} \\ =\left(4+3-4\times \sqrt{3}\right)+\frac{1}{\left(4+3-4\times \sqrt{3}\right)} \\ =\left(7-4\times \sqrt{3}\right)+\frac{1}{\left(7-4\times \sqrt{3}\right)} \\ =\frac{{\left(7-4\times \sqrt{3}\right)}^2}{\left(7-4\times \sqrt{3}\right)}+\frac{1}{\left(7-4\times \sqrt{3}\right)} \\ =\frac{7^2+{\left(4\sqrt{3}\right)}^2-2\times 7\times 4\sqrt{3}+1}{\left(7-4\times \sqrt{3}\right)} \\ =\frac{49+48-56\sqrt{3}+1}{\left(7-4\times \sqrt{3}\right)} \\ =\frac{98-56\sqrt{3}}{\left(7-4\times \sqrt{3}\right)} \\ =14\times \frac{\left(\left(7-4\times \sqrt{3}\right)\right)}{\left(7-4\times \sqrt{3}\right)} \\ =14 \end{gathered}$$