Sets

If A and B be two sets such that A x B consists of 6 elements. If three elements A x B are (1, 4), (2, 6) and (3, 6), find B x A.

• {(1, 4), (1, 6), (2, 4), (2, 6), (3, 4), (3, 6)}

• {(4, 1), (4, 2), (4, 3), (6, 1), (6, 2), (6, 3)}

• {(4, 4), (6, 6)}

• {(4, 1), (6, 2), (6, 3)}

If f (x) = $\sqrt{{\log }_{10}\left({\mathrm{x}}^2\right)}$. The set of all values of x for which f(x) is real, is

• [- 1, 1]

• $[1, \infty )$

• $(- \infty , - 1]$

• $(- \infty , - 1] \cup [1, \infty )$

The solution set contained in R of the inequation

3^x  + 3^{1 - x} - 4 < 0, is:

• (1, 3)

• (0, 1)

• (1, 2)

• (0, 2)

If f(x) = $$\begin{gathered} \left\{\left[\mathrm{x}\right], \mathrm{if} - 3 < \mathrm{x} \le - 1 \\ \left|\mathrm{x}\right|, \mathrm{if} - 1 < \mathrm{x} <\hspace{0.17em}1 \\ \left|\left[\mathrm{x}\right]\right|, \mathrm{if} 1 \le \mathrm{x} < 3\right\} \end{gathered}$$ then the set $\left(\mathrm{x} : \mathrm{f}\left(\mathrm{x}\right) \ge 0\right)$ is equal to

• (- 1, 3)

• [- 1, 3)

• (- 1, 3]

• [- 1, 3)

$\left\{\mathrm{x} \in \mathrm{R} : \left[\mathrm{x} - \left|\mathrm{x}\right|\right] = 5\right\}$ is equal to

• R, the set of all real numbers

• $\mathrm{\varphi }$, the empty set

• $\left\{\mathrm{x} \in \mathrm{x} < 0\right\}$

• $\left\{\mathrm{x} \in \mathrm{R} : \mathrm{x} \ge 0\right\}$

If N denotes the set of all positive integers and if f : $\mathrm{N} \to \mathrm{N}$ efined by f(n) = the sum of positive divisors of n then, f(2^k, 3), where k is a positive integers, is

• 2^{k + 1} - 1

• 2(2^{k + 1} - 1)

• 3(2^{k + 1} - 1)

• 4(2^{k + 1} - 1)

If f : R $\to$ R is defined by f(x)=x - [ x] - $\frac{1}{2}$ for x $\in$ R, where [x] is the greatest mteger not exceeding x, then $\left\{\mathrm{x} \in \mathrm{R} : \mathrm{f}\left(\mathrm{x}\right) = \frac{1}{2}\right\}$ is equal to :

• Z, the set of all integers

• N, the set of all natural numbers

• $\mathrm{\varphi }$, the empty set

• R

$$\begin{gathered} \mathrm{If} \mathrm{Q} \mathrm{denotes} \mathrm{the} \mathrm{set} \mathrm{of} \mathrm{all} \mathrm{rational} \mathrm{numbers} \mathrm{and} \mathrm{f}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) = \sqrt{{\mathrm{p}}^2 - {\mathrm{q}}^2} \mathrm{for} \\ \mathrm{any} \frac{\mathrm{p}}{\mathrm{q}} \in \mathrm{Q}, \mathrm{then} \mathrm{observe} \mathrm{the} \mathrm{following} \mathrm{statements}. \\ \mathrm{I}. \mathrm{f}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) \mathrm{is} \mathrm{real} \mathrm{for} \mathrm{each} \frac{\mathrm{p}}{\mathrm{q}} \in \mathrm{Q} \\ \mathrm{II}. \mathrm{f}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) \mathrm{is} \mathrm{a} \mathrm{complex} \mathrm{number} \mathrm{for} \mathrm{each} \frac{\mathrm{p}}{\mathrm{q}} \in \mathrm{Q}. \\ \mathrm{Which} \mathrm{of} \mathrm{the} \mathrm{following} \mathrm{is} \mathrm{correct} ? \end{gathered}$$

• Both I and II are true

• I is true, II is false

• I is false, II is true

• Both I and II are false

If a set A has 5 elements, then the number of ways of selecting two subsets P and Q from A such that P and Q are mutually disjoint, is

• 64

• 128

• 243

• 729

Let N be the set of all natural numbers, Z be
the set of all integers and $\mathrm{\sigma } : \mathrm{N} \to \mathrm{Z}$ be defined by

$$\begin{gathered} \mathrm{\sigma }\left(\mathrm{n}\right) = \left\{\frac{\mathrm{n}}{2}, \mathrm{if} \mathrm{n} \mathrm{is} \mathrm{even} \\ - \frac{\mathrm{n} - 1}{2}, \mathrm{if} \mathrm{n} \mathrm{is} \mathrm{odd}\right\}. \mathrm{Then}, \end{gathered}$$

• $\sigma$ is onto but not one-one

• $\mathrm{\sigma } \mathrm{is} \mathrm{one}-\mathrm{one} \mathrm{but} \mathrm{not} \mathrm{onto}$

• $\mathrm{\sigma } \mathrm{is} \mathrm{neither} \mathrm{one}-\mathrm{one} \mathrm{nor} \mathrm{onto}$

• $\mathrm{\sigma } \mathrm{is} \mathrm{one}-\mathrm{one} \mathrm{and} \mathrm{onto}$