Observe the following lists

Then the correct match for List I from List II is

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

$\mathrm{x} = \frac{1}{2}\left(\sqrt{3} + \frac{1}{\sqrt{3}}\right), \mathrm{then} \frac{\sqrt{{\mathrm{x}}^2 - 1}}{\mathrm{x} - \sqrt{{\mathrm{x}}^2 - 1}}$

• 1

• 2

• 3

• $\frac{1}{2}$

If a, b, c $\ne$ 0 and belong to the set to {0, 1, 2, 3, ..., 9}, then ${\log }_{10}\left(\frac{\mathrm{a} + 10\mathrm{b} + {10}^2\mathrm{c}}{{10}^{- 4}\mathrm{a} + {10}^{- 3}\mathrm{b} + {10}^{- 2}\mathrm{c}}\right)$ is equal to

• 1

• 2

• 3

• 4

$\left\{\mathrm{n}\left(\mathrm{n} + 1\right)\left(2\mathrm{n} + 1\right) : \mathrm{n} \in \mathrm{Z}\right\} \subset$

• $\left\{6\mathrm{k} : \mathrm{k} \in \mathrm{Z}\right\}$

• $\left\{12\mathrm{k} : \mathrm{k} \in \mathrm{Z}\right\}$

• $\left\{18\mathrm{k} : \mathrm{k} \in \mathrm{Z}\right\}$

• $\left\{24\mathrm{k} : \mathrm{k} \in \mathrm{Z}\right\}$

A three digit numbern is such that the last twodigits of it are equal and differ from the first. The number of such n's is

• 64

• 72

• 81

• 900

If (1 + x)¹⁵ = a₀ + a₁x + ... + a₁₅x¹⁵, then $\sum _{\mathrm{r} = 1}^{15}\mathrm{r}\frac{{\mathrm{a}}_{\mathrm{r}}}{{\mathrm{a}}_{\mathrm{r}} - 1}$ is equal to

• 110

• 115

• 120

• 135

The coefficient of x³y⁴z⁵ in the expansion of (xy + yz + xz)⁶ is

• 70

• 60

• 50

• None of these

If $\left|\mathrm{x}\right| < \frac{1}{2}$, then the coefficient of x^r in the expansion of $\frac{1 +\hspace{0.17em}2\mathrm{x}}{{\left(1 - 2\mathrm{x}\right)}^2}$, is

• r2^r

• (2r - 1)2^r

• r2^{2r + 1}

• (2r + 1)2^r