Let A and B be events in a sample space S suchthat P(A) = 0.5, P(B) = 0.4 andP(A $\cup$ B) = 0.6. Observe the following lists

	List I		List II
(i)	$\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right)$	(1)	0.4
(ii)	$\mathrm{P}\left(\mathrm{A} \cap \overline{\mathrm{B}}\right)$	(2)	0.2
(iii)	$\mathrm{P}\left(\overline{\mathrm{A}} \cap \mathrm{B}\right)$	(3)	0.3
(iv)	$\mathrm{P}\left(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\right)$	(4)	0.1

The correct match of List I from List II is

$$\begin{gathered} \mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = \left|\mathrm{x}\right| + \left|\sin \left(\mathrm{x}\right)\right| \mathrm{for} \mathrm{x} \in \left(- \frac{\mathrm{\pi }}{2}, \frac{\mathrm{\pi }}{2}\right), \\ \mathrm{then} \mathrm{its} \mathrm{left} \mathrm{hand} \mathrm{derivative} \mathrm{at} \mathrm{x} = 0 \mathrm{is} \end{gathered}$$

• 0

• - 1

• - 2

• - 3

If 1° = $\mathrm{\alpha }$. radians, then the approximate value of cos(60° 1') is

• $\frac{1}{2} + \frac{\mathrm{\alpha }\sqrt{3}}{120}$

• $\frac{1}{2} - \frac{\mathrm{\alpha }}{120}$

• $\frac{1}{2} - \frac{\mathrm{\alpha }\sqrt{3}}{120}$

• $\frac{1}{2} + \frac{\mathrm{\alpha }}{120}$

$\mathrm{If} \mathrm{u} = \mathrm{u}\left(\mathrm{x}, \mathrm{y}\right) = \sin \left(\mathrm{y} + \mathrm{ax}\right) - {\left(\mathrm{y} + \mathrm{ax}\right)}^2, \mathrm{then} \mathrm{it} \mathrm{implies}$

• ${\mathrm{u}}_{\mathrm{xx}} = {\mathrm{a}}^2 . {\mathrm{u}}_{\mathrm{yy}}$

• ${\mathrm{u}}_{\mathrm{yy}} = {\mathrm{a}}^2{\mathrm{u}}_{\mathrm{xx}}$

• ${\mathrm{u}}_{\mathrm{xx}} = - {\mathrm{a}}^2 . {\mathrm{u}}_{\mathrm{yy}}$

• ${\mathrm{u}}_{\mathrm{yy}} = - {\mathrm{a}}^2{\mathrm{u}}_{\mathrm{xx}}$

If a, b and n are natural numbers, then a^{2n - 1}  + b^{2n - 1} is divisible by

• a + b

• a - b

• a³ + b³

• a² + b²

6.

A bag contains n white and n black balls. Pairs of balls are drawn at random without replacement successively, until the bag is empty. If the number of ways in which each pair consists of one white and one black ballis 14400, then n is equal to

• 6

• 5

• 4

• 3

The number of five digit numbers divisible by 5 that can be formed using the numbers 0, 1, 2, 3, 4, 5 without repetition is

• 240

• 216

• 120

• 96

${}_{15}\mathrm{P}_8 = \mathrm{A} + 8 . {}_{14}\mathrm{P}_7 \Rightarrow \mathrm{A} =$

• ${}_{14}\mathrm{P}_6$

• ${}_{14}\mathrm{P}_8$

• ${}_{15}\mathrm{P}_7$

• ${}_{16}\mathrm{P}_9$

$\mathrm{If} {}^{\left(\mathrm{n} - 1\right)}\mathrm{C}_1 + {}^{\left(\mathrm{n} - 1\right)}\mathrm{C}_4 > {}^{\mathrm{n}}\mathrm{C}_3, \mathrm{then} \mathrm{the} \mathrm{minimum} \mathrm{value} \mathrm{of} \mathrm{n} \mathrm{is}$

• 5

• 6

• 7

• 8

If the coefficients of rth and (r + 1)th terms in the expansion of (3 + 7x)²⁹ are equal, then r is equal to

• 14

• 15

• 18

• 21