The number of all numbers having 5 digits, with distinct digits is

• 99999

• $9 \times {}^9\mathrm{P}_4$

• ${}^{10}\mathrm{P}_5$

• ${}^9\mathrm{P}_4$

The greatest integer which divides (p + 1) (p + 2) (p + 3) .... (p + q) for all p E N and fixed q $\in$ N is

• p!

• q!

• p

• q

Let ${\left(1 +\hspace{0.17em}\mathrm{x} + {\mathrm{x}}^2\right)}^9 = {\mathrm{a}}_0 +\hspace{0.17em}{\mathrm{a}}_1\mathrm{x} +\hspace{0.17em}{\mathrm{a}}_2{\mathrm{x}}^2 +\hspace{0.17em}... + {\mathrm{a}}_{18}{\mathrm{x}}^{18}$. Then,

• ${\mathrm{a}}_0 +\hspace{0.17em}{\mathrm{a}}_2 +\hspace{0.17em}... +\hspace{0.17em}{\mathrm{a}}_{18} = {\mathrm{a}}_0 +\hspace{0.17em}{\mathrm{a}}_3 + ... + {\mathrm{a}}_{17}$

• ${\mathrm{a}}_0 + {\mathrm{a}}_2 +\hspace{0.17em}... + {\mathrm{a}}_{18}$ is even

• ${\mathrm{a}}_0 + {\mathrm{a}}_2 +\hspace{0.17em}... + {\mathrm{a}}_{18}$ is divisible by 9

• ${\mathrm{a}}_0 + {\mathrm{a}}_2 +\hspace{0.17em}... + {\mathrm{a}}_{18}$ is divisible by 3 but not by 9

Let f : R ➔ R be such that f is injective and f(x)f(y) = f(x + y) for $⟇$ x, y $\in$ R. If f(x), f(y), f(z) are in G.P., then x, y, z are in

• AP always

• GP always

• AP depending on the value of x, y, z

• GP depending on the value of x, y, z

The probability that a non-leap year selected at random will have 53 Sunday is,

• 0

• 1/7

• 2/7

• 3/7

The equation sin x(sin(x) + cos(x)) = k has real solutions, where k is a real number. Then,

• $0 \le \mathrm{k} \le \frac{1 + \sqrt{2}}{2}$

• $2 - \sqrt{3} \le \mathrm{k} \le 2 + \sqrt{3}$

• $0 \le \mathrm{k} \le 2 - \sqrt{3}$

• $\frac{1 - \sqrt{2}}{2} \le \mathrm{k} \le \frac{1 + \sqrt{2}}{2}$

Transforming to parallel axes through a point (p, q), the equation

2x² + 3xy + 4y² + x + 18y + 25 = 0 becomes 2x² + 3xy + 4y² = 1. Then,

• p = - 2, q = 3

• p = 2, q = - 3

• p = 3, q = - 4

• p = - 4, q = 3

Let A(2,- 3) and B(- 2,1) be two angular points of AABC. If the centroid of the triangle moves on the line 2x + 3y = 1, then the locus of the angular point C is given by

• 2x + 3y = 9

• 2x - 3y = 9

• 3x + 2y = 5

• 3x + 2y = 3

The point P(3, 6) is first reflected on the line y =x and then the image point Q is again reflected on the line y = - x to get the image point Q'. Then, the circumcentre of the APQQ' is

• (6, 3)

• (6,- 3)

• (3,- 6)

• (0, 0)

Let d₁ and d₂ be the lengths of the perpendiculars drawn from any point of the line 7x - 9y + 10 = 0 upon the lines 3x + 4y = 5 and 12x +5y = 7, respectively. Then,

• d₁ > d₂

• d₁ = d₂

• d₁ < d₂

• d₁ = 2d₂