The greatest integer which divides (p + 1) (p + 2) (p + 3) .... (p + q) for all p E N and fixed q N is
p!
q!
p
q
Let f : R ➔ R be such that f is injective and f(x)f(y) = f(x + y) for x, y 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
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
B.
p = 2, q = - 3
Given equations are
2x2 + 3xy + 4y2 + x + 18y + 25 = 0 ...(i)
2x2 + 3xy + 4y2 = 1 ...(ii)
Let the origin be transferred to (p, q) axes being parallel to the previous axes; then the equation (i) becomes
From equation (ii), coefficient of x' and y' must be zero.
4p + 3q + 1 = 0 ...(iii)
3p + 8q + 18 = 0 ...(iv)
By solving equations (iii) and (iv), we get
p = 2, 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