If a, b and c are three non-coplanar vectors, then (a + b - c) . [(a - b) x (b - c)] equals

• 0

• a . b $\times$ c

• a . c $\times$ b

• 3a . b $\times$ c

The area of the region bounded by the curves x² + y² = 9 and x + y = 3 is

33.

For any three vectors a, b and c [a + b, b + c, c + a] is

• [a, b, c]

• 3[a, b, c]

• 2[a, b, c]

• 0

${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin \left(2\mathrm{x}\right) . \log \left(\tan \left(\mathrm{x}\right)\right)\mathrm{dx}$

• 0

• 2

• 4

• 7

If a plane passing through the point (2, 2, 1) and is perpendicular to the planes 3x + 2y + 4z + 1 = 0 and 2x + y + 3z + 2 = 0. Then, the equation of the plane is

• 2x - y - z - 1 = 0

• 2x + 3y + z - 1 = 0

• 2x + y + z + 3 = 0

• x - y + z - 1 = 0

From a city population, the probability of selecting a male or smoker $\frac{7}{10}$, a male smoker is $\frac{2}{5}$ and a male, if a smoker is already, selected, is $\frac{2}{3}$ Then, the probability of

• selecting a male is $\frac{3}{2}$

• selecting a smoker is $\frac{1}{5}$

• selecting a non-smoker is $\frac{2}{5}$

• selecting a smoker, if a male is first selected, is given by $\frac{8}{5}$

The solution of $\frac{d^2\mathrm{x}}{d{\mathrm{y}}^2} - \mathrm{x} = \mathrm{k}$ where k is a non-zero constant, vanishes when y = 0 and tends of finite limit as y tends to infinity, is 

• x = k(1 + e^{- y})

• x = k(e^y + e^{- y} - 2)

• x = k(e^{- y} - 1)

• x = k(e^y - 1)

The differential equation (3x + 4y + 1)dx + (4x + 5y + 1)dy = 0 represents a family of

• circles

• parabolas

• ellipses

• hyperbolas

If A, B, C are three events associated with a random experiment, then $\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{A}} \cap \mathrm{B}\right)$

• $\mathrm{P}\left(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}\right)$

• $\mathrm{P}\left(\mathrm{A}\hspace{0.17em}\cap \mathrm{B}\hspace{0.17em}\cap \mathrm{C}\right)$

• $\mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{A}} \cap \mathrm{B}\right)$

• $\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)$

The probability of atleast one double six being thrown in n throws with two ordinary dice is greater than 99%.

Then, the least numerical value of n is

• 100

• 164

• 170

• 184