For the circuit show below, the Boolean polynomial is

• $\left(~ \mathrm{p} \vee \mathrm{q}\right) \vee \left(\mathrm{p} \vee ~ \mathrm{q}\right)$

• $\left(~ \mathrm{p} \wedge \mathrm{q}\right) \vee \left(\mathrm{p} \wedge \mathrm{q}\right)$

• $\left(~ \mathrm{p} \wedge ~ \mathrm{q}\right) \vee \left(\mathrm{q} \wedge \mathrm{p}\right)$

• $\left(~ \mathrm{p} \wedge \mathrm{q}\right) \vee \left(\mathrm{p} \wedge ~ \mathrm{q}\right)$

In a Boolean Algebra B, for all x, y in B, $\mathrm{x} \wedge \left(\mathrm{x} \vee \mathrm{y}\right)$ is equal to

• y

• x

• 1

• 0

The value of $\left(1 + ∆\right)\left(1 - \nabla \right)$ is

• 0

• - 1

• 1

• None of the above

If u = a - b and v = a + b and $\left|\mathrm{a}\right| = \left|\mathrm{b}\right|$ = 2, then $\left|\mathrm{u} \times \mathrm{v}\right|$ is equal to

• $2\sqrt{16 - {\left(\mathrm{a} . \mathrm{b}\right)}^2}$

• $\sqrt{16 - {\left(\mathrm{a} . \mathrm{b}\right)}^2}$

• $\mathrm{2}\sqrt{4 - {\left(\mathrm{a} . \mathrm{b}\right)}^2}$

• $2\sqrt{4 + {\left(\mathrm{a} . \mathrm{b}\right)}^2}$

If the vectors a, b and c are coplanar, then

$\left|\begin{array}{ccc}\mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{a} . \mathrm{a}& \mathrm{a} . \mathrm{b}& \mathrm{a} . \mathrm{c}\\ \mathrm{b} . \mathrm{a}& \mathrm{b} . \mathrm{b}& \mathrm{b} . \mathrm{c}\end{array}\right|$ is equal to

• 1

• 0

• - 1

• None of these

A vector vis equally inclined to the x-axis, y-axis and z-axis respectively, its direction cosines are

• $< \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} >$

• $< - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}} >$

• $< \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} > \mathrm{or} < - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}} >$

• None of the above

A plane meets the axes in A, B and C such that centroid of the $∆$ABC is (1, 2, 3). The equation of the plane is

• x + y/2 + z/3 = 1

• x/3 + y/6 + z/9 = 1

• x + 2y + 3z = 1

• None of these

If $\mathrm{\alpha }, \mathrm{\beta } \mathrm{and} \mathrm{\gamma }$ are the angles which a half ray makes with the positive direction of the axes, then ${\sin }^2\left(\mathrm{\alpha }\right) + {\sin }^2\left(\mathrm{\beta }\right) + {\sin }^2\left(\mathrm{\gamma }\right)$ is equal to

• 1

• 2

• 0

• - 1

19.

The angle between a line with direction ratio 2 : 2 : 1 and a line joining (3, 1, 4) to (7, 2, 12) is

• ${\cos }^{-1}\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$

• ${\cos }^{-1}\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

• ${\tan }^{-1}\left(\raisebox{1ex}{$- 2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$

• None of the above

If x follows the Binomial distribution with parameters n = 6 and p and 9P(X = 4) = P(X = 2), then p is

• 1/4

• 1/3

• 1/2

• 2/3