1
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
If the Boolean expression
(p $$\oplus$$ q) $$\wedge$$ (~ p $$\odot$$ q) is equivalent
to p $$\wedge$$ q, where $$\oplus , \odot \in \left\{ { \wedge , \vee } \right\}$$, then the
ordered pair $$\left( { \oplus , \odot } \right)$$ is :
A
$$\left( { \vee , \wedge } \right)$$
B
$$\left( { \vee , \vee } \right)$$
C
$$\left( { \wedge , \vee } \right)$$
D
$$\left( { \wedge , \wedge } \right)$$
2
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
If p $$\to$$ ($$\sim$$ p$$\vee$$ $$\sim$$ q) is false, then the truth values of p and q are respectively :
A
F, F
B
T, F
C
F, T
D
T, T
3
JEE Main 2018 (Offline)
+4
-1
The Boolean expression

$$\sim \left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right)$$ is equvalent to
A
$${ \sim q}$$
B
$${ \sim p}$$
C
p
D
q
4
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
Consider the following two statements :

Statement p :
The value of sin 120o can be derived by taking $$\theta = {240^o}$$ in the equation
2sin$${\theta \over 2} = \sqrt {1 + \sin \theta } - \sqrt {1 - \sin \theta }$$

Statement q :
The angles A, B, C and D of any quadrilateral ABCD satisfy the equation
cos$$\left( {{1 \over 2}\left( {A + C} \right)} \right) + \cos \left( {{1 \over 2}\left( {B + D} \right)} \right) = 0$$

Then the truth values of p and q are respectively :
A
F, T
B
T, F
C
T, T
D
F, F
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