This chapter is currently out of syllabus
1
JEE Main 2020 (Online) 5th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
The negation of the Boolean expression x $$\leftrightarrow$$ ~ y is equivalent to :
A
$$\left( {x \wedge y} \right) \vee \left( { \sim x \wedge \sim y} \right)$$
B
$$\left( { \sim x \wedge y} \right) \vee \left( { \sim x \wedge \sim y} \right)$$
C
$$\left( {x \wedge y} \right) \wedge \left( { \sim x \vee \sim y} \right)$$
D
$$\left( {x \wedge \sim y} \right) \vee \left( { \sim x \wedge y} \right)$$
2
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Contrapositive of the statement :
‘If a function f is differentiable at a, then it is also continuous at a’, is:
A
If a function f is continuous at a, then it is not differentiable at a.
B
If a function f is not continuous at a, then it is differentiable at a.
C
If a function f is not continuous at a, then it is not differentiable at a.
D
If a function f is continuous at a, then it is differentiable at a.
3
JEE Main 2020 (Online) 4th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Given the following two statements:

$$\left( {{S_1}} \right):\left( {q \vee p} \right) \to \left( {p \leftrightarrow \sim q} \right)$$ is a tautology

$$\left( {{S_2}} \right): \,\,\sim q \wedge \left( { \sim p \leftrightarrow q} \right)$$ is a fallacy. Then:
A
both (S1) and (S2) are not correct
B
only (S1) is correct
C
only (S2) is correct
D
both (S1) and (S2) are correct
4
JEE Main 2020 (Online) 3rd September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Let p, q, r be three statements such that the truth value of
(p $$\wedge$$ q) $$\to$$ ($$\sim$$q $$\vee$$ r) is F. Then the truth values of p, q, r are respectively :
A
T, F, T
B
F, T, F
C
T, T, T
D
T, T, F
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