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JEE Main

Explore popular questions from Mathematical Reasoning for JEE Main. This collection covers Mathematical Reasoning previous year JEE Main questions hand picked by experienced teachers.

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Q 1. The statement {tex} \mathrm { p } \rightarrow ( \mathrm { q } \rightarrow \mathrm { p } ) {/tex} is equivalent

A

{tex} p \rightarrow ( p \rightarrow q ) {/tex}

{tex} p \rightarrow ( p \vee q ) {/tex}

C

{tex} p \rightarrow ( p \wedge q ) {/tex}

D

{tex} p \rightarrow ( p \leftrightarrow q ) {/tex}

Explanation

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Q 2. Let p be the statement "{tex} x {/tex} is an irrational number", {tex} q {/tex} be the statement "{tex} y {/tex} is a trascendental number", and r be the statement "{tex} x {/tex} is a rational number if {tex} y {/tex} is a transcendental number"
{tex}\mathrm {Statement- 1 } {/tex}: {tex}\mathrm { r } {/tex} is equivalent to either {tex} \mathrm { q } {/tex} or {tex} \mathrm { p } {/tex}.
{tex}\mathrm {Statement- 2 } {/tex}: {tex}\mathrm { r } {/tex} is equivalent to {tex} \left( \mathrm { p } \leftrightarrow ^ {\sim } \mathrm { q } \right) {/tex}

{tex}\mathrm {Statement- 1 } {/tex} is false, Statement {tex} - 2 {/tex} is true

B

{tex}\mathrm {Statement-1} {/tex} is true, {tex}\mathrm {Statement-2} {/tex} is false

C

{tex}\mathrm {Statement-1} {/tex} is true, {tex}\mathrm {Statement-2} {/tex} is true;
{tex}\mathrm {Statement-2} {/tex} is a correct explanation for
{tex}\mathrm {Statement-1} {/tex}

D

{tex}\mathrm {Statement-1} {/tex} is true, {tex}\mathrm {Statement-2} {/tex} is true;
{tex}\mathrm {Statement-2} {/tex} is not a correct explanation for
{tex}\mathrm {Statement-1} {/tex}

Explanation

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Q 3. {tex}\mathrm {Statement-1:} \quad {/tex} {tex}\sim (p{/tex} {tex} \leftrightarrow - \sim q){/tex} is equivalent to {tex} \mathrm { p } \leftrightarrow \mathrm { q } {/tex}.
{tex}\mathrm {Statement- 2 } {/tex} : {tex}\sim ( p{/tex} {tex} \leftrightarrow - {/tex} {tex}\sim q{/tex}) is a tautology.

{tex}\mathrm {Statement-1} {/tex} is true, {tex}\mathrm {Statement-2} {/tex} is false.

B

{tex}\mathrm {Statement-1} {/tex} is false, {tex}\mathrm {Statement-2} {/tex} is true.

C

{tex}\mathrm {Statement-1} {/tex} is true; {tex}\mathrm {Statement-2} {/tex} is true;
{tex}\mathrm {Statement-2} {/tex} is a correct explanation for
{tex}\mathrm {Statement-1.} {/tex}

D

{tex}\mathrm {Statement-1} {/tex} is true, {tex}\mathrm {Statement-2} {/tex} is true;
{tex}\mathrm {Statement-2} {/tex} is not a correct explanation for
{tex}\mathrm {statement-1} {/tex}

Explanation

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Q 4. Let {tex}\mathrm S {/tex} be a non-empty subset of {tex}\mathrm R. {/tex} Consider the following statement:
p: There is a rational number {tex} x \in S {/tex} such that {tex} x > 0 {/tex} which of the following statements is the negation of the statement p?

A

There is a rational number {tex} x \in S {/tex} such that {tex} x \leq 0 {/tex}

B

There is no rational number {tex} x \in S {/tex} such that {tex} x \leq 0 {/tex}

Every rational number {tex} x \in S {/tex} satisfies {tex} x \leq 0 {/tex}

D

{tex} \mathrm { x } \in \mathrm { S } {/tex} and {tex} \mathrm { x } \leq 0 \Rightarrow \mathrm { x } {/tex} is not rational

Explanation

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Q 5. The only statement among the followings that is a tautology is :

A

{tex} \mathrm { q } \rightarrow [ \mathrm { p } \wedge ( \mathrm { p } \rightarrow \mathrm { q } ) ] {/tex}

B

{tex} p \wedge ( p \vee q ) {/tex}

C

{tex} \mathrm { p } \vee ( \mathrm { p } \wedge \mathrm { q } ) {/tex}

{tex} [ p \wedge ( p \rightarrow q ) ] \rightarrow q {/tex}

Explanation

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Q 6. {tex} \left( ^ { ^\sim } \mathrm { p } \vee ^ { ^\sim } \mathrm { q } \right) {/tex} is logically equivalent to-

A

{tex} \mathrm { p } \wedge \mathrm { q } {/tex}

B

{tex} ^\sim \mathrm {p \rightarrow q} {/tex}

{tex} \mathrm {p \rightarrow ^\sim q} {/tex}

D

{tex} ^\sim \mathrm {p \rightarrow ^\sim q} {/tex}

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Q 7. The statement {tex} ( \mathrm {p \rightarrow ^\sim p} ) \wedge ( ^\sim \mathrm {p \rightarrow p} ) {/tex} is -

A

a tautology

a contradiction

C

neither a tautology nor a contradiction

D

None of these

Explanation

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Q 8. Which of the following is correct-

A

{tex} \left( ^\sim _ { \mathrm { p } } \vee ^\sim _ { \mathrm { q } } \right) \equiv ( \mathrm { p } \wedge \mathrm { q } ) {/tex}

{tex} ( \mathrm {p \rightarrow q} ) \equiv ( ^\sim \mathrm {q \rightarrow ^\sim p} ) {/tex}

C

{tex} ^\sim ( \mathrm {p \rightarrow ^\sim q} ) \equiv ( \mathrm {p \wedge ^\sim q} ) {/tex}

D

{tex} ( \mathrm {p \leftrightarrow q ) \equiv ( p \rightarrow q ) \vee ( q \rightarrow p} ) {/tex}

Explanation

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Q 9. The statement {tex} {\sim }( \mathrm {p \rightarrow q} ) \leftrightarrow \left( \sim { \mathrm {p } \vee \sim { q} } \right) {/tex} is -

A

a tautology

B

a contradiction

neither a tautology nor a contradiction

D

both tautology and contradiction

Explanation

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Q 10. Which of the following is wrong-

A

{tex} \mathrm { p } \vee ^\sim \mathrm { p } {/tex} is a tautology

B

{tex} ^\sim ( ^\sim \mathrm {p} ) \leftrightarrow \mathrm {p} {/tex} is a tautology

C

{tex} \mathrm { p } \wedge ^\sim \mathrm { p } {/tex} is a contradiction

{tex} ( ( \mathrm {p \wedge p ) \rightarrow q ) \rightarrow p} {/tex} is a tautology

Explanation

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Q 11. If statement {tex} \left( \mathrm { p } \vee ^ { \sim } \mathrm { r } \right) \rightarrow ( \mathrm { q } \wedge \mathrm { r } ) {/tex} is false and statement {tex} \mathrm { q } {/tex} is true then statement {tex} \mathrm { p } {/tex} is-

A

True

B

False

may be true or false

D

None of these

Explanation

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Q 12. Simplify (p∨q) ∧ (p ∨ ∼ q)

p

B

T

C

F

D

q

Explanation

(p∨q) ∧ (p ∨ ∼ q)

= p ∨ (q ∧ ∼ q) (distributive law)

= p ∨ 0 (complement law)

= p (0 is identify for v)

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Q 13. The statement p ⇒ p ∨ q is

A tautology

B

A contradiction

C

Both a tautology and contradiction

D

Neither a tautology nor a contradiction

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Q 14. p → q is logically equivalent to

A

p ∧ ∼ q

B

∼ p → ∼ q

∼ q → ∼ p

D

None of these

Explanation

We have

p → q ≅ ∼ p ∨ q

and, ∼ q → ∼ p ≅ ∼ (∼q) ∨ ∼ p ≅ q ∨ ∼ p ≅ ∼ p ∨ q ≅ p → q

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Q 15. Which of the following is logically equivalent to p ∧ q?

A

p → ∼ q

B

∼ p ∨ ∼ q

∼ (p → ∼ q)

D

∼ ( ∼ p ∧ ∼ q)

Explanation

We have,

p → q ≅ ∼ p ∨ q

∴ p → ∼ q ≅ ∼ p ∨ ∼ q ≅ ∼ (p ∧ q)

So, option (a) is not correct

∼ p ∨ ∼ q = ∼ (p ∧ q)

So, option (b) is not correct

∼ (p→∼q) = ∼ (∼p∨∼q) = p ∧ q

So, option (c) is incorrect

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Q 16. Some triangles are not isosceles. Identify the Venn diagram

A

C

D

Explanation

Some triangles are not isosceles.

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Q 17. Which of the following is contingency?

A

p ∨ ∼ p

B

p ∧ q ⇒ p ∨ q

p ∧ ∼ q

D

None of these

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Q 18. ∼ (p ∨ q) ∨ ( ∼ p ∧ q) is logically equivalent to

∼ p

B

p

C

q

D

∼ q

Explanation

∼ (p∨q) ∨ (∼p∧q)

≅ ( ∼ p ∧ ∼ q) ∨ ( ∼ p ∧ q)

≅ ∼ p ∧ ( ∼ q ∨ q) ≅ ∼ p ∨ t ≅ ∼ p

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Q 19. A compound sentence formed by two simple statements p and q using connective 'or' is called

A

Conjunction

Disjunction

C

Implication

D

None of these

Explanation

A compound sentence formed by two simple statements p and q using connective 'or' is called disjunction

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Q 20. If p and q are two statements, then p ∨ ∼ (p ⇒ ∼ q) is equivalent to

A

p ∧ ∼ q

p

C

q

D

∼ p ∧ q

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Q 21. Let p ∧ (q∨r) = (p∧q) ∨ (p ∧ r). Then, this law is known as

A

Commutative law

B

Associative law

C

De-Morgan's law

Distributive law

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Q 22. If p and q are two statements, then statement p ⇒ q ∧ ∼ q is

A

Tautology

B

Contradiction

Neither tautology not contradiction

D

None of the above

Explanation

By truth table


p

q

∼ q

q ∧ ∼ q

p ⇒ q ∧ ∼ q

T

T

F

F

T

F

T

F

F

T

F

T

F

F

F

F

F

F

T

T

Hence, it is neither a tautology nor contradiction

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Q 23. Which of the following is logically equivalent to ∼ ( ∼ p → q)?

A

p ∧ q

B

p ∧ ∼ q

C

∼ p ∧ q

∼ p ∧ ∼ q

Explanation

We have,

p → q ≅ ∼ p ∨ q

∴ ∼ ( ∼ p → q) ≅ ∼ (p ∨ q) ≅ ∼ p ∧ ∼ q

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Q 24. The statement (p⇒q) ⇔ ( ∼ p ∧ q) is a

Tautology

B

Contradiction

C

Neither (a) nor (b)

D

None of these

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Q 25. A compound sentence formed by two simple statements p and q using connective 'and' is called

Conjunction

B

Disjunction

C

Implication

D

None of these