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The given sentence is a compound statement in which components are

p: 2 is a factor of 12

q: 3 is a factor of 12

r: 6 is a factor of 12

Now, it can be represent in symbolic function as,

p ᴧ q ᴧ r: 2, 3 and 6 are factors of 12.

The given sentence is a compound statement in which components are

p: x is an even integer

q: y is an even integer

Now, it can be represent in symbolic function as,

p ᴧ q: x and y are even integers.

(i) If you access the website, then you pay a subscription fee.

(ii) If it rains, then there is a traffic jam.

(iii) If you log on the server, then you must have a passport.
(iv) If he is happy, then he is rich.

(v) If it is raining, then the game is canceled.

(vi) If it rains, then it is cold.

(vii) If it rains, then it is cold.

(viii) If it is cold, then it never rains.

The given sentence is a compound statement in which components are

p: Rahul passed in Hindi

q: Rahul passed in English

Now, it can be represent in symbolic function as,

p ᴧ q: Rahul passed in Hindi and English.

(i) Negation of statement p is "not p." The negation of p is symbolized by "~p." The truth value of ~p is the opposite of the truth value of p. 

The negation of the statement 

All the students complete their homework. 

is 

Some of the students did not complete their homework. 

(ii) The negation of the statement 

There exists a number which is equal to its square. is 

For every real number x, x²≠x.

(i) All the students complete their homework.

The negation of the statement is

“Some of the students did not complete their homework.”

(ii) There exists a number which is equal to its square.

The negation of the statement is

“For every real number x, x²≠x.”

(i) All birds sing.

The negation of the statement is:

It is false that “All birds sing.”

Or

“All birds do not sing.”

(ii) Some even integers are prime.

The negation of the statement is:

It is false that “even integers are prime.”

Or

“Not every even integers is prime.”

(iii) There is a complex number which is not a real number.

The negation of the statement is:

It is false that “complex numbers are not a real number.”

Or

“All complex number are real numbers.”

(iv) I will not go to school.

The negation of the statement is:

“I will go to school.”

(v) Both the diagonals of a rectangle have the same length.

The negation of the statement is:

“There is at least one rectangle whose both diagonals do not have the same length.”

(vi) All policemen are thieves.

The negation of the statement is:

“No policemen are thief”.

(i) prime factors of 6 are 2 and 3.

Statement is true.

(ii) All complex number are real numbers.

Statement: It has false values.

The given statement is compound statement then components are,

P: All real numbers are rational.

~p: All real numbers are not rational.

q: All real numbers are irrational.

~q: All real numbers are not irrational.

(p ᴧ q)= All real numbers are rationals or irrationals.

~(p ᴧ q)=~p v ~q= All real numbers are neither rationals nor irrationals.

The given sentence is a compound statement in which components are

p: Hindi is the optional paper

q: English is the optional paper

Now, it can be represent in symbolic function as,

p ᴧ q: Either Hindi or English is optional paper.

The given statement is compound statement then components are,

P: All rational numbers are real.

~p: All rational numbers are not real.

q: All rational numbers are complex.

~q: All rational numbers are not complex.

(p ᴧ q)= All rational numbers are real and complex.

~ (p ᴧ q)=~p v ~q= All rational numbers are neither real nor complex.

(i) Let p : A number is divisible by 9 

q : A number is divisible by 3 

Its converse is q ⇒ p and its contrapositive is ~q ⇒~P 

∴ contrapositive: ‘If a number is not divisible by 3, then it is not divisible by 9’. 

Converse: ‘If a number is divisible by 3 then it is divisible by 9. 

(ii) Contrapositive: ‘If you are not a citizen of India then you were not born in India’. 

Converse: ‘If you are a citizen of India then you are born in India’. 

(iii) Contrapositive: ‘If a triangle is not isosceles, then it is not equilateral’. 

Converse: ‘If a triangle is isosceles, then it is equilateral’.

(iv) Contrapositive: ‘If x is not odd, then x is not a prime number’. 

Converse: If x is odd then x is a prime number

(i) ‘If you get a job, then your credentials are good’. 

(ii) ‘If the banana tree stays warm for a month, then it will bloom’ 

(iii) ‘If the diagonals of a quadrilateral bisect each other, then it is a parallelogram’. 

(iv) ‘If you get an A⁺  in the class then you have done all exercises of the book’.

The compound statement is-

“30 is a multiple of 5 and 8.”.

The statement is not valid since 30 is not a multiple of 8.

Let n be even, n = 2m

∴ n³ − n = n(n² − 1) = 2m (4m² − 1), which is even.

If n is odd, n = 2m + 1.

Then, n³ − n = (2m + 1)³ − (2m + 1)

= (2m + 1){4m² + 4m + 1 − 1}

= 4(m² + m) (2m + 1), which is also even.

∴ For any integer n, n³ − n is always even.

Let us assume

p: n² is an even integer.

~p: n is not an even integer

q: n is also an even integer

~q=n is not an even integer.

Since, in the contrapositive, a conditional statement is logically equivalent to its contrapositive.

Therefore,

~q → ~p = If n is not an even integer then n² is not an even integer.

Hence, ~q is true → ~p is true.

Let p: x is an integer and x² is even.

q: x is even

For contrapositive,

~p = x is an integer and x² is not even.

~q = x is not even.

Now, the statement is: If x is an integer and x² is not even, then x is not even.

Proof:

Let x be an odd integer and x = 2n + 1

⇒x² = (2n + 1) ² = 4n² + 4n + 1 (odd integer)

Thus, if x is an integer and x² is not even, then x is not even.

Let p: n² is an even integer. q: n is also an even integer.

Let ~p is true i.e., n is not an even integer.

=> n is not an even integer. [Since square of an odd integer is odd]

=> ~ p is true.

Therefore, ~q is true which provides that ~p is true.

Hence proved.

(i) Mathematics is fun.

The given sentence is not statement because Mathematics is fun for some people not for all.

(ii) 6 + 1 = 7

The given sentence is a statement because it is true.

(i) We have,P : 125 is divisible by 5 and 7.

Let q: 125 is divisible by 5.

r: 125 is divisible by 7. q is true, r is false.

=> q ⋀ r is false.

[since, p ⋀ q has the truth value F (false) whenever eitherp or q or both have the truth value F]

Hence, p is not valid.

(ii) We have, p: 131 is a multiple of 3 or 11.

Let q: 131 is multiple of 3.

r: 131 is a multiple of 11.

p is true, r is false.

=> p v r is true.

[since, p v q has the truth value T (true) whenever either p or q or both have the truth value T]

Hence, q is valid.

Prime number definition, a number must only have itself and 1 as its factors.

Let us take an odd positive integer, n = 15

Since 15 is an odd positive integer but not prime number.

Thus, statement p is false.

(i) Let p: A line is straight.

And q: A line extends indefinitely in both the directions.

Both the sentences are True.

Therefore, the given sentence is TRUE.

(ii) Let p: A point occupies a position.

And q: Its location can be determined.

Both the sentences are True.

Therefore, the given sentence is TRUE.

(iii) Let p: The sand heats up quickly in the sun.

And q: The sand does not cool down fast at night.

Both the sentences are True.

Therefore, the given sentence is TRUE.

(iv) Let p: 32 is divisible by 8.

And q: 32 is divisible by 12.

The first sentence is True and the second sentence is False.

Therefore, the given sentence is FALSE.

(v) Let p: x = 1 is a root of the equation x² – x – 2 = 0

And q: x = 2 is a root of the equation x² – x – 2 = 0

The first sentence is False and the second sentence is True.

Therefore, the given sentence is FALSE.

(vi) Let p: 3 is rational.

And, q: √3 is irrational.

Both the sentences are True.

Therefore, the given sentence is TRUE.

(vii) Let p: All integers are rational numbers.

And, q: All rational numbers are not real numbers.

The first sentence is True and the second sentence is False.

Therefore, the given sentence is FALSE.

(viii) Let p: Lucknow is in Uttar Pradesh.

And q: Kanpur is in Uttarakhand.

The first sentence is True and the second sentence is False.

Therefore, the given sentence is FALSE.

(i) p: Each radius of a circle is a chord of the circle.

The Radius of the circle is not it chord.

Hence, this statement is False.

(ii) q: The centre of a circle bisect each chord of the circle.

A chord does not have to pass through the center.

Hence, this statement is False.

(iii) r: Circle is a particular case of an ellipse.

A circle can be an ellipse in a particular case when the circle has equal axes.

Hence, this statement is true.

(iv) s: If x and y are integers such that x > y, then – x < – y.

For any two integers, if x – y id positive then – (x - y) is negative.

Hence, this statement is true.

(v) t: √11 is a rational number.

Square root of prime numbers is irrational numbers.

Hence, this statement is False.

(i) Paris is in France, is a statement.

Paris is located in France, so the sentence is true.

So, the statement is true.

(ii) Each prime number has exactly two factors, is a statement.

This is a mathematically proven fact.

So, the statement is true.

(iii) The equation x² + 5|x| + 6 = 0 has no real roots.

Find the roots of x² + 5|x| + 6 = 0:

Case 1: x ≥ 0

x² + 5x + 6 = (x + 2)(x + 3) = 0 ⇒ x = −2, −3 but we already assumed x ≥ 0, which is a contradiction.

Case 2: x < 0

x² − 5x + 6 = (x − 2)(x − 3) = 0 ⇒ x = (2,3) but we already assumed x < 0, which is a contradiction.

So, equation x² + 5|x| + 6 = 0 has no real roots.

Therefore, the given sentence is true, and it is a statement.

(iv) (2 + √3) is a complex number, is a statement.

Complex numbers are in the form ‘a+ib’.

(2 + √3) cannot be expressed in ‘a+ib’ form,.

2 + √3 is not a complex number.

The given sentence is a statement, and it is false.

(v) Is 6 a positive integer?

This is an interrogative sentence, so it is not a statement.

(vi) The product of -3 and -2 is -6, is a statement.

Product of -3 and -2 = -3 x -2 = 6 ≠ -6

This statement is false.

(vii) The angles opposite the equal sides of an isosceles triangle are equal, is a statement.

It is a mathematically proven result.

So the given sentence is true.

(viii) Oh! it is too hot.

This is an exclamatory sentence, so it is not a statement.

(ix) Monika is a beautiful girl, is not a statement.

The given sentence is an opinion, can be true for some cases, false for some other case.

(x) Every quadratic equation has at least one real root, is a statement.

Because not every quadratic equation will have a real root.

So the given sentence is false.

(i) This statement is False 

Because the Radius of the circle is not it chord 

(ii) This statement is False 

Because A chord does not have to pass through the center. 

(iii) This statement is true, 

Because a circle can be an ellipse in a particular case when the circle has equal axes. 

(iv) This statement is true, 

Because for any two integer, if x – y id positive then –(x-y) is negative 

(v) This statement is False 

Because square root of prime numbers are irrational numbers.

(i) p: √11 is an irrational number is a TRUE statement. 

An irrational number is any number which cannot be expressed as a fraction of two integers. √11 cannot be expressed as a fraction of two integers, so √11 is an irrational number. 

(ii) q: Circle is a particular case of an ellipse is a TRUE statement. 

A circle is a particular case of an ellipse with the same radius in all points.

The equation of an ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1

When a = b, we will get the equation of the circle, x² + y² = 1 

(iii) r: Each radius of a circle is a chord of the circle is a FALSE statement. 

A chord intersects the circle at two points, but radius intersects the circle only at one point. So the radius is not a chord of the circle. 

(iv) S: The center of a circle bisects each chord of the circle is a FALSE statement. 

The only diameter of a circle is bisected by the center of the circle. Except for diameter, no other chords are bisected the center of the circle. The only center lies on the diameter of the circle. 

(v) t: If a and b are integers such that a < b, then –a > -b is a TRUE statement. 

a < b, then –a > -b, is TRUE by the rule of inequality.

(vi) y: The quadratic equation x² + x + 1 = 0 have no real roots is a TRUE statement. 

General form of a quadratic equation is ax² + bx + c = 0. 

If b² – 4ac < 0, there is no real solution. 

In the given equation; x² + x + 1 = 0 

a = 1; b = 1; c = 1 

b² – 4ac = 1 – 4 x 1 x 1 = -3 < 0 

So, there is no real root.

(i) p: √11 is an irrational number.

True statement.

Reason:

An irrational number is any number which cannot be expressed as a fraction of two integers.

√11 cannot be expressed as a fraction of two integers, so √11 is an irrational number.

(ii) q: Circle is a particular case of an ellipse.

True statement.

Reason:

The equation of an ellipse is x²/a² + y²/b² = 1

Special case: When a = b

Then x² + y² = 1, which is an equation of circle.

So, we can say that, a circle is a particular case of an ellipse with the same radius in all points.

(iii) r: Each radius of a circle is a chord of the circle.

False statement.

Reason:

A chord intersects the circle at two points, but radius intersects the circle only at one point.

So the radius is not a chord of the circle.

(iv) S: The center of a circle bisects each chord of the circle.

False statement.

Reason:

The only diameter of a circle is bisected by the center of the circle. Except for diameter, no other chords are passes through the center of a circle.

(v) t: If a and b are integers such that a < b, then –a > -b.

True statement.

Reason:

a < b, then –a > - b [By rule of inequality]

(vi) y: The quadratic equation x² + x + 1 = 0 has no real roots.

True statement.

Reason:

General form of a quadratic equation, ax² + bx + c = 0, has no real roots if discriminant, D < 0.

Where D= b² – 4ac < 0.

Given equation; x² + x + 1 = 0

Here, a= 1, b = 1 and c = 1

Now, b² – 4ac = 1 – 4 x 1 x 1 = -3 < 0

So, there is no real root.

(c) As we know that a statement is a sentence which is either true or false.

6 is a natural number; this is true.

Hence, it is a statement

(C) 6 is a natural number.

Explanation:

A statement is an assertive (declarative) sentence if it is either true or false but not both.

Here, 6 is a natural number is true

(d) No sentence can be called a statement, if it is an order. So, ‘Come here’ is not a statement.

B. or

In the statement “2 + 7 > 9 or 2 + 7 < 9”

Since, Or is connecting two statement.

(D) Come here.

Explanation:

To given order like Come here, Go there are not statements.

B. Earth

In the statement “Earth revolves round the Sun and Moon is a satellite of earth” And is connective.

Correct option  (a) a contradiction

Explanation:

(p ∧ ~q) ∧ (~p ∧ q) ≡ (p ∧ ~p) ∧ (~q ∧ q) 

≡ (f ∧ f) 

≡ f

(f → false)

(By using associative laws and commutative laws)

(p ∴ ~q) ∧ (~p ∧ q) is a contradiction.

(i) You watch television if and only if your mind is free. 

(ii) A quadrilateral is a rectangle if and only if it is equiangular. 

(iii) You get an A grade if and only if you do all the homework regularly. 

(iv) A tumbler is half empty if and only if it is half full.

The negation of statement is as follows.

(i) There exists real number x and y for which x + y ≠ y + x. This is not the same as statement 

(ii). Thus, the given statements are not the negation of each other.

In order to test the validity of the argument (S₁ , S₂ ; S), we first construct the truth table for the conditional statement.

S₁ ∧ S₂ → S i.e [(p ∨ q) ∧ ~p] →  q The truth table is as given below:

The truth table is as given below:

We observe that the last comumn of the truth table for S₁ ∧ S₂ ∧ S₂ → contains T only.

No Negation of 

(i) is ‘there exists real number x and y for which x + y & y + x’, which is not same as (ii).

Let p: x = y; x, y∈ R

On squaring both sides, we get

x² = y² : q

p ⟹q

Hence, proved.

Argument Used: 

x² = π² is irrational, therefore x = π is irrational. 

p: “If x² is irrational, then x is rational.” 

Let us take an irrational number given by x = √k, where k is a rational number. 

Squaring both sides, we get, 

x² = k 

Therefore, x² is a rational number and contradicts our statement. 

Hence, the given argument is wrong.

Given for any real number x, y if x=y

Now we have to find x² = y²

Let us assume

p: x=y where x and y are real number

On squaring both sides we get

x² = y² : q (Assumption)

Therefore,

p = q

Hence, Proved

Argument Used: x² = π² is irrational, therefore x = π is irrational.

p: “If x² is irrational, then x is rational.”

Let us take an irrational number given by x = √k, where k is a rational number.

Squaring both sides, we get,

x² = k

x² is a rational number and contradicts our statement.

Hence, the given argument is wrong.

Let the statements, 

p: Integer n is even 

q: If n² is even 

Let p be true. 

Then 

Let n = 2k 

Squaring both the sides, we get, 

n² = 4k² 

n² = 2.2k² 

Therefore, n² is an even number. 

So, q is true when p is true. 

Hence, the statement is true.

Let us Assume that q and r be the statements given 

q: x is an integer and x² is odd. 

r: x is an odd integer. 

since the given statement can be written as : 

p: if q, then r. 

Let r be false . then, 

x is not an odd integer, then 

x is an even integer 

x = (2n) for some integer n 

x² = 4n² 

x² is an even integer 

Thus, q is False 

Therefore, r is false ⇒ q is false 

Hence, p: “ if q, then r” is a true statement.

(i) Direct Method:

Let us assume that ‘q’ and ‘r’ be the statements given by

q: x is a real number such that x³ + x = 0.

r: x is 0.

The given statement can be written as:

if q, then r.

Let q be true. Then, x is a real number such that x³ + x = 0

x is a real number such that x(x² + 1) = 0

x = 0

r is true

Thus, q is true

Therefore, q is true and r is true.

Hence, p is true.

(ii) Method of Contrapositive:

Let r be false. Then,

R is not true

x ≠ 0, x∈R

x(x² + 1) ≠ 0, x ∈ R

q is not true

Thus, -r = -q

Hence, p : q and r is true

(iii) Method of Contradiction:

If possible, let p be false. Then,

P is not true

-p is true

-p (p => r) is true

q and –r is true

x is a real number such that x³ + x = 0 and x ≠ 0

x = 0 and x ≠ 0

This is a contradiction.

Hence, p is true.

(i) p: If x and y are odd integers, then x + y is an even integer.

Let us assume that ‘p’ and ‘q’ be the statements given by

p: x and y are odd integers.

q: x + y is an even integer

the given statement can be written as :

if p, then q.

Let p be true. Then, x and y are odd integers

x = 2m+1, y = 2n+1 for some integers m, n

x + y = (2m+1) + (2n+1)

x + y = (2m+2n+2)

x + y = 2(m+n+1)

x + y is an integer

q is true.

So, p is true and q is true.

Hence, “if p, then q “is a true statement.”

(ii) q: if x, y are integer such that xy is even, then at least one of x and y is an even integer.

Let us assume that p and q be the statements given by

p: x and y are integers and xy is an even integer.

q: At least one of x and y is even.

Let p be true, and then xy is an even integer.

So,

xy = 2(n + 1)

Now,

Let x = 2(k + 1)

Since, x is an even integer, xy = 2(k + 1). y is also an even integer.

Now take x = 2(k + 1) and y = 2(m + 1)

xy = 2(k + 1).2(m + 1) = 2.2(k + 1)(m + 1)

So, it is also true.

Hence, the statement is true.

Let us consider a triangle ABC with all angles equal.

Then, each angle of the triangle is equal to 60.

So, ABC is not an obtuse angle triangle.

Hence, the statement “p: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle” is False.

(i) Given: ‘If a natural number is odd, then its square is also odd. 

Let p : a natural number is odd 

q : a natural number square is also odd. Then, if p then q is same as 

• p ⇒q (p implies q) 
• i.e., A natural number is odd implies that its square is also odd. 
• p is sufficient condition for q e., for the square of a natural number to be odd it is sufficient that the number itself is odd. 
• p only if q e., A natural number is odd only if its square is odd. 
• q is a necessary condition for p e., for a natural number to be odd it is necessary that its square must be odd. 
• ~q implies ~p 
• If the square of a natural number is not odd, then the number itself is also not odd.

The given statement can be written in five different ways as follows.

(i) A triangle is equiangular implies that it is an obtuse-angled triangle.

(ii) A triangle is equiangular only if it is an obtuse-angled triangle.

(iii) For a triangle to be equiangular, it is necessary that the triangle is an obtuse-angled triangle.

(iv) For a triangle to be an obtuse-angled triangle, it is sufficient that the triangle is equiangular.

(v) If a triangle is not an obtuse-angled triangle, then the triangle is not equiangular.

(i) Here ‘or’ is exclusive 

∵ ‘sun rises’ and ‘moon sets’ cannot be true simultaneously 

(ii) Here ‘or’ is inclusive 

∵ For license, one can have both ration card and a passport. 

(iii) Here ‘or’ is exclusive 

∵ Integer cannot be both positive as well as negative.