Let p:n² is an even integer
and q: n is also an even integer.
Let\(\sim \)p is true, i.e. n is not an even integer.
\(\Rightarrow \)n² is not an integer.
\(\Rightarrow \) \(\sim \) p is true.

Let p:x is even integer.
q:x is prime
We have to show that,p \(\Rightarrow \) q is false, i.e. p-> \(\sim \)q.
For this, we have to given an example.
Now, take a number greater than x which is even, but not a prime. So, x=10 is counter example.
Thus, we conclude that given statement is false.

Valid

The compound statement is '100 is multiple of 5 and 4'.
The compound statement 'p and q' is a valid statement.

The given statement is of the form "q if p", where
p : Mohan studies hard.
q : He will be a good students.
It is an equivalent form of the statement "if p then q". So the equivalent formulation of the given statement is "if Mohan studies hard, then he will be a good student".
(Here, note that in p, he is replaced by Mohan and in q, Mohan is replaced by he.)

The conjunction of the statement p and q is given by 
\(p\wedge q\): Mohan is honest and qualified
The disjunction of the statement p and q is given by
\(p\vee q\) :Mohan is honest or qualified

(i) The negation of the given statements is It is false that sum of 2 and 3 is 4
or
The sum of 2 and 3 is not equal to 4
This statements is true because 2+3= 5 \(\neq \) 4
(ii) The negation of the given statements is
It is false that \(\sqrt { 3 } \) is rational
or
\(\sqrt { 3 } \) is not rational
This statements is true because \(\sqrt { 3 } \) is rational
(iii) The negation of the given statements is It is false thet New Delhi is a city
or
New Delhi is not a city
This statements is false.
(iv) The negation of the given staement is 
It is false that every natural number is greater than 0.
This is a false statement.

Here 'OR' is exclusive, because a lady cannot gave birth to a baby who is both boy and a girl

(i) The component statements are
p: It is raining
q: The sun is shinning
The connecting word is 'or'
(ii) The component statements are 
p: All primes are even
q: All primes are odd
The connecting word is 'or'
(iii) The component statements are
p:Two lines intersect at a point
q:Two lines are parallel.
The connecting word is 'or'

2 + 4 \(\le \)5 and 3 + 4 \(\ge \)6

Let given statement be false.i.e \(\sqrt { 11 } \) is rational.
It means \(\sqrt { 11 } \)=\(\frac { p }{ q } \), where p and q are coprime and q \(\neq \) 0
\(\Rightarrow \)   \(11=\frac { { p }^{ 2 } }{ { q }^{ 2 } } \Rightarrow { p }^{ 2 }=11{ q }^{ 2 }\)
It means 11 divides p.
Thus, there exists an integer r such that
p = 11r \(\Rightarrow \)p² = 121r²
It means 11 divides q.
It means 11 is a common factor of p and q which contradicts our assumption that p and q have no common factor.

Compound statement is false and its component statements are
p: Every rectangle is a square
q: Every square is a rectangle

(i) A chord of a circle is a line whose two endpoints lie on the circle and all the points on the line lie inside the circle. So the radius of a circle is not a chord of the circle. Thus the given statement is false.
(ii) The centre of a circle bisects chord of circle when the chord is diameter of circle. When the chord is other than diameter then centre of circle does not lie on the chord. Thus the given statement is false.
(iii) In the equation of an ellipse is we put a = b then we get an equation of circle. Thus the given statement is true.
(iv) It is given that x, y\(\in \) Z such that x > y. Multiplying both sides by negative sign, we have x,y\(\in \) Z such that - x < - y.
Thus the given statement is true.
(v) Since \(\sqrt { 11 } \) cannot be expressed in the form  \(\frac { a }{ b } \) where a and b are integers and b\(\neq \)0
Thus the given statement is false.

The given compound statement is of the form "if p then q".
p: x \(\in \) R such that.x³ + 4x = 0
q: x= 0
(i) Direct method:
We assume that p is true then
x \(\in \) R such that x³ + 4x = 0
\(\Rightarrow \) x \(\in \) R such x (x²+ 4) = 0
\(\Rightarrow \) x \(\in \) R such that x = 0 or x²+ 4 = 0
\(\Rightarrow \)x = 0
\(\Rightarrow \)q is true
So when p is true, q is true.
Thus the given compound statement is true.
(ii) Method of contradiction:
We assume that p is true and q is false. then
x \(\in \) R such that x³+4x=0
\(\Rightarrow \) x \(\in \) R such that x(x² + 4) = 0
\(\Rightarrow \) x \(\in \) R such that x=0 or x² +4=0
\(\Rightarrow \) x = 0
which is a contradiction. So our assumption that x \(\neq \) 0 is false. Thus the given compound statement is true.
(iii) Method of contrapositive:
We assume that q is false, then
X \(\neq \) 0
\(\Rightarrow \) x \(\in \) R such that x³+ 4 x \(\neq \) 0
\(\Rightarrow \) p is false
So when q is false, p is false.
Thus the given compound statement is true.

The three examples of sentences, which are not statements, are as follows.
(i) He is a doctor.
It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement.
(ii) Geometry is difficult.
This is not a statement because for some people, geometry can be easy and for some others, it can be difficult.
(iii) Where is she going?
This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is not a statement.