**RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6: **We’ll look at the validity of statements in this exercise. Students who want to do well in their board exams should look over RD Sharma Class 11 Solutions. The **RD Sharma Class 11 Solutions** covers the methods in detail, without skipping any important aspects of the problem-solving process. The pdf of **RD Sharma Solutions Class 11 Maths Chapter 31** Exercise 31.6 can be downloaded by students using the links provided below.

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## Download RD Sharma Class 11 Solutions Chapter 31 Exercise 31.6 Free PDF

RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6

### Access answers to RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6- Important Question with Answers

**1. Check the validity of the following statements:(i) p: 100 is a multiple of 4 and 5.(ii) q: 125 is a multiple of 5 and 7.(iii) r: 60 is a multiple of 3 or 5.**

**Solution:**

**(i)** p: 100 is a multiple of 4 and 5.

We know that 100 is a multiple of 4 as well as 5. So, the given statement is true.

Hence, the statement is true.

**(ii)** q: 125 is a multiple of 5 and 7

We know that 125 is a multiple of 5 and not a multiple of 7. So, the given statement is false.

Hence, the statement is false.

**(iii)** r: 60 is a multiple of 3 or 5.

We know that 60 is a multiple of 3 as well as 5. So, the given statement is true.

Hence, the statement is true.

**2. Check whether the following statement is true or not:(i) p: If x and y are odd integers, then x + y is an even integer.**

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

**Solution:**

**(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.

**3. Show that the statementp : “If x is a real number such that x ^{3} + x = 0, then x is 0” is true by(i) Direct method(ii) method of Contrapositive(iii) method of contradiction**

**Solution:**

**(i)** Direct Method:

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

q: x is a real number such that x^{3 }+ 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 x^{3 }+ x = 0

x is a real number such that x(x^{2 }+ 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(x^{2}+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 x^{3}+x = 0and x≠ 0

x =0 and x≠0

This is a contradiction.

Hence, p is true.

**4. Show that the following statement is true by the method of the contrapositivep: “If x is an integer and x ^{2} is odd, then x is also odd.”**

**Solution:**

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

q: x is an integer and x^{2} is odd.

r: x is an odd integer.

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

x^{2} = 4n^{2}

x^{2} is an even integer

Thus, q is False

Therefore, r is false and q is false

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

**5.** **Show that the following statement is true“The integer n is even if and only if n ^{2} is even”**

**Solution:**

Let the statements,

p: Integer n is even

q: If n^{2} is even

Let p be true. Then,

Let n = 2k

Squaring both the sides, we get,

n^{2} = 4k^{2}

n^{2} = 2.2k^{2}

n^{2} is an even number.

So, q is true when p is true.

Hence, the given statement is true.

**RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 Q6.**

**6. By giving a counter example, show that the following statement is not true.p: “If all the angles of a triangle are equal, then the triangle is an obtuse-angled triangle.”**

**Solution:**

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.

**RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 Q7.**

**7. Which of the following statements are true and which are false? In each case give a valid reason for saying so(i) p: Each radius of a circle is a chord of the circle.(ii) q: The centre of a circle bisects each chord of the circle.(iii) r: Circle is a particular case of an ellipse.(iv) s: If x and y are integers such that x > y, then – x < – y.(v) t: √11 is a rational number.**

**Solution:**

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

The Radius of the circle is not its chord.

Hence, this statement is False.

**(ii)** q: The centre of a circle bisects 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.

The square root of prime numbers is irrational numbers.

Hence, this statement is False.

**RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6 Q8.**

**8. Determine whether the argument used to check the validity of the following statement is correct:p: “If x ^{2} is irrational, then x is rational.”The statement is true because the number x^{2} = π^{2} is irrational, therefore x = π is irrational.**

**Solution:**

Argument Used: x^{2} = π^{2} is irrational, therefore x = π is irrational.

p: “If x^{2} 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,

x^{2} = k

x^{2} is a rational number and contradicts our statement.

Hence, the given argument is wrong.

We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 31 Exercise 31.6. If you have any queries related to **CBSE** Class 11, feel free to ask us in the comment section below.

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