Related Articles
Class 11 RD Sharma Solutions – Chapter 31 Derivatives – Exercise 31.6
• Last Updated : 28 Dec, 2020

### Question 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) 100 is completely divisible by 4 and 5 completely without leaving any remainder.Hence, the given statement is true.

(ii) 125 is a multiple of 5 and not divisible perfectly by 7. Hence, the given statement is false.

(iii) The integer 60 is both a multiple of 3 as well as 5. Hence, the given statement is true.

### Question 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 A and B be the components of this statement, which are given by

A: x and y are odd integers.

B: x + y is an even integer

The given statement can be written as :

If A, then B.

Let us assume that A is true. We have, x and y are odd integers.

Let us assume the values to be :

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

Adding the values , we get

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

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

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

x + y is an integer, which is divisible by 2.

Then, B holds true.

Therefore, A is true and B is true.

Hence, if A, then B 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.

### Question 3. Show that the statement

p : “If x is a real number such that x3 + x = 0, then x is 0” is true by

(i) Direct method

(ii) method of Contrapositive

Solution:

(i) Direct Method:

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

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

r: x is 0.

We can write this statement 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

Therefore, q is true and r is true.

Hence, p is true.

(ii) Method of Contrapositive:

Let r be false. Then,

x ≠ 0, x∈R

x(x2+1)≠0, x∈R

q is not true

Thus, -r = -q

Hence, p : q and r is true

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

Hence, p is true.

### Question 4. Show that the following statement is true by the method of the contrapositive

p: “If x is an integer and x2 is odd, then x is also odd.”

Solution:

Let us assume that q and r be the component statements of the given compound statement p,

q: x is an integer and x2 is odd.

r: x is an odd integer.

We can rewrite this statement as,

p: if q, then r.

Let us assume r to be false. Then,

If x is not an odd integer, then x is an even integer

x = (2n) for some integer n

Squaring both sides,

x2 = 4n2

x2 is an even integer, since, it is divisible by 2

Therefore, r is false and q is also false.

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

### Question 5. Show that the following statement is true

“The integer n is even if and only if n2 is even”

Solution:

Let the component statements,

p: Integer n is even

q: If n2 is even

Let us assume p to be true. Then,

Let n = 2k

On squaring both the sides,

n2 = 4k2

n2 = 2 x 2 x k2

n2 is an even number, since it is divisible by 2.

So, q is also true when p is true.

Therefore, the given statement is true.

### Question 6. By giving a counterexample, 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 assume any triangle ABC with all angles equal.

Sum of angles of a triangle is 180 degrees, therefore, each angle of the triangle is equal to 60.

So, this triangle ABC is not an obtuse angled triangle.

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

### Question 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 bisect 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.

A chord of the circle is a line segment joining two end points on its circumference. But radius joins the centre with the end point.

Hence, this statement is False.

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

A circle may have many chords. A chord does not necessarily have to pass through the centre of the circle.

Hence, this statement is False.

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

If the circle has equal axes, then it is equivalent to an ellipse.

Hence, this statement is true.

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

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

If we negate the equation, we get

– x < – y

Hence, this statement is true.

(v) t: √11 is a rational number.

Square root of all prime numbers are irrational numbers.

Hence, this statement is False.

### Question 8. Determine whether the argument used to check the validity of the following statement is correct:

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

The statement is true because the number x2 = π2 is irrational, therefore x = π is irrational.

Solution:

We have the following argument, x2 = π2 is irrational, therefore x = π is irrational.

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

Supposedly, we have an irrational number of the form k, where  x = √k, where k is a rational number.

Squaring both sides of the equations, we get,

x2 = k

Now, since k is rational (given), x2 is also rational. This contradicts our statement.

Therefore, the given argument is wrong.

My Personal Notes arrow_drop_up