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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

(iii) method of contradiction

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

(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 shows a contradiction.

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.

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