NCERT Solutions for Exercise 14.5 Class 11 Maths Chapter 14 - Mathematical Reasoning

# NCERT Solutions for Exercise 14.5 Class 11 Maths Chapter 14 - Mathematical Reasoning

Edited By Ravindra Pindel | Updated on Jul 13, 2022 06:30 PM IST

In the previous exercises of this chapter, you have already learned about the mathematical statements, the negation of a statement, compound statements, quantifiers, implications, contrapositive and converse, etc. In the NCERT solutions for Class 11 Maths chapter 12 exercise 14.5, you will learn about validating mathematical statements. There are three methods to validate any mathematical statement.

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• Direct method
• Method of contrapositive

Exercise 14.5 Class 11 Maths is very brief where you will get questions related to validation of mathematical statements with the reason for validation the statement. This NCERT syllabus exercise is very important for the CBSE Exam as well as engineering entrance exams like JEE Main, SRMJEE, etc. Generally, one question from this chapter is asked in JEE Main exam. You can click on the NCERT Solutions link where you will get NCERT exercise solutions for Classes 6 to 12 at one place.

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## Mathematical Reasoning Class 11 Chapter 14-Exercise: 14.5

If $x$ is a real number such that $x^3 + 4x = 0$, then $x$ is 0 : (if p then q)

p: x is a real number such that $\dpi{100} x^3 + 4x = 0$.

q: x is 0.

In order to prove the statement “if p then q”

Direct Method: By assuming that p is true, prove that q must be true.

So,

p is true:There exists a real number x such that $\dpi{100} x^3 + 4x = 0 \implies x(x^2 + 4) = 0$

$\dpi{100} \implies x = 0\ or\ (x^2 + 4)= 0$

$\dpi{100} \implies x = 0\ or\ x^2 = -4\ (not\ possible)$

Hence, x = 0

Therefore q is true.

If $x$ is a real number such that $x^3 + 4x = 0$, then $x$ is 0 : (if p then q)

p: x is a real number such that $\dpi{100} x^3 + 4x = 0$.

q: x is 0.

In order to prove the statement “if p then q”

Contradiction: By assuming that p is true and q is false.

So,

p is true: There exists a real number x such that $x^3 + 4x = 0$

q is false: $\dpi{100} x \neq 0$

Now, $\dpi{100} x^3 + 4x = 0 \implies x(x^2 + 4) = 0$

$\dpi{100} \implies x = 0\ or\ (x^2 + 4)= 0$

$\dpi{100} \implies x = 0\ or\ x^2 = -4\ (not\ possible)$

Hence, x = 0

But we assumed $\dpi{100} x \neq 0$. This contradicts our assumption.

Therefore q is true.

If $x$ is a real number such that $x^3 + 4x = 0$, then $x$ is 0 : (if p then q)

p: x is a real number such that $\dpi{100} x^3 + 4x = 0$.

q: x is 0.

In order to prove the statement “if p then q”

Contrapositive Method: By assuming that q is false, prove that p must be false.

So,

q is false: $\dpi{100} x \neq 0$

$\dpi{100} \implies$ x.(Positive number) $\dpi{100} \neq$ 0.(Positive number)

$\dpi{100} \implies x(x^2 + 4) \neq 0(x^2 + 4)$

$\dpi{100} \implies x(x^2 + 4) \neq 0 \implies x^3 + 4x \neq 0$

Therefore p is false.

Given,

For any real numbers a and b, $a^2 = b^2$ implies that $a = b$.

Let a = 1 & b = -1

Now,

$\dpi{100} a^2 = (1)^2$= 1

$\dpi{100} b^2 = (-1)^2$ = 1

$\implies a^2 =1= b^2$

But a $\dpi{80} \neq$ b

Hence $a^2 = b^2$ does not imply that $a = b$.

Hence the given statement is not true.

p: If x is an integer and $x^2$ is even, then $x$ is also even.

Given, If x is an integer and $x^2$ is even, then $x$ is also even.

Let, p : x is an integer and $x^2$ is even

q: $x$ is even

In order to prove the statement “if p then q”

Contrapositive Method: By assuming that q is false, prove that p must be false.

So,

q is false: x is not even $\dpi{80} \implies$ x is odd $\dpi{80} \implies$ x = 2n+1 (n is a natural number)

$\\ \therefore x^2 = (2n+1)^2 \\ \implies x^2 = 4n^2 + 4n + 1 \\ \implies x^2 = 2.2(n^2 + n) + 1 = 2m + 1$

Hence $x^2$ is odd $\dpi{80} \implies$$x^2$ is not even

Hence p is false.

Hence the given statement is true.

We know, Sum of all the angles of a triangle = $180^{\circ}$

If all the three angles are equal, then each angle is $60^{\circ}$

But $60^{\circ}$ is not an obtuse angle, and hence none of the angles of the triangle is obtuse.

Hence the triangle is not an obtuse-angled triangle.

Hence the given statement is not true.

q: The equation $x^2 - 1 = 0$ does not have a root lying between 0 and 2.

Given,

The equation $x^2 - 1 = 0$ does not have a root lying between 0 and 2.

Let x = 1

$\therefore (1)^2 - 1 = 1 -1 =0$

Hence 1 is a root of the equation $x^2 - 1 = 0$.

But 1 lies between 0 and 2.

Hence the given statement is not true.

p: Each radius of a circle is a chord of the circle

The statement is False.

By definition, A chord is a line segment intersecting the circle in two points. But a radius is a line segment joining any point on circle to its centre.

q: The centre of a circle bisects each chord of the circle.

The statement is False.

A chord is a line segment intersecting the circle in two points. But it is not necessary for a chord to pass through the centre.

r: Circle is a particular case of an ellipse.

The statement is True.

In the equation of an ellipse if we put a = b, then it is a circle.

s: If $x$ and $y$ are integers such that $x > y$, then $-x < -y$.

The statement is True.

Give, x>y

Multiplying -1 both sides

(-1)x<(-1)y $\implies$ -x < -y

(When -1 is multiplied to both L.H.S & R.H.S, sign of inequality changes)

By the rule of inequality.

t : $\sqrt{11}$ is a rational number.

The statement is False.

Since 11 is a prime number, therefore $\sqrt{11}$ is irrational.

## More About NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.5:-

Class 11 Maths chapter 14 exercise 14.5 consists of questions related to validating statements using different methods. The theory, rules, and methods for validation of statements given before the Class 11 Maths chapter 14 exercise 14.5. You can go through the theory and examples given before this exercise. It will help you to get conceptual clarity.

Also Read| Mathematical Reasoning Class 11 Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.5:-

• NCERT book Class 11 Maths chapter 14 exercise 14.5 solutions are helpful for the students in the board exams as well as in competitive exams like the engineering entrance exams.
• The solution for Class 11th Maths chapter 14 exercise 14.5 are created by experts who knows how to write answer in the board exams.

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## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. What is a mathematical acceptable statement ?

Any sentence which is either true or false but not both is a mathematical acceptable statement.

2. The negation of a statement p is denoted by ?

The negation of a statement p is denoted by ~p.

3. What is a compound statement ?

A compound statement is a statement made up of two or more smaller statements.

4. What are the methods for checking validity of an statement ?

Methods to check validity of statements:

(i) direct method (ii) contrapositive method (iii) method of contradiction (iv) using a counter example

5. Write an example of mathematical acceptable statement ?

The sum of two positive numbers is positive is an example of a mathematical acceptable statement.

6. Write an example which is not acceptable for mathematical statement ?

The sum of x and y is greater than 0 is not a mathematical acceptable statement.

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