Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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

ffImage
Last updated date: 19th May 2024
Total views: 581.1k
Views today: 5.81k
MVSAT offline centres Dec 2023

NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.5 (Ex 14.5) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 14 Mathematical Reasoning Exercise 14.5 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 14 - Mathematical Reasoning

Exercise:

Exercise - 14.5

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

Popular Vedantu Learning Centres Near You
centre-image
Mithanpura, Muzaffarpur
location-imgVedantu Learning Centre, 2nd Floor, Ugra Tara Complex, Club Rd, opposite Grand Mall, Mahammadpur Kazi, Mithanpura, Muzaffarpur, Bihar 842002
Visit Centre
centre-image
Anna Nagar, Chennai
location-imgVedantu Learning Centre, Plot No. Y - 217, Plot No 4617, 2nd Ave, Y Block, Anna Nagar, Chennai, Tamil Nadu 600040
Visit Centre
centre-image
Velachery, Chennai
location-imgVedantu Learning Centre, 3rd Floor, ASV Crown Plaza, No.391, Velachery - Tambaram Main Rd, Velachery, Chennai, Tamil Nadu 600042
Visit Centre
centre-image
Tambaram, Chennai
location-imgShree Gugans School CBSE, 54/5, School road, Selaiyur, Tambaram, Chennai, Tamil Nadu 600073
Visit Centre
centre-image
Avadi, Chennai
location-imgVedantu Learning Centre, Ayyappa Enterprises - No: 308 / A CTH Road Avadi, Chennai - 600054
Visit Centre
centre-image
Deeksha Vidyanagar, Bangalore
location-imgSri Venkateshwara Pre-University College, NH 7, Vidyanagar, Bengaluru International Airport Road, Bengaluru, Karnataka 562157
Visit Centre
View More
Competitive Exams after 12th Science

Access NCERT Solutions for Class 11 Maths Chapter 14 – Mathematical Reasoning

Exercise 14.5

1. Show that the statement ${\text{p}}$: ‘If $x$ is a real number such that ${x^3} + 4x = 0$, then $x$ is 0’ is true by

  1. Direct method

  2. Method of contradiction

  3. Method of contrapositive

Ans: Write the different component statements of the given statement ${\text{p}}$.

We will now determine the component statements.

Let the first component statement be ${\text{q}}$.

${\text{q}}$: $x$ is a real number such that ${x^3} + 4x = 0$.

Let the second component statement be ${\text{r}}$.

${\text{r}}$: $x$ is 0.

(i). The direct method is as follows,

To show that the statement ${\text{p}}$ is true, we will assume that the statement ${\text{q}}$ is true and then show that ${\text{r}}$ is true. 

Therefore, let the statement ${\text{q}}$ be true. 

${x^3} + 4x = 0$

$x\left( {{x^2} + 4} \right) = 0$

$x = 0$ or ${x^2} + 4 = 0$

However, since $x$ is a real number, therefore, it is 0. 

Thus, the statement ${\text{r}}$ is true. 

Therefore, the given statement is true.

(ii). The method of contradiction is as follows,

To show the statement ${\text{p}}$ to be true by contradiction, we assume that 

${\text{p}}$ is not true and then prove our assumption to be false. 

Let $x$ be a real number such that ${x^3} + 4x = 0$ and let $x$ is not 0. 

Therefore, ${x^3} + 4x = 0$

$x\left( {{x^2} + 4} \right) = 0$

$x = 0$ or ${x^2} + 4 = 0$

$x = 0$ or ${x^2} =  - 4$

However, $x$ is real. Therefore, $x = 0$, which is a contradiction since we have 

assumed that $x$ is not 0.

Thus, the given statement ${\text{p}}$ is true.

(iii). The method of contrapositive is as follows,

To prove statement ${\text{p}}$ to be true by contrapositive method, we assume that the statement${\text{r}}$ is false and prove that the statement${\text{q}}$ must also be false. 

Here, the statement ${\text{r}}$ is false. This implies that it is required to consider the negation of statement ${\text{r}}$. This obtains the following statement. 

$ \sim {\text{r}}$: $x$ is not 0

Also, we can be observe that ${x^2} + 4$ will always be positive 

Now $x = 0$ implies that the product of any positive real number with $x$ is not zero. 

Let us consider the product of $x$ with ${x^2} + 4$

$x\left( {{x^2} + 4} \right) = 0$

${x^3} + 4x = 0$

This shows that statement ${\text{q}}$ is not true. 

Hence, it has been proved that

$ \sim {\text{r}} \Rightarrow  \sim {\text{q}}$

Therefore, the given statement ${\text{p}}$ is true.


2. Show that the statement ‘For any real numbers $a$ and $b$, ${a^2} = {b^2}$ implies that $a = b$’ is not true by giving a counter-example.

Ans: We will write the given statement in the form of a conditional statement as follows,

If $a$ and $b$ are real numbers such that ${a^2} = {b^2}$, then $a = b$.

We will now determine the component statements.

Let the first component statement be ${\text{p}}$.

${\text{p}}$: $a$ and $b$ are real numbers such that ${a^2} = {b^2}$.

Let the second component statement be ${\text{q}}$.

${\text{q}}$: $a = b$

Now it is required to prove the given statement false. For this purpose, it has to be proved that if ${\text{p}}$, then $ \sim {\text{q}}$.

To show this, we need two real numbers $a$ and $b$, with ${a^2} = {b^2}$ such that $a \ne b$.

Assume the two numbers as follows,

Let $a = 1$ and $b =  - 1$.

${a^2} = {1^2} = 1$ and ${b^2} = {\left( { - 1} \right)^2} = 1$

Hence, it is proved that ${a^2} = {b^2}$.

However, $a \ne b$.

Therefore, the given statement is proved false.


3. Show that the following statement is true by the method of contrapositive. 

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

Ans: Write the different component statements of the given statement ${\text{p}}$.

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

We will now determine the component statements.

Let the first component statement be ${\text{q}}$.

${\text{q}}$: $x$ is an integer and ${x^2}$is even.

Let the second component statement be ${\text{r}}$.

${\text{r}}$: $x$ is even.

The method of contrapositive is as follows,

To prove statement ${\text{p}}$ to be true by contrapositive method, we assume that the statement ${\text{r}}$ is false and prove that the statement ${\text{q}}$ must also be false. 

Consider that $x$ is not even.

To prove that ${\text{q}}$ is false, we need to prove that $x$ is not an integer so ${x^2}$ is not even. 

Since $x$ is not an even number implies that ${x^2}$ will also not be even. 

Therefore, the statement ${\text{q}}$ is false. 

Thus, the given statement ${\text{p}}$ is true by the method of contrapositive. 


4. By giving a counter example, show that the following statements are not true. 

  1. ${\text{p}}$: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle. 

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

Ans: 

(i). Analyse the statement ${\text{p}}$.

The given statement is of the form ‘If $A$ then $B$’. Thus, it is a conditional statement. 

Write the different component statements of the given statement ${\text{p}}$.

We will now determine the component statements.

Let the first component statement be ${\text{q}}$.

${\text{q}}$:All the angles of a triangle are equal.

Let the second component statement be ${\text{r}}$.

${\text{r}}$:The triangle is an obtuse-angled triangle.

Now we need to prove that the given statement ${\text{p}}$false. For this purpose, it has to be proved that if ${\text{q}}$, then $ \sim {\text{r}}$

To show this, we require angles of a triangle such that none of them is an obtuse angle. It is known that the sum of all the angles of a triangle is $180^\circ $ from the Triangle sum theorem. 

Therefore, if all the three angles of the triangle are equal, then each of them will be of measure $60^\circ $so it is an equilateral triangle. And $60^\circ $ is not an obtuse angle. In an equilateral triangle, the measure of all angles is equal. Hence, the triangle is not an obtuse-angled triangle. 

Thus, it can be concluded that the given statement ${\text{p}}$ is false.

(ii). Analyse the statement ${\text{q}}$. The statement ${\text{q}}$ is as follows,

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

We need to prove this statement false. To do so, we will take the help of a counter-example.

Consider the given equation and solve it.

${x^2} - 1 = 0$

${x^2} = 1$

$x =  \pm \sqrt 1 $

$x =  \pm 1$

From the above obtained solutions, we can see that one of the roots of the given equation ${x^2} - 1 = 0$ is $x = 1$, which lies between 0 and 2. 

Therefore, the given statement is false.


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

Opting for the NCERT solutions for Ex 14.5 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 14.5 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 14 Exercise 14.5 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 14 Exercise 14.5, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 14 Exercise 14.5 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

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

1. Is the Reasoning (Ex. 14.5) chapter in the math book for 11th grade important?

Class 11 Mathematical Reasoning When doing your assignment, NCERT Solutions is of great assistance. NCERT Solutions for Class 11 Mathematical Reasoning, Chapter 14 Expert Vedantu Teachers prepared each exercise.

2.  What is Direct Variation?

A differential equation links the derivative of one or more illustrated functions. Applications usually involve a differential equation that establishes a relationship between the three components: functions that represent physical quantities, derivatives that show the rates at which those quantities change, and derivatives that indicate the rates at which those quantities change.

3. Why should a person choose the Vedantu NCERT Solutions for Class 11 Math, Chapter 14 Differential Equations Exercise 14.5?

The NCERT solutions for Ex. 14.5 from Class 11 Maths from Vedantu are regarded as the best option for CBSE students when it comes to exam preparation. This chapter has many exercises. Exercise 14.5 for Class 11 Maths NCERT solutions are available in PDF format on this page. You can view and study this solution immediately from the Vedantu website or app, or you can download it as needed.

4. What is the contradiction? With Example.

A contradiction exists in mathematics when a proposition p is true and its negation p is also true. Now let's use an example to better understand the idea of contradiction. Consider the words "p" and "q." With respect to prime numbers a and b, X is equal to a divided by b.


Think about the sentences p and q.

Statement p: x = a/b, where a and b are prime numbers.

Statement q divides both A and B: 2.

5. What is contrapositive?

When you switch a statement's hypothesis and conclusion and reject both of them, you have a contrapositive statement. If we reverse the hypothesis and the conclusion and reject both, the conclusion is: If it's not a triangle, it's not a polygon.