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

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

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