# NCERT Solutions For Class 11 Maths Chapter 14 Exercise 14.5

Sets

## Chapter 14 Ex.14.5 Question 1

Show that the statement

\(p:\) “If \(x\) is a real number such that \({x^3} + 4x = 0\), then \(x\) is 0” is true by

(i) direct method

(ii) method of contradiction

(iii) method of contrapositive

**Solution**

It is given that, \(p:\) “If \(x\) is a real number such that \({x^3} + 4x = 0\), then \(x\) is \(0\)”

Let, \(q:x\) is a real number such that, \(r:x\) is \(0.\)

(i) To show that statement \(p\) is true, we assume that \(q\) is true and then show that \(r\) is true.

Therefore, let statement \(q\) be true.

\[\begin{align}{x^3} + 4x &= 0\\x\left( {{x^2} + 4} \right) &= 0\\x = 0{\text{ or } }{\rm{ }}{x^2} + 4 &= 0\\x = 0{\text{ or }}{x^2} &= -4\end{align}\]

Since \(x\) is real, \(x = 0\).

Thus, statement is true.

Therefore, the given statement is true.

(ii) To show statement \(p\) to be true by contradiction, we assume that \(p\) is not true.

Let \(x\) be a real number such that \({x^3} + 4x = 0{\rm{ }}\) and \(x \ne 0\)

Therefore,

\[\begin{align}{x^3} + 4x &= 0\\x\left( {{x^2} + 4} \right) &= 0\\x = 0{\text{ or }}{x^2} + 4 &= 0\\

x = 0{\text{ or }}{x^2} &= -4\end{align}\]

Since, \(x\) is real.

Therefore, \(x = 0\) which is a contradiction since we have assumed that \(x \ne 0\).

Thus, the given statement \(p\) is true.

(iii) To prove statement \(p\) to be true by contrapositive method, we assume that \(r\) is false and prove that \(q\) must be false.

Here, \(r\) is false implies that it is required to consider the negation of statement \(r.\) This obtains the following statement.

\(r:x\) is not \(0.\)

It can be seen that \(\left( {{x^2} + 4} \right)\) will always be positive.

\(x \ne 0\) implies that the product of any positive real number with \(x\) is not zero.

Let us consider the product of \(x\) with \(\left( {{x^2} + 4} \right)\)

\[x\left( {{x}^{2}}+4 \right)=~{{x}^{3}}+4x\]

It is given that \({x^3} + 4x = 0\)

This shows that statement \(q\) is not true.

Thus, it has been proved that \( \sim r \Rightarrow \sim p\)

Therefore, the given statement \(p\) is true.

## Chapter 14 Ex.14.5 Question 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.

**Solution**

The given statement can be written in the form of ‘if-then’ as follows. If \(a\) and \(b\) are real numbers such that \({a^2} = {b^2}\), then \(a = b\).

Let \(p:a\) and \(b\) are real numbers such that \(a = b\).

Hence, \(q:a = b\)

The given statement has to be proved false.

Two real numbers \(a\) and \(b\) with \({a^2} = {b^2}\) are required such that \(a \ne b\).

Let \(a = 1\) and \(b = - 1\)

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

Here, \({a^2} = {b^2}\)

However, \(a = b\)

Thus, it can be concluded that the given statement is false.

## Chapter 14 Ex.14.5 Question 3

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

*\(p:\) If *\(x\)* is an integer and *\({x^2}\)* is even, then *\(x\)* is also even.*

**Solution**

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

Let \(q:x\) is an integer and \({x^2}\) is even.

\(r:x\) is even.

To prove that \(p\) is true by contrapositive method, we assume that \(r\) is false, and prove that \(q\) is also false.

Let \(x\) is not even.

To prove that \(q\) is false, it has to be proved that \(x\) is not an integer or \({x^2}\) is not even. \(x\) is not even implies that \({x^2}\) is also not even.

Therefore, statement \(q\) is false. Thus, the given statement \(p\) is true.

## Chapter 14 Ex.14.5 Question 4

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

(i) \(p:\) If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.

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

**Solution**

(i) The given statement is of the form ‘if \(q\) then \(r\)’.

\(q:\) All the angles of a triangle are equal.

\(r:\) The triangle is an obtuse-angled triangle.

The given statement \(p\) has to be proved false. For this purpose, it has to be proved that if \(q\) then \( \sim r\).

To show this, angles of a triangle are required such that none of them is an obtuse angle.

It is known that the sum of all angles of a triangle is \(180^\circ \).

Therefore, if all the three angles are equal, then each of them is of measures \(60^\circ \), which is not an obtuse angle.

In an equilateral triangle, the measure of all angles is equal.

However, the triangle is not an obtuse-angled triangle.

Thus, it can be concluded that the given statement \(p\) is false.

(ii) The given statement is as follows.

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

This statement has to be proved false. To show this, a counter example is required.

Consider, \({x^2}-1 = 0\)

\[\begin{align} \Rightarrow \;&{{x}^{2}}=1 \\ \Rightarrow \;& x=\pm 1\end{align}\]

One root of the equation , i.e. the root \(x = 1\), lies between \(0\) and \(2.\)

Thus, the given statement is false.

## Chapter 14 Ex.14.5 Question 5

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 center of a circle bisects 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) \(t:\) \(\sqrt {11} \) is a rational number.

**Solution**

(i) The given statement \(p\) is false.

According to the definition of chord, it should intersect the circle at two distinct points.

(ii) The given statement \(q\) is false.

If the chord is not the diameter of the circle, then the Centre will not bisect that chord.

In other words, the centre of a circle only bisects the diameter, which is the chord of the circle.

(iii) The equation of an ellipse is,

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]

If we put \(a = b = 1\), then we obtain

\({x^2} + {y^2} = 1\), which is an equation of a circle.

Therefore, circle is a particular case of an eclipse.

Thus, statement \(r\) is true.

(iv) \(x > y\)

By a rule of inequality \(-x < -y\)

Thus, the given statement s is true.

(v) \(11\) is a prime number and we know that the square root of any prime number is an irrational number.

Therefore, \(\sqrt {11} \) is an irrational number.

Thus, the given statement \(t\) is false.