NCERT Solutions For Class 11 Maths Chapter 14 Exercise 14.5

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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.

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