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

Question:1.(ii) Show that the statement
p: “If $x$ is a real number such that $x^3 + 4x = 0$, then $x$ is 0” is true by

method of contradiction

Answer:

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 $x^3 + 4x = 0$.

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: $x \neq 0$

Now, $x^3 + 4x = 0 \implies x(x^2 + 4) = 0$

$\implies x = 0\ or\ (x^2 + 4)= 0$

$\implies x = 0\ or\ x^2 = -4\ (not\ possible)$

Hence, x = 0

But we assumed $x \neq 0$. This contradicts our assumption.

Therefore q is true.

Question:1(iii) Show that the statement p: “If $x$ is a real number such that $x^3 + 4x = 0$, then $x$ is 0” is true by method of contrapositive

Answer:

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 $x^3 + 4x = 0$.

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: $x \neq 0$

$\implies$ x.(Positive number) $\neq$ 0.(Positive number)

$\implies x(x^2 + 4) \neq 0(x^2 + 4)$

$\implies x(x^2 + 4) \neq 0 \implies x^3 + 4x \neq 0$

Therefore p is false.

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.

Answer:

Given,

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

Let a = 1 & b = -1

Now,

$a^2 = (1)^2$= 1

$b^2 = (-1)^2$ = 1

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

But a $\neq$ b

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

Hence the given statement is not true.

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.

Answer:

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 $\implies$ x is odd $\implies$ 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 $\implies$$x^2$ is not even

Hence p is false.

Hence the given statement is true.

Question:4.(i) By giving a counter example, show that the following statement is not true.
p: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.

Answer:

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.

Question:4.(ii) By giving a counter example, show that the following statement is not true.

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

Answer:

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.

Question:5.(i) Is the following statement true or false? Give a valid reason for saying so.

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

Answer:

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.

Question:5.(ii) Is the following statement true or false? Give a valid reason for saying so.

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

Answer:

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.

Question:5.(iii) Is the following statement true or false? Give a valid reason for saying so.

r: Circle is a particular case of an ellipse.

Answer:

The statement is True.

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

Question:5.(iv) Is the following statement true or false? Give a valid reason for saying so.

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

Answer:

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.

Question:5.(v) Is the following statement true or false? Give a valid reason for saying so.

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

Answer:

The statement is False.

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