## NCERT Solutions for Class 11 Maths Chapter 14 – Mathematical Reasoning Ex 14.5

#### Page No 342:

#### Question 1:

Show that the statement

*p*: “If

*x*is a real number such that

*x*

^{3}+ 4

*x*= 0, then

*x*is 0” is true by

(i) direct method

(ii) method of contradiction

(iii) method of contrapositive

#### Answer:

*p*: “If

*x*is a real number such that

*x*

^{3}+ 4

*x*= 0, then

*x*is 0”.

Let

*q*:*x*is a real number such that*x*^{3}+ 4*x*= 0*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.
∴

*x*^{3}+ 4*x*= 0*x*(

*x*

^{2}+ 4) = 0

⇒

*x*= 0 or*x*^{2 }+ 4 = 0
However, since

*x*is real, it is 0.
Thus, statement

*r*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}+ 4*x*= 0 and let*x*is not 0.
Therefore,

*x*^{3}+ 4*x*= 0*x*(

*x*

^{2}+ 4) = 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

*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 (

*x*^{2}+ 4) will always be positive.*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*(*x*^{2}+ 4) ≠ 0
⇒

*x*^{3}+ 4*x*≠ 0
This shows that statement

*q*is not true.
Thus, it has been proved that

∼

*r*⇒ ∼*q*
Therefore, the given statement

*p*is true.#### 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:

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*^{2}=*b*^{2}.*q*:

*a*=

*b*

The given statement has to be proved false. For this purpose, it has to be proved that if

*p*, then ∼*q*. To show this, two real numbers,*a*and*b*, with*a*^{2}=*b*^{2}are required such that*a*≠*b*.
Let

*a*= 1 and*b*= –1*a*

^{2}= (1)

^{2}= 1 and

*b*

^{2}= (– 1)

^{2}= 1

∴

*a*^{2}=*b*^{2}
However,

*a*≠*b*
Thus, it can be concluded that the given statement is false.

#### 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:

*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.#### 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.#### Answer:

(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 ∼*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°. Therefore, if all the three angles are equal, then each of them is of measure 60°, 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*x*

^{2}= 1

*x*= ± 1

One root of the equation

*x*^{2}– 1 = 0, i.e. the root*x*= 1, lies between 0 and 2.
Thus, the given statement is false.

#### Page No 343:

#### 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 centre 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*: is a rational number.#### Answer:

(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,

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 ellipse.

Thus, statement

*r*is true.
(iv)

*x*>*y*
⇒ –

*x*< –*y*(By a rule of inequality)
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, is an irrational number.

Thus, the given statement

*t*is false._{}

^{}