# Mathematical Reasoning – Exercise 14.5 – Class XI

1. Show that the statement p: “If x is a real number such that x3+ 4x = 0, then x is 0” is true by

(i) direct method

(iii)method of contrapositive

Solution:

p: “If x is a real number such that x3 + 4x = 0, then x is 0”.

q: x is a real number such that x3 + 4x = 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.

∴ x3 + 4x = 0 x (x2 + 4) = 0

⇒ x = 0 or x2+ 4 = 0

Since x is real, it is 0.

Thus, statement r is true. i.e., x is 0.

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 x3 + 4x = 0 and let x is not 0.

Therefore, x3 + 4x = 0

x (x2 + 4) = 0

Then, x = 0 or x2 + 4 = 0

⇒ x = 0 or x2 = – 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 (x2 + 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 (x2 + 4).

∴ x (x2 + 4) ≠ 0

⇒ x3 + 4x ≠ 0

This shows that statement q is not true.

Thus, it has been proved that ∼r ⇒∼q

Therefore, the given statement p is true.

1. Show that the statement “For any real numbers a and b, a2 = b2 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 a2 = b2, then a = b.

Let p: a and b are real numbers such that a2 = b2

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 a2 = b2 are required such that a ≠ b.

Let a = 1 and b = –1 a2 = (1)2 = 1 and b2 = (– 1)2 = 1

∴ a2 = b2

However, a ≠ b

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

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

p: If x is an integer and x2 is even, then x is also even.

Solution:

p: If x is an integer and x2 is even, then x is also even.

Let q: x is an integer and x2 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 x2 is not even.

x is not even implies that x2 is also not even.

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

1. 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 x2 – 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 ∼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 x2 – 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 x2 – 1 = 0 x2 = 1 x = ± 1

One root of the equation x2 – 1 = 0, i.e. the root x = 1, lies between 0 and 2.

Thus, the given statement is false.

1. 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: √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, (x2)/(a2) + (y2)/(b2) = 1

If we put a = b = 1, then we obtain x2 + y2 = 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, √11 is an irrational number.

Thus, the given statement t is false.