11th NCERT/CBSE Mathematical Reasoning Exercise 14.1

Hi

Question (1)

Show that the statementp: “If x is a real number such that x

(i) direct method

(ii) method of contradiction

(iii) method of contrapositive

Solution

p: “If x is a real number such that xLet q: x is a real number such that x

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

x (x

⇒ x = 0 or x

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

Therefore, x

x (x

x = 0 or x

x = 0 or x

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

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

∴ x (x

⇒ x

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, aSolution

The given statement can be written in the form of “if-then” as follows.If a and b are real numbers such that a

Let p: a and b are real numbers such that a

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

a

∴ a

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

Solution

p: If x is an integer and xLet q: x is an integer and x

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 x

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

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 &t°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

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

Consider x

x

x = ± 1 One root of the equation x

Thus, the given statement is false.

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.

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

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, $\sqrt {11}$ is an irrational number.

Thus, the given statement t is false.