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Exercise - 14.5

Mathematical Reasoning

**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

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 x^{3}+ 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. Hence, x^{3}+ 4x = 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}+ 4x = 0 and let x is not 0. Therefore, x^{3}+ 4x = 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). Therefore, x (x^{2}+ 4) ≠ 0 ⇒ x^{3}+ 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.

**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.

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 Therefore, a² = b². 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.

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.

**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.

(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.

**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 : √11 is a rational number.

(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 anirrational number. Therefore, √11 is an irrational number. Thus, the given statement t is false.

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