**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.

**Solution:**

The given compound statement is of the form “if p then q”

p: x ϵ R such that x^{3} + 4x = 0

q: x = 0

**(i) Direct method:**

We assume that p is true, then

x ϵ R such that x^{3} + 4x = 0

⇒ x ϵ R such that x(x^{2} + 4) = 0

⇒ x ϵ R such that x = 0 or x^{2} + 4 = 0

⇒ x = 0 => q is true.

So, when p is true, q is true.

Thus, the given compound statement is true.

**(ii) Method of contradiction :**

We assume that p is true and q is false, then

x ϵ R such that x^{3} + 4x = 0

⇒ x ϵ R such that x(x^{2} + 4) = 0

⇒ x ϵ R such that x = 0 or x^{2} + 4 = 0

⇒ x = 0.

which is a contradiction. So, our assumption that x ≠ 0 is false. Thus, the given compound statement is true.

**(iii) Method of contrapositive:**

We assume that q is false, then x ≠ 0

x ϵ R such that x^{3} + 4x = 0

⇒ x ϵ R such that x = 0 or x^{2} + 4 = 0

∴ statement q is false, so x ≠ 0. So, we have,

x ϵ R such that x^{2} = -2

Which is not true for any x ϵ R.

⇒ p is false

So, when q is false, p is false.

Thus, the given compound statement 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.**

**Solution:**

The given compound statement is of the form “if p then q”

We assume that p is true, then a, b ⍷ R such that a^{2} = b^{2}

Let us take a = -3 and b = 3

Now, a^{2} = b^{2}, but a ≠ b

So, when p is true, q is false.

Thus, the given compound 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.**

**Solution:**

The given compound statement is of the form “if p then q”

p: x ϵ Z and x^{2} is even.

q: x is an even integer.

We assume that q is false, then x is not an even integer.

⇒ x is an odd integer.

⇒ x^{2} is an odd integer.

⇒ p is false

So, when q is false, p is false.

Thus, the given compound statement 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.

**Solution:**

**(i)** Since the triangle is an obtuse-angled triangle then 0 > 90°.

Let 0 = 100°

Also, all the angles of the triangle are equal.

∴ The Sum of all angles of the triangle is 300°, which is not possible.

Thus, the given compound statement is not true,

**(ii)** We see that x = 1 is a root of the equation x^{2} – 1 = 0, which lies between 0 and 2. Thus, the given compound statement is not true.

**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 center 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)** A chord of a circle is a line whose two endpoints lie on the circle and all the points on the line lie inside the circle. So, the radius of a circle is not a chord of the circle. Thus, the given statement is false.

**(ii)** The center of a circle bisects the chord of the circle when the chord is the diameter of the circle. When the chord is other than diameter then the center of the circle does not lie on the chord. Thus, the given statement is false.

**(iii)** In the equation of an ellipse if we put a = b, then we get an equation of a circle.

Thus, the given statement is true.

**(iv)** It is given that x, y ϵ Z such that x > y. Multiplying both sides by a negative sign, we have

x, y ϵ Z such that -x < -y.

Thus, the given statement is true.

**(v)** Since cannot be expressed in the form , where a and b are integers and b ≠ 0. Thus, the given statement is false.