**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

^{3}+ 4x = 0, then x is 0’ is true by

**(i) direct method**

**Solution:**

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 = 0r: 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}+ 4x = 0x(x

^{2}+ 4) = 0x = 0 or x

^{2}+ 4 = 0However, since x is real, it is 0.

Thus, statement r is true.

Therefore, the given statement is true.

**(ii) method of contradiction**

**Solution:**

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 = 0x (x

^{2}+ 4) = 0x = 0 or x

^{2}+ 4 = 0x = 0 or x

^{2}= –4However, x is real. Therefore, x = 0, which is a contradiction since we have assumed that x is not 0.

**(iii) method of contrapositive**

**Solution:**

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

I can be seen that (x

^{2}+ 4) will always be positivex = 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) = 0x

^{3}+ 4x = 0This 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.

^{2}= b

^{2}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 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}and b^{2}= (–1)^{2}= 1a

^{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.

^{2}is even, then x is also even.

**Solution:**

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 **counterexample**, 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.**

**Solution:**

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) q: The equation x**^{2} – 1 = 0 does not have a root lying between 0 and 2.

^{2}– 1 = 0 does not have a root lying between 0 and 2.

**Solution:**

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 = 0x

^{2}= 1x = ±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.

**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. **

**Solution:**

The given statement p is false.

According to the definition of chord, it should intersect the circle at two distinct points.

**(ii) q: The centre of a circle bisects each chord of the circle. **

**Solution:**

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) r: Circle is a particular case of an ellipse.**

**Solution:**

The equation of an ellipse is,

X

^{2}/a^{2}+ Y^{2}/b^{2}= 1If we put a = b = 1, then we obtain

x

^{2}+ y^{2}= 1,which is an equation of a circleTherefore, circle is a particular case of an eclipse.

Thus, statement r is true.

**(iv) s: If x and y are integers such that x > y, then – x < – y. **

**Solution:**

x > y

⇒ – x < – y (By a rule of inequality)

Thus, the given statement s is true.

**(v) t: √11 is a rational number.**

**Solution:**

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.