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Class 11 NCERT Solutions- Chapter 14 Mathematical Reasoning – Exercise 14.5
• Last Updated : 05 May, 2021

### (i) direct method

Solution:

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

Let 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

However, since x is real, it is 0.

Thus, statement r is true.

Therefore, the given statement is true.

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

Therefore, x3 + 4x = 0

x (x2 + 4) = 0

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.

### (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 (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.

### Question 2: 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 and b2 = (–1)2 = 1

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 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.

### (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 x2  – 1 = 0 does not have a root lying between 0 and 2.

Solution:

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.

### (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,

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 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.

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