Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5

Question 1.

Show that the statement

p : “If x is real number such that x^{3} + 4x = 0, then x = 0 is true by

- direct method
- method of contradiction
- method of contrapositive.

Solution:

1. Direct method

x^{3} + 4x = 0 or x(x^{2} + 4) = 0

But x^{2} + 4 ≠ 0, x ∈ R. So, x = 0.

2. Method of contradiction

Let x ≠ 0 and let it be x = p, p ∈ R, p is a root of x^{3} + 4x = 0.

∴ p^{3} + 4p = 0

⇒ p(p^{2} + 4) = 0

p ≠ 0 and also p^{2} + 4 ≠ 0 ⇒ p(p^{2} + 4) = 0,

p ≠ 0 and also p^{2} + 4 ≠ 0 ⇒ p(p^{2} + 4) ≠ 0, which is a contradication. So, x = 0.

3. Contrapositive

q is not true.

⇒ Let x = 0 is not true ⇒ Let x = p ≠ 0.

∴ p^{3} + 4p = 0, p being the root of x^{2} + 4 < 0

or p(p^{2} + 4) = 0, Now p ≠ 0. Also, p^{2} + 4 ≠ 0

⇒ p(p^{2} + 4) ≠ 0, if q is not true.

∴ x = 0 is the root of x^{3} + 4x = 0.

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 counter example?

Solution:

Let a = 1, b = – 1, a^{2} = b^{2} = 1. But a ≠ b.

The given 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:

Let x is not even, i.e., x = 2n + 1

∴ x = (2n + 1)^{2} = 4n^{2} + 4n + 1 = 4(n^{2} + n) + 1

4(n^{2} + n) + 1 is odd.

i.e., “If q is not true, then p is not true” is proved.

Hence, the given statement is true.

Question 4.

By giving counter examples show that the following statements are not true:

- p : If all the angles of a triangles are equal, then the triangle is obtuse.
- q : The equation x
^{2}– 1 = 0 does not have a root lying between 0 and 2.

Solution:

1. Let an angle of triangle be 90° + θ.

∴ Sum of the angles = 3(90° + θ) = 270° + 3θ, which is greater than 180°.

∴ A triangle having equal angles cannot be an obtuse angled triangle.

2. The equation x^{2} – 1 = 0 has the root x = 1, which lies between 0 and 2. The given 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.

- p : Each radius of a circle is a chord of the circle.
- q : The centre of a circle bisects each chord of the circle.
- r : Circle is a particular case of an ellipse.
- s = If x andy are integers such that x > y, then – x < – y.
- t : \(\sqrt{11}\) is a rational numbers.

Solution:

1. False:

The end points of radius do not lie on the circle. Therefore it is not a chord.

2. False:

Only diameters are bisected at the centre. Other chords do not pass through the centre. Therefore, centre cannot bisect them.

3. True:

Equation of ellipse is

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.

When b = a, the equation becomes

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}\) = 1 or x^{2} + y^{2} = a^{2}, which is the equation of the circle.

4. True:

If x and y are integers and x > y, then – x <- y.

By rule of inequality.

5. False:

11 is a prime number.

∴ \(\sqrt{11}\) is an irrational.