GSEB Solutions Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5

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 x³ + 4x = 0, then x = 0 is true by

1. direct method
2. method of contradiction
3. method of contrapositive.

Solution:
1. Direct method
x³ + 4x = 0 or x(x² + 4) = 0
But x² + 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 x³ + 4x = 0.
∴ p³ + 4p = 0
⇒ p(p² + 4) = 0
p ≠ 0 and also p² + 4 ≠ 0 ⇒ p(p² + 4) = 0,
p ≠ 0 and also p² + 4 ≠ 0 ⇒ p(p² + 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.
∴ p³ + 4p = 0, p being the root of x² + 4 < 0
or p(p² + 4) = 0, Now p ≠ 0. Also, p² + 4 ≠ 0
⇒ p(p² + 4) ≠ 0, if q is not true.
∴ x = 0 is the root of x³ + 4x = 0.

Question 2.
Show that the statement “For any real numbers a and b, a² = b² implies that a = b” is not true by giving counter example?
Solution:
Let a = 1, b = – 1, a² = b² = 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 x² is even, then x is also even.
Solution:
Let x is not even, i.e., x = 2n + 1
∴ x = (2n + 1)² = 4n² + 4n + 1 = 4(n² + n) + 1
4(n² + 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:

1. p : If all the angles of a triangles are equal, then the triangle is obtuse.
2. q : The equation x² – 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 x² – 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.

1. p : Each radius of a circle is a chord of the circle.
2. q : The centre of a circle bisects each chord of the circle.
3. r : Circle is a particular case of an ellipse.
4. s = If x andy are integers such that x > y, then – x < – y.
5. 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 x² + y² = a², 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.