# 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.