RBSE Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5 Textbook Exercise Questions and Answers.

RBSE Class 11 Maths Solutions Chapter 14 Mathematical Reasoning Ex 14.5

Question 1.
Show that the statement
p : “If x is a real number such that x³ + 4x = 0, then x is 0” is true by
(i) direct method
(ii) method of contradiction
(iii) method of contrapositive
Answer:
Given statement : If x is a real number such that x³ + 4x = 0, then x = 0.
Component statement are given by:
p : x is a real number such that x³ + 4x = 0 q : x = 0

(i) By direct method :
Since, x is a real number and x³ + 4x = 0 Then, x(x² + 4) = 0
or either x = 0, or x² + 4 = 0
From x² + 4 = 0
x² = -4
which is not defined, since x is a real number and square of any real number is not negative.
Then, x = 0

(ii) By contradiciton:
Let p is not true.
i. e. x is not a real number.
Then x will be an imaginary number.
Then, x = ai (v i = √-1)
Again, x³ + 4x = 0
⇒ x(x² +4) = 0
⇒ a.i(a²i² + 4) = 0
⇒ a.i(-a² + 4) = 0
⇒ -a² =4
⇒ a² = 4
⇒ a =± 2
⇒ a ≠ 0
Then, if x is imaginary number then x = ±2i and x ≠ 0
when x ≠ 0 then x³ ≠ 0 and 4x ≠ 0 so, x³ +4x ≠ 0
which contradicts the given statement.
Thus, assumption that p is not true is wrong. Then, p is true.

(iii) By contrapositive method :
Let q in not true, i.e. x ≠ 0 then, let x = a (a 0) thus, from x³ + 4x = 0
a³ + 4a = 0
⇒ a(a³ + 4)=0
Since, a ≠ 0, then a² + 4 = 0
⇒ a² = -4
it implies that a is imaginary
when a is imaginary then x will be imaginary i.e. x will not be real.
Thus, if q is false, p will be false.
Thus, if p is true.
i.e. x³ + 4x = 0 for real number x
then q is true
i.e., x = 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 a counter-example.
Answer:
Given statement : “For any real numbers a and b, a² = b² implies that a = b"
Let a = 4, and b = -4
i.e. a ≠ bn
Then a² =16 and b² =16
Thus, a² = b²
whereas a² = b² but a ≠ b
Thus, 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.
Answer:
Given statement: If x is an integer and x² is even then x is also even.
The component statement are :
p : x is an integer and x² is even.
q : x is an even integer.
Let component statement q is not true. i.e x is not an even integer.
Then, integer x will be odd.
Now, let x = (2m + 1) where m is integer
Thus, x² = (2m + 1)²
= 4m² + 4m + 1
or x² = 2(2m² +2m + 1/2)
It implies x² is an odd integer.
Thus, x² is not an even integer.
Then, statement p is not true.
Then, if component q is not true then component p is also not true.
Thus, given statement is true.

Question 4.
By giving a counter example, 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.
(ii) q : The equation x² - 1 = 0 does not have a root lying between 0 and 2.
Answer:
(i) Given statement p : If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.
Let any ΔPQR is obtuse angled triangle and its angles are same.
Then each angle of this triangle will be more then 90°.
Thus, ∠P = 90° + x
∠Q = 90° + y
and ∠R= 110°
Again, ∠P + ∠Q + ∠R = 90° + x + 90° + y + 110°
= 290° + (x + y)
But, ∠P + ∠Q + ∠R = 180°
Then, no triangle is possible in which
∠P + ∠Q + ∠R > 290°
Thus, statement P is not true.

(ii) Given statement q: Root of equation x² - 1 = 0 does not lie between 0 and 2.
Let one root of equation x² - 1=0 be 1 which does not lie between 0 and 2.
Then value of (1)² -1 = 0 is zero.
i.e. Assumed root satisfy the given equation.
It implies that 1 is such root of equation x² - 1 = 0 which lies between 0 and 2 Thus, statement q is not true.

Question 5.
Which of the following statement 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.
(ii) q : The centre of a circle bisects each chord of the circle.
(iii) r : Circle is a particular case of an ellipse.
(iv) s : if x and y are integers such that x > y, then - x < - y.
(v) t: \(\sqrt{11}\) is a rational number.
Answer:
(i) Statement p : Each radius of any circle is chord of circle. Since, chord of circle meets the circle at two distinct points whereas radius cannot cut the circle at two points.
No radius can be chord of circle.
Thus, given statement is false.

(ii) Statement q : The centre of a circle bisects each chord of the circle.
Since, diameter of a circle is a chord which passes through centre of the circle, then centre of circle will not lie on other chords and cannot divide them.
Thus, given statement is false.

(iii) Statement r : Circle is a particular case of an ellipse.
Equation of ellipse : \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
If a = b, then \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
⇒ x² + y² = a²
which is a equation of circle.
Thus, in special case, circle is an ellipse, or circle is special case of any ellipse.
Thus, given statement is true.

(iv) Statement s : If x andy are integers such that x > y,
then - x < - y under the law of inequality, x > y then -x < -y [As 4 > 3 then -4 < -3]
Thus, given statement is true.

(v) Statement t: \(\sqrt{11}\) is a rational number.
If \(\sqrt{11}\) is a rational number, then it can be written in the form a
\(\frac{a}{b}\), whereas a and b are integers. b
and there is not common factor between them.
Then, = \(\sqrt{11}=\frac{\sqrt{11}}{1}=\frac{a}{b}\)
Since, p = \(\sqrt{11}\) is not an integer
Thus, \(\sqrt{11}\) is not a rational number
Thus, given statement is false.