Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Ex 14.5 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 11 Maths 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.
Let q and r be the statements given by q: x is a real number such that x3 + 4x =0.
r : x = 0
Then p : if q, there r.
(i) Direct Method:
Let q be true, then q is true.
⇒ x is real number such that x3 + 4x = 0.
⇒ x is a real number such that x (x2 + 4) = 0.
⇒ x = 0, [∵ n ∈ R, ∴ x2 + 4 ≠ 0]
⇒ r is true.
Thus, q is true
⇒ r is true.
Hence, p is true.
(ii) Method of contradiction :
If possible, let p be not true, then p is not true.
⇒ ~ p is true.
⇒ ~ (q ⇒ r) is true, [∵ p = q ⇒ r]
⇒ q and ~ r is true, [∵ ~ (q ⇒ r) ~ q and ~ r]
⇒ x is a real number such that x3 + 4x = 0 and x ≠ 0.
⇒ x = 0 and x ≠ 0.
This is a contradiction.
Hence, p is true.
(iii) Method of contrapositive :
Let r be not true, then r is not true.
⇒ x ≠ 0, x ∈ R.
⇒ x (x2 + 4) ≠ 0, n ∈ R ⇒ q is not true.
Thus, ~ x ⇒ ~ q.
Hence, p : q ⇒ r 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.
Answer.
The given compound statement is of the form “if p then q”.
We assume that p is true then a, b efi such that a2 = b2
Let us take a = – 3 and b = 3
Now a2 = b2 but a ≠ b
So when p is true, q is false.
Thus the given compound 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.
Let q and r be the statements given by
q : if x is an integer and x2 is even,
r : x is an even integer.
Thus, p : “If q, then r”.
If possible, let r be false. Thus, r is false.
⇒ x is not an even integer.
⇒ x is an odd integer.
⇒ x = (2x +1) for some integer 4.
⇒ x2 = 4x2 + 4x + 1 = 4n(n + 1) + 1
⇒ x2 is an odd integer, [∵ 4x (x + 1) is even]
⇒ q is false.
Thus, r is false ⇒ q is false.
Hence, p : “If q, then r” is a true statement.
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) Since the triangle is obtuse angled triangle then θ > 90°
Let θ = 100°
Also all the angles of the triangle are equal.
Sum of all angles of the triangle are 300°, which is not possible.
Thus the given compound statement is not true.
(ii) We see that x = 1 is a root of the equation x2 – 1 = 0, which lies between 0 and 2.
Thus the given compound 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.
(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 : √11 is a rational number.
Answer.
(i) False : The end points of radius do not lie on the circle, therefore, it is not a chord.
(ii) False : Only diameters are bisected at the centre. Other chords do not pass through the centre. Therefore centre can not bisect them.
(iii) True : Equation of ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\)
When b = a. The equation becomes \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}\) = 1 or x2 + y2 = a2, which is equation of the circle.
(iv) True : If x and y are integers and x > y then – x < – y. By rule of inequality.
(v) False : 11 is a prime number.
∴ √11 is irrational.