Ncert Math Solution Class 6 Exercise 14.5 All Solution
NCERT Solutions for Class 11 Maths chapter 14-Mathematical Reasoning Exercise 14.5
NCERT Solutions for Class-11 Maths Chapter-14 Mathematical Reasoning
NCERT Solutions for Class 11 Maths Chapter-14 Mathematical Reasoning Exercise 14.5 prepared by the expert of Entrancei score more with Entrancei NCERT Class 11 maths solutions. You can download and share NCERT Solutions for Class 11 Maths.
NCERT Solutions for Class-11 Maths Exercise 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
Solution :
p: "If x is a real number such that x3 + 4x = 0, then x is 0".
Let q: x is a real number such that x3 + 4x = 0 r: x is 0.
(i) To show that statement p is true, we assume that q is true and then show that r is true.
Therefore, let statement q be true.
∴ x3 + 4x = 0
x (x2 + 4) = 0
⇒ x = 0 or x2 + 4 = 0
However, since x is real, it is 0.
Thus, statement r is true.
Therefore, the given statement is true.
(ii) To show statement p to be true by contradiction, we assume that p is not true.
Let x be a real number such that x3 + 4x = 0 and let x is not 0.
Therefore, x3 + 4x = 0
x (x2 + 4) = 0
x = 0 or x2 + 4 = 0
x = 0 or x2 = – 4
However, x is real. Therefore, x = 0, which is a contradiction since we have assumed that x is not 0.
Thus, the given statement p is true.
(iii) To prove statement p to be true by contrapositive method, we assume that r is false and prove that q must be false.
Here, r is false implies that it is required to consider the negation of statement r. This obtains the following statement. ∼r: x is not 0.
It can be seen that (x2 + 4) will always be positive.
x ≠ 0 implies that the product of any positive real number with x is not zero.
Let us consider the product of x with (x2 + 4).
∴ x (x2 + 4) ≠ 0
⇒ x3 + 4x ≠ 0
This shows that statement q is not true.
Thus, it has been proved that ∼r ⇒ ∼q
Therefore, the given statement p 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.
Solution :
The given statement can be written in the form of "if-then" as follows.
If a and b are real numbers such that a2 = b2, then a = b.
Let p: a and b are real numbers such that a2 = b2.
q: a = b
The given statement has to be proved false. For this purpose, it has to be proved that if p, then ∼q.
To show this, two real numbers, a and b, with a2 = b2 are required such that a ≠ b.
Let a = 1 and b = –1
a2 = (1)2 = 1 and b2 = (– 1)2 = 1
∴ a2 = b2
However, a ≠ b
Thus, it can be concluded that the given statement is false.
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 :
p: If x is an integer and x2 is even, then x is also even.
Let q: x is an integer and x2 is even. r: x is even.
To prove that p is true by contrapositive method, we assume that r is false, and prove that q is also false.
Let x is not even.
To prove that q is false, it has to be proved that x is not an integer or x2 is not even.
x is not even implies that x2 is also not even.
Therefore, statement q is false.
Thus, the given statement p 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.
Solution :
(i) The given statement is of the form "if q then r".
q: All the angles of a triangle are equal.
r: The triangle is an obtuse-angled triangle.
The given statement p has to be proved false. For this purpose, it has to be proved that if q, then ∼r.
To show this, angles of a triangle are required such that none of them is an obtuse angle.
It is known that the sum of all angles of a triangle is 180°. Therefore, if all the three angles are equal, then each of them is of measure 60°, which is not an obtuse angle.
In an equilateral triangle, the measure of all angles is equal. However, the triangle is not an obtuse-angled triangle.
Thus, it can be concluded that the given statement p is false.
(ii) The given statement is as follows.
q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2.
This statement has to be proved false. To show this, a counter example is required.
Consider x2 – 1 = 0
x2 = 1
x = ± 1
One root of the equation x2 – 1 = 0, i.e. the root x = 1, lies between 0 and 2.
Thus, the given statement is false.
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:
Solution :
(i) The given statement p is false.
According to the definition of chord, it should intersect the circle at two distinct points.
(ii) The given statement q is false.
If the chord is not the diameter of the circle, then the centre will not bisect that chord.
In other words, the centre of a circle only bisects the diameter, which is the chord of the circle.
(iii) The equation of an ellipse is,
If we put a = b = 1, then we obtain x2 + y2 = 1, which is an equation of a circle
Therefore, circle is a particular case of an ellipse.
Thus, statement r is true.
(iv) x > y
⇒ –x < –y (By a rule of inequality)
Thus, the given statement s is true.
(v) 11 is a prime number and we know that the square root of any prime number is an irrational number. Therefore,
Thus, the given statement t is false.
Ncert Math Solution Class 6 Exercise 14.5 All Solution
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